Where 子句中的未知列

Question

I have a simple query:

SELECT u_name AS user_name FROM users WHERE user_name = "john";

I get Unknown Column 'user_name' in where clause . Can I not refer to 'user_name' in other parts of the statement even after select 'u_name as user_name' ?

Answer 1

SQL is evaluated backwards, from right to left. So the where clause is parsed and evaluate prior to the select clause. Because of this the aliasing of u_name to user_name has not yet occurred.

Answer 2

关于什么：

SELECT u_name AS user_name FROM users HAVING user_name = "john";

Answer 3

See the following MySQL manual page: http://dev.mysql.com/doc/refman/5.0/en/select.html

"A select_expr can be given an alias using AS alias_name. The alias is used as the expression's column name and can be used in GROUP BY, ORDER BY, or HAVING clauses."

(...)

It is not permissible to refer to a column alias in a WHERE clause, because the column value might not yet be determined when the WHERE clause is executed. See Section B.5.4.4, “Problems with Column Aliases”.

Answer 4

select u_name as user_name from users where u_name = "john";

Think of it like this, your where clause evaluates first, to determine which rows (or joined rows) need to be returned. Once the where clause is executed, the select clause runs for it.

To put it a better way, imagine this:

select distinct(u_name) as user_name from users where u_name = "john";

You can't reference the first half without the second. Where always gets evaluated first, then the select clause.

Answer 5

If you're trying to perform a query like the following (find all the nodes with at least one attachment) where you've used a SELECT statement to create a new field which doesn't actually exist in the database, and try to use the alias for that result you'll run into the same problem:

SELECT nodes.*, (SELECT (COUNT(*) FROM attachments 
WHERE attachments.nodeid = nodes.id) AS attachmentcount 
FROM nodes
WHERE attachmentcount > 0;

You'll get an error "Unknown column 'attachmentcount' in WHERE clause".

Solution is actually fairly simple - just replace the alias with the statement which produces the alias, eg:

SELECT nodes.*, (SELECT (COUNT(*) FROM attachments 
WHERE attachments.nodeid = nodes.id) AS attachmentcount 
FROM nodes 
WHERE (SELECT (COUNT(*) FROM attachments WHERE attachments.nodeid = nodes.id) > 0;

You'll still get the alias returned, but now SQL shouldn't bork at the unknown alias.

Answer 6

Your defined alias are not welcomed by the WHERE clause you have to use the HAVING clause for this

SELECT u_name AS user_name FROM users HAVING user_name = "john";

OR you can directly use the original column name with the WHERE

SELECT u_name AS user_name FROM users WHERE u_name = "john";

Same as you have the result in user defined alias as a result of subquery or any calculation it will be accessed by the HAVING clause not by the WHERE

SELECT u_name AS user_name ,
(SELECT last_name FROM users2 WHERE id=users.id) as user_last_name
FROM users  WHERE u_name = "john" HAVING user_last_name ='smith'

Answer 7

Either:

SELECT u_name AS user_name
FROM   users
WHERE  u_name = "john";

or:

SELECT user_name
from
(
SELECT u_name AS user_name
FROM   users
)
WHERE  u_name = "john";

The latter ought to be the same as the former if the RDBMS supports predicate pushing into the in-line view.

Answer 8

更正：

SELECT u_name AS user_name FROM users WHERE u_name = 'john';

Answer 9

No you need to select it with correct name. If you gave the table you select from an alias you can use that though.

Answer 10

No you cannot. user_name is doesn't exist until return time.

Answer 11

由第 1 行和第 2 行引起并由第 3 行解决的WHERE子句中的未知列：

1. $sql = "SELECT * FROM users WHERE username =".$userName;
2. $sql = "SELECT * FROM users WHERE username =".$userName."";
3. $sql = "SELECT * FROM users WHERE username ='".$userName."'";

Answer 12

May be it helps.

You can

SET @somevar := '';
SELECT @somevar AS user_name FROM users WHERE (@somevar := `u_name`) = "john";

It works.

BUT MAKE SURE WHAT YOU DO!

• Indexes are NOT USED here
• There will be scanned FULL TABLE - you hasn't specified the LIMIT 1 part
• So, - THIS QUERY WILL BE SLLLOOOOOOW on huge tables.

But, may be it helps in some cases

Answer 13

While you can alias your tables within your query (ie, "SELECT u.username FROM users u;"), you have to use the actual names of the columns you're referencing. AS only impacts how the fields are returned.

Answer 14

Not as far as I know in MS-SQL 2000/5. I've fallen foul of this in the past.

Answer 15

Just had this problem.

Make sure there is no space in the name of the entity in the database.

eg ' user_name' instead of 'user_name'

Answer 16

SELECT user_name
FROM
(
SELECT name AS user_name
FROM   users
) AS test
WHERE  user_name = "john"

Answer 17

try your task using IN condition or OR condition and also this query is working on spark-1.6.x

 SELECT  patient, patient_id FROM `patient` WHERE patient IN ('User4', 'User3');

or

SELECT  patient, patient_id FROM `patient` WHERE patient = 'User1' OR patient = 'User2';

Answer 18

For me the root of the problem was a number which I copied to use in a WHERE clause. The number had "invisible" symbol, at least for MySQL Workbench. I placed the number in the Chrome console it was clearly visible.

Answer 19

I had the same problem, I found this useful.

mysql_query("SELECT * FROM `users` WHERE `user_name`='$user'");

remember to put $user in ' ' single quotes.