CODEWITHSUNDEEP
mysqlsqlmysql-error-1054
James
James

Reputation: 2943

Using column alias in WHERE clause of MySQL query produces an error

The query I'm running is as follows, however I'm getting this error:

#1054 - Unknown column 'guaranteed_postcode' in 'IN/ALL/ANY subquery'

SELECT `users`.`first_name`, `users`.`last_name`, `users`.`email`,
SUBSTRING(`locations`.`raw`,-6,4) AS `guaranteed_postcode`
FROM `users` LEFT OUTER JOIN `locations`
ON `users`.`id` = `locations`.`user_id`
WHERE `guaranteed_postcode` NOT IN #this is where the fake col is being used
(
 SELECT `postcode` FROM `postcodes` WHERE `region` IN
 (
  'australia'
 )
)

My question is: why am I unable to use a fake column in the where clause of the same DB query?

Upvotes: 280

Views: 275241

Answers (8)

Mrugesh Vyas
Mrugesh Vyas

Reputation: 1

use HAVING for using alias column in where clause

Upvotes: 0

victor hugo
victor hugo

Reputation: 35838

You can only use column aliases in GROUP BY, ORDER BY, or HAVING clauses.

Standard SQL doesn't allow you to refer to a column alias in a WHERE clause. This restriction is imposed because when the WHERE code is executed, the column value may not yet be determined.

Copied from MySQL documentation

As pointed in the comments, using HAVING instead may do the work. Make sure to give a read at this question too: WHERE vs HAVING.

Upvotes: 543

Sameera Prasad Jayasinghe
Sameera Prasad Jayasinghe

Reputation: 1800

You can use SUBSTRING(locations.raw,-6,4) for where conditon

SELECT `users`.`first_name`, `users`.`last_name`, `users`.`email`,
SUBSTRING(`locations`.`raw`,-6,4) AS `guaranteed_postcode`
FROM `users` LEFT OUTER JOIN `locations`
ON `users`.`id` = `locations`.`user_id`
WHERE SUBSTRING(`locations`.`raw`,-6,4) NOT IN #this is where the fake col is being used
(
SELECT `postcode` FROM `postcodes` WHERE `region` IN
(
 'australia'
)
)

Upvotes: 2

Joni
Joni

Reputation: 111219

Standard SQL (or MySQL) does not permit the use of column aliases in a WHERE clause because

when the WHERE clause is evaluated, the column value may not yet have been determined.

(from MySQL documentation). What you can do is calculate the column value in the WHERE clause, save the value in a variable, and use it in the field list. For example you could do this:

SELECT `users`.`first_name`, `users`.`last_name`, `users`.`email`,
@postcode AS `guaranteed_postcode`
FROM `users` LEFT OUTER JOIN `locations`
ON `users`.`id` = `locations`.`user_id`
WHERE (@postcode := SUBSTRING(`locations`.`raw`,-6,4)) NOT IN
(
 SELECT `postcode` FROM `postcodes` WHERE `region` IN
 (
  'australia'
 )
)

This avoids repeating the expression when it grows complicated, making the code easier to maintain.

Upvotes: 23

themhz
themhz

Reputation: 8424

I am using mysql 5.5.24 and the following code works: 

select * from (
SELECT `users`.`first_name`, `users`.`last_name`, `users`.`email`,
SUBSTRING(`locations`.`raw`,-6,4) AS `guaranteed_postcode`
FROM `users` LEFT OUTER JOIN `locations`
ON `users`.`id` = `locations`.`user_id`
) as a
WHERE guaranteed_postcode NOT IN --this is where the fake col is being used
(
 SELECT `postcode` FROM `postcodes` WHERE `region` IN
 (
  'australia'
 )
)

Upvotes: 2

Hett
Hett

Reputation: 3823

You can use HAVING clause for filter calculated in SELECT fields and aliases

Upvotes: 8

George Khouri
George Khouri

Reputation: 354

Maybe my answer is too late but this can help others.

You can enclose it with another select statement and use where clause to it. 

SELECT * FROM (Select col1, col2,...) as t WHERE t.calcAlias > 0

calcAlias is the alias column that was calculated.

Upvotes: 30

rodion
rodion

Reputation: 6115

As Victor pointed out, the problem is with the alias. This can be avoided though, by putting the expression directly into the WHERE x IN y clause:

SELECT `users`.`first_name`,`users`.`last_name`,`users`.`email`,SUBSTRING(`locations`.`raw`,-6,4) AS `guaranteed_postcode`
FROM `users` LEFT OUTER JOIN `locations`
ON `users`.`id` = `locations`.`user_id`
WHERE SUBSTRING(`locations`.`raw`,-6,4) NOT IN #this is where the fake col is being used
(
 SELECT `postcode` FROM `postcodes` WHERE `region` IN
 (
  'australia'
 )
)

However, I guess this is very inefficient, since the subquery has to be executed for every row of the outer query.

Upvotes: 24

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