Subquery in from must have an alias ошибка

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.

var google_conversion_label = «owonCMyG5nEQ0aD71QM»;

Я делаю быстрый запрос с использованием Postgresql 8.2, и я делал подобные запросы тысячу раз раньше, но я не могу понять, почему я получаю эту ошибку. Я, вероятно, упускаю что-то очевидное, но он говорит, что мой «подзапрос в FROM должен иметь псевдоним». У меня есть псевдоним для моего подзапроса «внутренний», и я не могу понять, почему еще я получу ошибку.

    SELECT "Branch", "Zip_5", "CountofStops", avg("EarlyTime") As 
    "Average_Arrival_Time"
    FROM
    (SELECT branch_id as "Branch", substring(stop_zip_postal_code, 1, 5) as 
    "Zip_5", count(stop_name) as "CountofStops", min(actual_arrival_time) as 
    "EarlyTime"

    FROM distribution_stop_information

    WHERE company_no = '001' AND route_date > '3/13/2017'

    GROUP BY branch_id, stop_zip_postal_code)
    inner

    GROUP BY "Branch", "Zip_5"

    ORDER BY Zip_5

********** Error **********

ERROR: subquery in FROM must have an alias
SQL state: 42601
Hint: For example, FROM (SELECT ...) [AS] foo.

inner является зарезервированным ключевым словом. Используйте другое имя в качестве псевдонима.

inner. , , думать «внутреннее соединение». Вам нужен лучший псевдоним, чем этот.

SELECT Branch, Zip_5, CountofStops, avg(EarlyTime) As Average_Arrival_Time
FROM (SELECT branch_id as Branch, left(stop_zip_postal_code, 5) as Zip_5, 
             count(stop_name) as CountofStops,
             min(actual_arrival_time) as EarlyTime
      FROM distribution_stop_information
      WHERE company_no = '001' AND route_date > '2017-03-13'
      GROUP BY branch_id, stop_zip_postal_code
     ) b
GROUP BY Branch, Zip_5
ORDER BY Zip_5;

Примечания:

• Не заключайте имена столбцов в двойные кавычки, если в этом нет необходимости. Они просто лишние.
• Используйте стандартные форматы для констант даты.
• LEFT() является удобным сокращением для substring( . . ., 1, . . .)

class FromMultiTableTestCase(ModelTestCase):
    requires = [Blog, Comment, User]

    def setUp(self):
        super(FromMultiTableTestCase, self).setUp()

        for u in range(2):
            user = User.create(username='u%s' % u)
            for i in range(3):
                b = Blog.create(user=user, title='b%s-%s' % (u, i))
                for j in range(i):
                    Comment.create(blog=b, comment='c%s-%s' % (i, j))

    def test_from_multi_table(self):
        q = (Blog
             .select(Blog, User)
             .from_(Blog, User)
             .where(
                 (Blog.user == User.id) &
                 (User.username == 'u0'))
             .order_by(Blog.pk)
             .naive())

        qc = len(self.queries())
        blogs = [b.title for b in q]
        self.assertEqual(blogs, ['b0-0', 'b0-1', 'b0-2'])

        usernames = [b.username for b in q]
        self.assertEqual(usernames, ['u0', 'u0', 'u0'])
        self.assertEqual(len(self.queries()) - qc, 1)

    def test_subselect(self):
        inner = User.select(User.username)
        self.assertEqual(
            [u.username for u in inner.order_by(User.username)], ['u0', 'u1'])

        # Have to manually specify the alias as "t1" because the outer query
        # will expect that.
        outer = (User
                 .select(User.username)
                 .from_(inner.alias('t1')))
        sql, params = compiler.generate_select(outer)
        self.assertEqual(sql, (
            'SELECT users."username" FROM '
            '(SELECT users."username" FROM "users" AS users) AS t1'))

        self.assertEqual(
            [u.username for u in outer.order_by(User.username)], ['u0', 'u1'])