Ошибка 42601 ошибка синтаксиса в конце - ErrorsMaster.ru - большая энциклопедия ошибок и их решений

Syntax errors are quite common while coding.

But, things go for a toss when it results in website errors.

PostgreSQL error 42601 also occurs due to syntax errors in the database queries.

At Bobcares, we often get requests from PostgreSQL users to fix errors as part of our Server Management Services.

Today, let’s check PostgreSQL error in detail and see how our Support Engineers fix it for the customers.

What causes error 42601 in PostgreSQL?

PostgreSQL is an advanced database engine. It is popular for its extensive features and ability to handle complex database situations.

Applications like Instagram, Facebook, Apple, etc rely on the PostgreSQL database.

But what causes error 42601?

PostgreSQL error codes consist of five characters. The first two characters denote the class of errors. And the remaining three characters indicate a specific condition within that class.

Here, 42 in 42601 represent the class “Syntax Error or Access Rule Violation“.

In short, this error mainly occurs due to the syntax errors in the queries executed. A typical error shows up as:

Here, the syntax error has occurred in position 119 near the value “parents” in the query.

How we fix the error?

Now let’s see how our PostgreSQL engineers resolve this error efficiently.

Recently, one of our customers contacted us with this error. He tried to execute the following code,

CREATE OR REPLACE FUNCTION prc_tst_bulk(sql text)
RETURNS TABLE (name text, rowcount integer) AS
$$
BEGIN
WITH m_ty_person AS (return query execute sql)
select name, count(*) from m_ty_person where name like '%a%' group by name
union
select name, count(*) from m_ty_person where gender = 1 group by name;
END
$$ LANGUAGE plpgsql;

But, this ended up in PostgreSQL error 42601. And he got the following error message,

ERROR: syntax error at or near "return"
LINE 5: WITH m_ty_person AS (return query execute sql)

Our PostgreSQL Engineers checked the issue and found out the syntax error. The statement in Line 5 was a mix of plain and dynamic SQL. In general, the PostgreSQL query should be either fully dynamic or plain. Therefore, we changed the code as,

RETURN QUERY EXECUTE '
WITH m_ty_person AS (' || sql || $x$)
SELECT name, count(*)::int FROM m_ty_person WHERE name LIKE '%a%' GROUP BY name
UNION
SELECT name, count(*)::int FROM m_ty_person WHERE gender = 1 GROUP BY name$x$;

This resolved the error 42601, and the code worked fine.

[Need more assistance to solve PostgreSQL error 42601?- We’ll help you.]

Conclusion

In short, PostgreSQL error 42601 occurs due to the syntax errors in the code. Today, in this write-up, we have discussed how our Support Engineers fixed this error for our customers.

PREVENT YOUR SERVER FROM CRASHING!

Never again lose customers to poor server speed! Let us help you.

Our server experts will monitor & maintain your server 24/7 so that it remains lightning fast and secure.

GET STARTED

var google_conversion_label = «owonCMyG5nEQ0aD71QM»;

Источник

I have used the next SQL statement in both MySQL and PostgreSQL, but it fails in PostgreSQL

db.Query(`SELECT COUNT(*) as N FROM email WHERE address = ?`, email)

with this error:

pq: F:"scan.l" M:"syntax error at end of input" S:"ERROR" C:"42601" P:"50" R:"scanner_yyerror" L:"993"

What’s the problem? The error messages in PostgreSQL are very cryptic.

asked Oct 29, 2012 at 10:26

You haven’t provided any details about the language/environment, but I’ll try a wild guess anyway:

MySQL’s prepared statements natively use ? as the parameter placeholder, but PostgreSQL uses $1, $2 etc. Try replacing the ? with $1 and see if it works:

WHERE address = $1

The error messages in PostgreSQL are very cryptic.

In general, I’ve found that Postgres error messages are better than competing products (ahem, MySQL and especially Oracle), but in this instance you’ve managed to confuse the parser beyond sanity.

answered Oct 29, 2012 at 10:32

intgrintgr

19.6k5 gold badges59 silver badges68 bronze badges

In golang for queries we have use

MySQL uses the ? variant
PostgreSQL uses an enumerated $1, $2, etc bindvar syntax
SQLite accepts both ? and $1 syntax
Oracle uses a :name syntax

answered Feb 8, 2020 at 0:01

refrazulrefrazul

1111 silver badge3 bronze badges

You are using Go right?

try:

db.Query(`SELECT COUNT(*) as N FROM email WHERE address = $1`, email)

answered Oct 29, 2012 at 10:32

simonmenkesimonmenke

2,81919 silver badges28 bronze badges

In my case, it was due to using a — line comment where the program that was responsible for interacting with the database read in the multiple lines of my query all as one giant line. This meant that the line comment corrupted the remainder of the query. The fix was to use a /* block comment */ instead.

answered Apr 7, 2020 at 22:40

calamaricalamari

3292 silver badges8 bronze badges

Another possibility — check for any missing END directives for IF, LOOP, etc.

I’d been working in a loop so long, that I didn’t realize I never put in an END LOOP;.

The message that Postgres displays:

ERROR:  syntax error at end of input
LINE 123: $$;

is terribly unhelpful at telling you if some construct was never closed.

answered Apr 12 at 21:38

RandallRandall

2,8101 gold badge21 silver badges24 bronze badges

Источник

Syntax errors are quite common while coding.

But, things go for a toss when it results in website errors.

PostgreSQL error 42601 also occurs due to syntax errors in the database queries.

At Bobcares, we often get requests from PostgreSQL users to fix errors as part of our Server Management Services.

Today, let’s check PostgreSQL error in detail and see how our Support Engineers fix it for the customers.

What causes error 42601 in PostgreSQL?

PostgreSQL is an advanced database engine. It is popular for its extensive features and ability to handle complex database situations.

Applications like Instagram, Facebook, Apple, etc rely on the PostgreSQL database.

But what causes error 42601?

PostgreSQL error codes consist of five characters. The first two characters denote the class of errors. And the remaining three characters indicate a specific condition within that class.

Here, 42 in 42601 represent the class “Syntax Error or Access Rule Violation“.

In short, this error mainly occurs due to the syntax errors in the queries executed. A typical error shows up as:

Here, the syntax error has occurred in position 119 near the value “parents” in the query.

How we fix the error?

Now let’s see how our PostgreSQL engineers resolve this error efficiently.

Recently, one of our customers contacted us with this error. He tried to execute the following code,

CREATE OR REPLACE FUNCTION prc_tst_bulk(sql text)
RETURNS TABLE (name text, rowcount integer) AS
$$
BEGIN
WITH m_ty_person AS (return query execute sql)
select name, count(*) from m_ty_person where name like '%a%' group by name
union
select name, count(*) from m_ty_person where gender = 1 group by name;
END
$$ LANGUAGE plpgsql;

But, this ended up in PostgreSQL error 42601. And he got the following error message,

ERROR: syntax error at or near "return"
LINE 5: WITH m_ty_person AS (return query execute sql)

RETURN QUERY EXECUTE '
WITH m_ty_person AS (' || sql || $x$)
SELECT name, count(*)::int FROM m_ty_person WHERE name LIKE '%a%' GROUP BY name
UNION
SELECT name, count(*)::int FROM m_ty_person WHERE gender = 1 GROUP BY name$x$;

This resolved the error 42601, and the code worked fine.

[Need more assistance to solve PostgreSQL error 42601?- We’ll help you.]

Conclusion

In short, PostgreSQL error 42601 occurs due to the syntax errors in the code. Today, in this write-up, we have discussed how our Support Engineers fixed this error for our customers.

PREVENT YOUR SERVER FROM CRASHING!

Never again lose customers to poor server speed! Let us help you.

Our server experts will monitor & maintain your server 24/7 so that it remains lightning fast and secure.

GET STARTED

var google_conversion_label = «owonCMyG5nEQ0aD71QM»;

восстановить базу из дампа:

-- -- PostgreSQL database dump -- -- Dumped from database version 10.19 (Ubuntu 10.19-0ubuntu0.18.04.1) -- Dumped by pg_dump version 10.19 (Ubuntu 10.19-0ubuntu0.18.04.1) SET statement_timeout = 0; SET lock_timeout = 0; SET idle_in_transaction_session_timeout = 0; SET client_encoding = 'UTF8'; SET standard_conforming_strings = on; SELECT pg_catalog.set_config('search_path', '', false); SET check_function_bodies = false; SET xmloption = content; SET client_min_messages = warning; SET row_security = off; -- -- Name: plpgsql; Type: EXTENSION; Schema: -; Owner: -- CREATE EXTENSION IF NOT EXISTS plpgsql WITH SCHEMA pg_catalog; -- -- Name: EXTENSION plpgsql; Type: COMMENT; Schema: -; Owner: -- COMMENT ON EXTENSION plpgsql IS 'PL/pgSQL procedural language'; -- -- Name: attribute_id_seq; Type: SEQUENCE; Schema: public; Owner: bender -- CREATE SEQUENCE public.attribute_id_seq START WITH 1 INCREMENT BY 1 NO MINVALUE NO MAXVALUE CACHE 1; ALTER TABLE public.attribute_id_seq OWNER TO bender; SET default_tablespace = ''; SET default_with_oids = false; -- -- Name: attribute; Type: TABLE; Schema: public; Owner: bender -- CREATE TABLE public.attribute ( attribute_id integer DEFAULT nextval('public.attribute_id_seq'::regclass) NOT NULL, name character varying(30) NOT NULL, attribute_type_id integer NOT NULL ); ALTER TABLE public.attribute OWNER TO bender; -- -- Name: attribute_type_id_seq; Type: SEQUENCE; Schema: public; Owner: bender -- CREATE SEQUENCE public.attribute_type_id_seq START WITH 1 INCREMENT BY 1 NO MINVALUE NO MAXVALUE CACHE 1; ALTER TABLE public.attribute_type_id_seq OWNER TO bender; -- -- Name: attribute_type; Type: TABLE; Schema: public; Owner: bender -- CREATE TABLE public.attribute_type ( attribute_type_id integer DEFAULT nextval('public.attribute_type_id_seq'::regclass) NOT NULL, name character varying(50) NOT NULL ); ALTER TABLE public.attribute_type OWNER TO bender; -- -- Name: film_id_seq; Type: SEQUENCE; Schema: public; Owner: bender -- CREATE SEQUENCE public.film_id_seq START WITH 1 INCREMENT BY 1 NO MINVALUE NO MAXVALUE CACHE 1; ALTER TABLE public.film_id_seq OWNER TO bender; -- -- Name: film; Type: TABLE; Schema: public; Owner: bender -- CREATE TABLE public.film ( film_id integer DEFAULT nextval('public.film_id_seq'::regclass) NOT NULL, name character varying(50) NOT NULL ); ALTER TABLE public.film OWNER TO bender; -- -- Name: film_attributes_id_seq; Type: SEQUENCE; Schema: public; Owner: bender -- CREATE SEQUENCE public.film_attributes_id_seq START WITH 1 INCREMENT BY 1 NO MINVALUE NO MAXVALUE CACHE 1; ALTER TABLE public.film_attributes_id_seq OWNER TO bender; -- -- Name: film_attributes; Type: TABLE; Schema: public; Owner: bender -- CREATE TABLE public.film_attributes ( film_attributes_id integer DEFAULT nextval('public.film_attributes_id_seq'::regclass) NOT NULL, attribute_id integer NOT NULL, film_id integer NOT NULL, value_text character varying, value_integer integer, value_float double precision, value_boolean boolean, value_timestamp timestamp with time zone ); ALTER TABLE public.film_attributes OWNER TO bender; -- -- Name: film_attributes_values; Type: VIEW; Schema: public; Owner: bender -- CREATE VIEW public.film_attributes_values AS SELECT NULL::character varying(50) AS name, NULL::character varying(50) AS attribute_type, NULL::character varying(30) AS attribute_name, NULL::character varying AS attribute_value; ALTER TABLE public.film_attributes_values OWNER TO bender; -- -- Name: film_tasks; Type: VIEW; Schema: public; Owner: bender -- CREATE VIEW public.film_tasks AS SELECT NULL::character varying(50) AS name, NULL::character varying[] AS today_tasks, NULL::character varying[] AS twenty_days_tasks; ALTER TABLE public.film_tasks OWNER TO bender; -- -- Data for Name: attribute; Type: TABLE DATA; Schema: public; Owner: bender -- COPY public.attribute (attribute_id, name, attribute_type_id) FROM stdin; 1 Рецензии 3 3 Премия Оскар 2 4 Премия Ника 2 5 Премия Золотой Глобус 2 10 Описание фильма 3 11 Длительность (мин.) 1 12 Длительность проката (дней) 1 2 Рейтинг 7 6 Премьера в мире 6 7 Премьера в России 6 8 Старт продажи билетов 6 9 Старт проката 6 13 Окончание проката 6 . -- -- Data for Name: attribute_type; Type: TABLE DATA; Schema: public; Owner: bender -- COPY public.attribute_type (attribute_type_id, name) FROM stdin; 1 integer 2 boolean 3 text 4 date 5 numeric 6 timestamp 7 float . -- -- Data for Name: film; Type: TABLE DATA; Schema: public; Owner: bender -- COPY public.film (film_id, name) FROM stdin; 1 Spoiler-man: No Way 2 Matrix 4 . -- -- Data for Name: film_attributes; Type: TABLE DATA; Schema: public; Owner: bender -- COPY public.film_attributes (film_attributes_id, attribute_id, film_id, value_text, value_integer, value_float, value_boolean, value_timestamp) FROM stdin; 1 1 1 Годный фильм, распинаюсь про сюжет, пишу про игру актеров, все круто N N N N 2 1 2 Джон Уик уже не тот, сестры Вачовски сбрендили, полная фигня N N N N 5 3 1 f N N N N 7 6 2 N N N N 2021-12-10 00:00:00+03 9 7 2 N N N N 2021-12-30 00:00:00+03 10 8 1 N N N N 2021-12-10 00:00:00+03 11 8 2 N N N N 2021-12-07 00:00:00+03 12 12 1 N 21 N N N 13 12 2 N 14 N N N 14 9 1 N N N N 2021-12-15 00:00:00+03 15 9 2 N N N N 2021-12-15 00:00:00+03 16 13 1 N N N N 2022-01-04 00:00:00+03 17 13 2 N N N N 2022-01-04 00:00:00+03 18 3 2 t N N N N 6 6 1 N N N N 2021-12-15 00:00:00+03 8 7 1 N N N N 2022-01-04 00:00:00+03 . -- -- Name: attribute_id_seq; Type: SEQUENCE SET; Schema: public; Owner: bender -- SELECT pg_catalog.setval('public.attribute_id_seq', 13, true); -- -- Name: attribute_type_id_seq; Type: SEQUENCE SET; Schema: public; Owner: bender -- SELECT pg_catalog.setval('public.attribute_type_id_seq', 6, true); -- -- Name: film_attributes_id_seq; Type: SEQUENCE SET; Schema: public; Owner: bender -- SELECT pg_catalog.setval('public.film_attributes_id_seq', 18, true); -- -- Name: film_id_seq; Type: SEQUENCE SET; Schema: public; Owner: bender -- SELECT pg_catalog.setval('public.film_id_seq', 2, true); -- -- Name: attribute attribute_pkey; Type: CONSTRAINT; Schema: public; Owner: bender -- ALTER TABLE ONLY public.attribute ADD CONSTRAINT attribute_pkey PRIMARY KEY (attribute_id); -- -- Name: attribute_type attribute_type_name_key; Type: CONSTRAINT; Schema: public; Owner: bender -- ALTER TABLE ONLY public.attribute_type ADD CONSTRAINT attribute_type_name_key UNIQUE (name); -- -- Name: attribute_type attribute_type_pkey; Type: CONSTRAINT; Schema: public; Owner: bender -- ALTER TABLE ONLY public.attribute_type ADD CONSTRAINT attribute_type_pkey PRIMARY KEY (attribute_type_id); -- -- Name: attribute attribute_unq; Type: CONSTRAINT; Schema: public; Owner: bender -- ALTER TABLE ONLY public.attribute ADD CONSTRAINT attribute_unq UNIQUE (name); -- -- Name: film_attributes film_attributes_pkey; Type: CONSTRAINT; Schema: public; Owner: bender -- ALTER TABLE ONLY public.film_attributes ADD CONSTRAINT film_attributes_pkey PRIMARY KEY (film_attributes_id); -- -- Name: film film_pkey; Type: CONSTRAINT; Schema: public; Owner: bender -- ALTER TABLE ONLY public.film ADD CONSTRAINT film_pkey PRIMARY KEY (film_id); -- -- Name: film film_unq; Type: CONSTRAINT; Schema: public; Owner: bender -- ALTER TABLE ONLY public.film ADD CONSTRAINT film_unq UNIQUE (name); -- -- Name: attribute_index; Type: INDEX; Schema: public; Owner: bender -- CREATE INDEX attribute_index ON public.attribute USING btree (name COLLATE "C.UTF-8" varchar_ops); -- -- Name: film_index; Type: INDEX; Schema: public; Owner: bender -- CREATE INDEX film_index ON public.film USING btree (name COLLATE "C.UTF-8"); -- -- Name: attribute attribute_type_fkey; Type: FK CONSTRAINT; Schema: public; Owner: bender -- ALTER TABLE ONLY public.attribute ADD CONSTRAINT attribute_type_fkey FOREIGN KEY (attribute_type_id) REFERENCES public.attribute_type(attribute_type_id) NOT VALID; -- -- Name: film_attributes film_attribute_attribute_fkey; Type: FK CONSTRAINT; Schema: public; Owner: bender -- ALTER TABLE ONLY public.film_attributes ADD CONSTRAINT film_attribute_attribute_fkey FOREIGN KEY (attribute_id) REFERENCES public.attribute(attribute_id); -- -- Name: film_attributes film_attribute_film_fkey; Type: FK CONSTRAINT; Schema: public; Owner: bender -- ALTER TABLE ONLY public.film_attributes ADD CONSTRAINT film_attribute_film_fkey FOREIGN KEY (film_id) REFERENCES public.film(film_id); -- -- PostgreSQL database dump complete --

ERROR: ОШИБКА:  ошибка синтаксиса (примерное положение: "1")
LINE 180: 1 Рецензии 3
^
SQL state: 42601
Character: 4115

Пользователь заполняет форму, где вводит значения в textBox-ы.

NpgsqlCommand com = new NpgsqlCommand("INSERT INTO 'Tip' (code_tip, 
name_tip) VALUES (@p1, @p2)", con);
com.Parameters.Add("code_tip", NpgsqlTypes.NpgsqlDbType.Bigint).Value = 
textBox1;
com.Parameters.Add("name_tip", NpgsqlTypes.NpgsqlDbType.Char, 40).Value = 
textBox2;
com.ExecuteNonQuery();

На этом моменте visual-studio выдает мне ошибку:

Npgsql.NpgsqlException: "ОШИБКА: 42601: ошибка синтаксиса (примерное положение: "'Tip'")"

Подскажите, что не так?

задан 21 сен 2018 в 11:03

INSERT INTO ‘Tip’

По синтаксису (и стандарту SQL) insert запроса после ключевого слова into должно идти имя таблицы. Вы указали строковой литерал. Парсер соответственно удивляется и отвечает, что вы написали непонятно что.

в одинарных кавычках 'Tip' — строковой литерал.
без кавычек Tip — имя объекта, принудительно приводимое парсером к нижнему регистру, т.е. tip
в двойных кавычках "Tip" — регистрозависимое имя объекта

Если у вас таблица именно Tip, то единственным корректным способом к ней обращаться будут двойные кавычки:

INSERT INTO "Tip" ...

ответ дан 21 сен 2018 в 11:56

МелкийМелкий

20.8k3 золотых знака26 серебряных знаков52 бронзовых знака

Содержание

PostgreSQL error 42601- How we fix it
What causes error 42601 in PostgreSQL?
How we fix the error?
Conclusion
PREVENT YOUR SERVER FROM CRASHING!
10 Comments

PostgreSQL error 42601- How we fix it

by Sijin George | Sep 12, 2019

Syntax errors are quite common while coding.

But, things go for a toss when it results in website errors.

PostgreSQL error 42601 also occurs due to syntax errors in the database queries.

At Bobcares, we often get requests from PostgreSQL users to fix errors as part of our Server Management Services.

Today, let’s check PostgreSQL error in detail and see how our Support Engineers fix it for the customers.

What causes error 42601 in PostgreSQL?

PostgreSQL is an advanced database engine. It is popular for its extensive features and ability to handle complex database situations.

Applications like Instagram, Facebook, Apple, etc rely on the PostgreSQL database.

But what causes error 42601?

PostgreSQL error codes consist of five characters. The first two characters denote the class of errors. And the remaining three characters indicate a specific condition within that class.

Here, 42 in 42601 represent the class “Syntax Error or Access Rule Violation“.

In short, this error mainly occurs due to the syntax errors in the queries executed. A typical error shows up as:

Here, the syntax error has occurred in position 119 near the value “parents” in the query.

How we fix the error?

Now let’s see how our PostgreSQL engineers resolve this error efficiently.

Recently, one of our customers contacted us with this error. He tried to execute the following code,

But, this ended up in PostgreSQL error 42601. And he got the following error message,

This resolved the error 42601, and the code worked fine.

[Need more assistance to solve PostgreSQL error 42601?- We’ll help you.]

Conclusion

In short, PostgreSQL error 42601 occurs due to the syntax errors in the code. Today, in this write-up, we have discussed how our Support Engineers fixed this error for our customers.

PREVENT YOUR SERVER FROM CRASHING!

Never again lose customers to poor server speed! Let us help you.

Our server experts will monitor & maintain your server 24/7 so that it remains lightning fast and secure.

SELECT * FROM long_term_prediction_anomaly WHERE + “‘Timestamp’” + ‘”BETWEEN ‘” +
2019-12-05 09:10:00+ ‘”AND’” + 2019-12-06 09:10:00 + “‘;”)

Hello Joe,
Do you still get PostgreSQL errors? If you need help, we’ll be happy to talk to you on chat (click on the icon at right-bottom).

У меня ошибка drop table exists “companiya”;

CREATE TABLE “companiya” (
“compania_id” int4 NOT NULL,
“fio vladelca” text NOT NULL,
“name” text NOT NULL,
“id_operator” int4 NOT NULL,
“id_uslugi” int4 NOT NULL,
“id_reklama” int4 NOT NULL,
“id_tex-specialist” int4 NOT NULL,
“id_filial” int4 NOT NULL,
CONSTRAINT “_copy_8” PRIMARY KEY (“compania_id”)
);

CREATE TABLE “filial” (
“id_filial” int4 NOT NULL,
“street” text NOT NULL,
“house” int4 NOT NULL,
“city” text NOT NULL,
CONSTRAINT “_copy_5” PRIMARY KEY (“id_filial”)
);

CREATE TABLE “login” (
“id_name” int4 NOT NULL,
“name” char(20) NOT NULL,
“pass” char(20) NOT NULL,
PRIMARY KEY (“id_name”)
);

CREATE TABLE “operator” (
“id_operator” int4 NOT NULL,
“obrabotka obrasheniya” int4 NOT NULL,
“konsultirovanie” text NOT NULL,
“grafick work” date NOT NULL,
CONSTRAINT “_copy_2” PRIMARY KEY (“id_operator”)
);

CREATE TABLE “polsovateli” (
“id_user” int4 NOT NULL,
“id_companiya” int4 NOT NULL,
“id_obrasheniya” int4 NOT NULL,
“id_oshibka” int4 NOT NULL,
CONSTRAINT “_copy_6” PRIMARY KEY (“id_user”)
);

CREATE TABLE “reklama” (
“id_reklama” int4 NOT NULL,
“tele-marketing” text NOT NULL,
“soc-seti” text NOT NULL,
“mobile” int4 NOT NULL,
CONSTRAINT “_copy_3” PRIMARY KEY (“id_reklama”)
);

CREATE TABLE “tex-specialist” (
“id_tex-specialist” int4 NOT NULL,
“grafik” date NOT NULL,
“zarplata” int4 NOT NULL,
“ispravlenie oshibok” int4 NOT NULL,
CONSTRAINT “_copy_7” PRIMARY KEY (“id_tex-specialist”)
);

CREATE TABLE “uslugi” (
“id_uslugi” int4 NOT NULL,
“vostanavlenia parola” int4 NOT NULL,
“poterya acaunta” int4 NOT NULL,
CONSTRAINT “_copy_4” PRIMARY KEY (“id_uslugi”)
);

ALTER TABLE “companiya” ADD CONSTRAINT “fk_companiya_operator_1” FOREIGN KEY (“id_operator”) REFERENCES “operator” (“id_operator”);
ALTER TABLE “companiya” ADD CONSTRAINT “fk_companiya_uslugi_1” FOREIGN KEY (“id_uslugi”) REFERENCES “uslugi” (“id_uslugi”);
ALTER TABLE “companiya” ADD CONSTRAINT “fk_companiya_filial_1” FOREIGN KEY (“id_filial”) REFERENCES “filial” (“id_filial”);
ALTER TABLE “companiya” ADD CONSTRAINT “fk_companiya_reklama_1” FOREIGN KEY (“id_reklama”) REFERENCES “reklama” (“id_reklama”);
ALTER TABLE “companiya” ADD CONSTRAINT “fk_companiya_tex-specialist_1” FOREIGN KEY (“id_tex-specialist”) REFERENCES “tex-specialist” (“id_tex-specialist”);
ALTER TABLE “polsovateli” ADD CONSTRAINT “fk_polsovateli_companiya_1” FOREIGN KEY (“id_companiya”) REFERENCES “companiya” (“compania_id”);

ERROR: ОШИБКА: ошибка синтаксиса (примерное положение: “”companiya””)
LINE 1: drop table exists “companiya”;
^

Источник

Здравствуйте! Довольно распространенный вопрос, находил решения, но все же компилятор выводит ошибку. Как-то неправильно задаю первичный ключ.

Использую Postgresql

Java

1 2 3 4 5 6 7 8 9 10 11 12

@Entity @Table(name="users") public class Users{     @Id     @GeneratedValue(strategy=GenerationType.IDENTITY)     @Column(columnDefinition="serial")     private Long id;     @Column(name="login")     private String login;     @Column(name="password")     private String password; }

SQL

1 2 3 4 5

CREATE TABLE IF NOT EXISTS users (   id  SERIAL PRIMARY KEY NOT NULL,   login VARCHAR(20) UNIQUE NOT NULL,   password VARCHAR(20) NOT NULL );

Ошибка:
org.hibernate.tool.schema.spi.CommandAcceptanceExc eption: Error executing DDL via JDBC Statement
Caused by: org.postgresql.util.PSQLException: ОШИБКА: ошибка синтаксиса (примерное положение: «auto_increment»)

Добавлено через 27 минут
SQL-запрос пишу в консоли postgresql. Таблица без записей.
Также в проекте имеется обычный конфигурационный файл для JPA. Думаю, что роли он здесь не играет

__________________
Помощь в написании контрольных, курсовых и дипломных работ, диссертаций здесь

@YohDeadfall — I understand that part about it, but this is not script that I am creating or even code that I am creating. This is all created under the hood by Npsql/EntityFramework. My quick guess is that I am extending my DbContext from IdentityDbContext<IdentityUser> which wants to create all of the tables for roles, users, claims, etc. If I change this to just extend from DbContext, then everything works as advertised.

Below is the script that EF is trying to use created from dotnet ef migrations script — please be aware that I have removed my custom part of the script for brevity.

You can see there are two specific calls that are being made where [NormalizedName] and [NormalizedUserName] are being used.

CREATE TABLE IF NOT EXISTS "__EFMigrationsHistory" ( "MigrationId" varchar(150) NOT NULL, "ProductVersion" varchar(32) NOT NULL, CONSTRAINT "PK___EFMigrationsHistory" PRIMARY KEY ("MigrationId") ); CREATE TABLE "AspNetRoles" ( "Id" text NOT NULL, "ConcurrencyStamp" text NULL, "Name" varchar(256) NULL, "NormalizedName" varchar(256) NULL, CONSTRAINT "PK_AspNetRoles" PRIMARY KEY ("Id") ); CREATE TABLE "AspNetUsers" ( "Id" text NOT NULL, "AccessFailedCount" int4 NOT NULL, "ConcurrencyStamp" text NULL, "Email" varchar(256) NULL, "EmailConfirmed" bool NOT NULL, "LockoutEnabled" bool NOT NULL, "LockoutEnd" timestamptz NULL, "NormalizedEmail" varchar(256) NULL, "NormalizedUserName" varchar(256) NULL, "PasswordHash" text NULL, "PhoneNumber" text NULL, "PhoneNumberConfirmed" bool NOT NULL, "SecurityStamp" text NULL, "TwoFactorEnabled" bool NOT NULL, "UserName" varchar(256) NULL, CONSTRAINT "PK_AspNetUsers" PRIMARY KEY ("Id") ); CREATE TABLE "AspNetRoleClaims" ( "Id" int4 NOT NULL, "ClaimType" text NULL, "ClaimValue" text NULL, "RoleId" text NOT NULL, CONSTRAINT "PK_AspNetRoleClaims" PRIMARY KEY ("Id"), CONSTRAINT "FK_AspNetRoleClaims_AspNetRoles_RoleId" FOREIGN KEY ("RoleId") REFERENCES "AspNetRoles" ("Id") ON DELETE CASCADE ); CREATE TABLE "AspNetUserClaims" ( "Id" int4 NOT NULL, "ClaimType" text NULL, "ClaimValue" text NULL, "UserId" text NOT NULL, CONSTRAINT "PK_AspNetUserClaims" PRIMARY KEY ("Id"), CONSTRAINT "FK_AspNetUserClaims_AspNetUsers_UserId" FOREIGN KEY ("UserId") REFERENCES "AspNetUsers" ("Id") ON DELETE CASCADE ); CREATE TABLE "AspNetUserLogins" ( "LoginProvider" text NOT NULL, "ProviderKey" text NOT NULL, "ProviderDisplayName" text NULL, "UserId" text NOT NULL, CONSTRAINT "PK_AspNetUserLogins" PRIMARY KEY ("LoginProvider", "ProviderKey"), CONSTRAINT "FK_AspNetUserLogins_AspNetUsers_UserId" FOREIGN KEY ("UserId") REFERENCES "AspNetUsers" ("Id") ON DELETE CASCADE ); CREATE TABLE "AspNetUserRoles" ( "UserId" text NOT NULL, "RoleId" text NOT NULL, CONSTRAINT "PK_AspNetUserRoles" PRIMARY KEY ("UserId", "RoleId"), CONSTRAINT "FK_AspNetUserRoles_AspNetRoles_RoleId" FOREIGN KEY ("RoleId") REFERENCES "AspNetRoles" ("Id") ON DELETE CASCADE, CONSTRAINT "FK_AspNetUserRoles_AspNetUsers_UserId" FOREIGN KEY ("UserId") REFERENCES "AspNetUsers" ("Id") ON DELETE CASCADE ); CREATE TABLE "AspNetUserTokens" ( "UserId" text NOT NULL, "LoginProvider" text NOT NULL, "Name" text NOT NULL, "Value" text NULL, CONSTRAINT "PK_AspNetUserTokens" PRIMARY KEY ("UserId", "LoginProvider", "Name"), CONSTRAINT "FK_AspNetUserTokens_AspNetUsers_UserId" FOREIGN KEY ("UserId") REFERENCES "AspNetUsers" ("Id") ON DELETE CASCADE ); CREATE INDEX "IX_AspNetRoleClaims_RoleId" ON "AspNetRoleClaims" ("RoleId"); CREATE UNIQUE INDEX "RoleNameIndex" ON "AspNetRoles" ("NormalizedName") WHERE [NormalizedName] IS NOT NULL; CREATE INDEX "IX_AspNetUserClaims_UserId" ON "AspNetUserClaims" ("UserId"); CREATE INDEX "IX_AspNetUserLogins_UserId" ON "AspNetUserLogins" ("UserId"); CREATE INDEX "IX_AspNetUserRoles_RoleId" ON "AspNetUserRoles" ("RoleId"); CREATE INDEX "EmailIndex" ON "AspNetUsers" ("NormalizedEmail"); CREATE UNIQUE INDEX "UserNameIndex" ON "AspNetUsers" ("NormalizedUserName") WHERE [NormalizedUserName] IS NOT NULL; INSERT INTO "__EFMigrationsHistory" ("MigrationId", "ProductVersion") VALUES ('20180514204732_initial', '2.0.3-rtm-10026');

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При попытке восстановления дампа под Windopws 7 столкнулся с ошибкой:

COPY carriers (business_entity_id, name) FROM stdin;
8	Arriva
50000	ASEAG
.

[Err] ОШИБКА: ошибка синтаксиса (примерное положение: «8»)
LINE 2: 8 Arriva
^

Мы получаем простую синтаксическую ошибку, потому что Postgres получает данные как код SQL.

Пример ниже не поддерживается утилитой pgAdmin.

COPY tablel FROM STDIN;

Как сделать резервную копию базы в Postgress?

pg_dump -U user database > fileName.sql

где:

pg_dump — это программа для создания резервных копий базы данных Postgres Pro;
postgres — имя пользователя БД (совпадает с именем базы данных);
transactions — имя базы к которой есть доступ у нашего пользователя postgres;
transactions.sql — имя создаваемого файла дампа;
hostname — имя сервера БД, это pg.sweb.ru;
format — формат дампа (может быть одной из трех букв: ‘с’ (custom — архив .tar.gz), ‘t’ (tar — tar-файл), ‘p’ (plain — текстовый файл). В команде букву надо указывать без кавычек.);
dbname — имя базы данных.

pg_dump -U postgres transactions > transactions.sql

Как сделать restore в Postgress?

Тут все несколько запутанней, поэтому выкладываю все 3 варианта начну с того который решил мою проблему:

psql -U postgres -d belgianbeers -a -f beers.sql

pg_restore -h localhost -U postgres -F t -d transactions «D:/transactions.sql»

pg_restore —host localhost —port 5432 —username postgres —dbname transactions —clean —verbose «D:transactions.sql»

Не забывайте если указываете полный путь брать его в двойные кавычки!!!

when I am using this command to update table in PostgreSQL 13:

UPDATE rss_sub_source 
SET sub_url = SUBSTRING(sub_url, 1, CHAR_LENGTH(sub_url) - 1) 
WHERE sub_url LIKE '%/'
limit 10

but shows this error:

SQL Error [42601]: ERROR: syntax error at or near "limit"
Position: 111

why would this error happen and what should I do to fix it?

asked Jul 22, 2021 at 14:09

LIMIT isn’t a valid keyword in an UPDATE statement according to the official PostgreSQL documentation:

[ WITH [ RECURSIVE ] with_query [, ...] ]
UPDATE [ ONLY ] table_name [ * ] [ [ AS ] alias ]
SET { column_name = { expression | DEFAULT } |
( column_name [, ...] ) = [ ROW ] ( { expression | DEFAULT } [, ...] ) |
( column_name [, ...] ) = ( sub-SELECT )
} [, ...]
[ FROM from_item [, ...] ]
[ WHERE condition | WHERE CURRENT OF cursor_name ]
[ RETURNING * | output_expression [ [ AS ] output_name ] [, ...] ]

^{Reference: UPDATE (PostgreSQL Documentation )}

Solution

Remove LIMIT 10 from your statement.

answered Jul 22, 2021 at 14:32

John K. N.John K. N.

15.7k10 gold badges45 silver badges100 bronze badges

You could make something like this

But a Limit without an ORDER BY makes no sense, so you must choose one that gets you the correct 10 rows

UPDATE rss_sub_source t1
SET t1.sub_url = SUBSTRING(t1.sub_url, 1, CHAR_LENGTH(t1.sub_url) - 1) 
FROM (SELECT id FROM rss_sub_source WHERE sub_url LIKE '%/' ORDER BY id LIMIT 10) t2 
WHERE t2.id = t1.id

answered Jul 22, 2021 at 14:51

nbknbk

7,7295 gold badges12 silver badges27 bronze badges

when I am using this command to update table in PostgreSQL 13:

UPDATE rss_sub_source SET sub_url = SUBSTRING(sub_url, 1, CHAR_LENGTH(sub_url) - 1) WHERE sub_url LIKE '%/' limit 10

but shows this error:

SQL Error [42601]: ERROR: syntax error at or near "limit" Position: 111

why would this error happen and what should I do to fix it?

asked Jul 22, 2021 at 14:09

LIMIT isn’t a valid keyword in an UPDATE statement according to the official PostgreSQL documentation:

[ WITH [ RECURSIVE ] with_query [, ...] ]
UPDATE [ ONLY ] table_name [ * ] [ [ AS ] alias ]
SET { column_name = { expression | DEFAULT } |
( column_name [, ...] ) = [ ROW ] ( { expression | DEFAULT } [, ...] ) |
( column_name [, ...] ) = ( sub-SELECT )
} [, ...]
[ FROM from_item [, ...] ]
[ WHERE condition | WHERE CURRENT OF cursor_name ]
[ RETURNING * | output_expression [ [ AS ] output_name ] [, ...] ]

^{Reference: UPDATE (PostgreSQL Documentation )}

Solution

Remove LIMIT 10 from your statement.

answered Jul 22, 2021 at 14:32

John K. N.John K. N.

15.7k10 gold badges45 silver badges100 bronze badges

You could make something like this

But a Limit without an ORDER BY makes no sense, so you must choose one that gets you the correct 10 rows

UPDATE rss_sub_source t1
SET t1.sub_url = SUBSTRING(t1.sub_url, 1, CHAR_LENGTH(t1.sub_url) - 1) 
FROM (SELECT id FROM rss_sub_source WHERE sub_url LIKE '%/' ORDER BY id LIMIT 10) t2 
WHERE t2.id = t1.id

answered Jul 22, 2021 at 14:51

nbknbk

7,7295 gold badges12 silver badges27 bronze badges

Источник

Добрый вечер. Есть выражение:

$this->insertStmt = $this->connection->getPdo()->prepare("
    INSERT INTO files (
           real_name, 
           virtual_name, 
           album,
           size,
           resolution, 
           duration, 
           comment,
           path,
           user
    ) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)
");

Которое вызывается как обычно:

protected function doInsert(object $object)
{
    $values = [
        $object->getRealName(),
        $object->getVirtualName(),
        $object->getAlbum(),
        $object->getSize(),
        $object->getResolution(),
        $object->getDuration(),
        $object->getComment(),
        $object->getPath(),
        $object->getUser(),
    ];
       
    $this->insertStmt->execute($values);
}

Примерное содержание $values:

array(9) { 
    [0]=> string(15) "BvrK9z6UPxY.jpg" 
    [1]=> string(16) "1265dde1c67abc1c" 
    [2]=> string(23) "По умолчанию" 
    [3]=> int(54973) 
    [4]=> string(7) "720x430" 
    [5]=> NULL 
    [6]=> string(0) "" 
    [7]=> string(108) "files/id5cd487313a93a/По умолчанию/2019-05-10/1265dde1c67abc1c.jpg" 
    [8]=> string(15) "id5cd487313a93a" 
}

Сообщение ошибки:

Type: PDOException
Code: 42601
Message: SQLSTATE[42601]: Syntax error: 7 ОШИБКА: ошибка синтаксиса (примерное положение: "user") LINE 11: user ^

С точки зрения синтаксиса вроде все верно, много раз перепроверил, IDE ни на что не ругается. В чем трабл, господа?

Источник

Я использовал следующий оператор SQL как в MySQL, так и в PostgreSQL, но он не работает в PostgreSQL.

db.Query(`SELECT COUNT(*) as N FROM email WHERE address = ?`, email)

с этой ошибкой:

pq: F:"scan.l" M:"syntax error at end of input" S:"ERROR" C:"42601" P:"50" R:"scanner_yyerror" L:"993"

В чем проблема? Сообщения об ошибках в PostgreSQL очень загадочные.

person
Community

schedule
29.10.2012

source
источник

Ответы (4)

Вы не предоставили никаких подробностей о языке / среде, но я все равно попробую предположить:

Подготовленные операторы MySQL изначально используют ? в качестве заполнителя параметра, но PostgreSQL использует $1, $2 и т. Д. Попробуйте заменить ? на $1 и посмотрите, работает ли это:

WHERE address = $1

Сообщения об ошибках в PostgreSQL очень загадочные.

В общем, я обнаружил, что сообщения об ошибках Postgres лучше, чем у конкурирующих продуктов (кхм, MySQL и особенно Oracle), но в этом случае вам удалось запутать синтаксический анализатор сверх разумности.

person
intgr

schedule
29.10.2012

Вы используете Go, верно?

пытаться:

db.Query(`SELECT COUNT(*) as N FROM email WHERE address = $1`, email)

person
simonmenke

schedule
29.10.2012

В golang для запросов мы используем

MySQL использует? вариант
PostgreSQL использует синтаксис bindvar с перечислением $ 1, $ 2 и т. Д.
SQLite принимает оба? и синтаксис $ 1
Oracle использует синтаксис: name

person
refrazul

schedule
08.02.2020

В моем случае это произошло из-за использования строчного комментария, когда программа, отвечающая за взаимодействие с базой данных, считывала несколько строк моего запроса как одну гигантскую строку. Это означало, что строковый комментарий испортил оставшуюся часть запроса. Исправление заключалось в использовании вместо этого / * блочного комментария * /.

person
calamari

schedule
07.04.2020

Источник