Question

How can I use the if statement in MYSQL query to return yes if it’s true and no if it’s not?

See the example :

IF(table_name.field_name_value = 1, 'yes','no') AS active

I’m trying something like this but with no success

Answer 1

From the documentation on IF() function:

mysql> SELECT IF(1<2,'yes','no');
        -> 'yes'

So, basically, the syntax is as mentioned above.

EDIT:

You are saying my example is incorrect. Thus, I provide you a proof (which proves the example from the documentation I mentioned above):

I have a table, which has uid column, containing ID of the user. The table is part of the database on some site (which is not relevant). When I make the following query:

SELECT `uid`, IF(`uid`=3, 'yes', 'no') AS `active` FROM `mysite_users`;

I receive the following result:

+-----+--------+
| uid | active |
+-----+--------+
|   0 | no     |
|   1 | no     |
|   3 | yes    |
|   8 | no     |
|   9 | no     |
|  10 | no     |
|  11 | no     |
|  12 | no     |
|  13 | no     |
|  14 | no     |
|  15 | no     |
+-----+--------+
11 rows in set (0,00 sec)

Which is exactly what I would expect (and what should be expected after reading the documentation). Is it still not working for you?