Question

I’ve got the following query:

SELECT name, GROUP_CONCAT(job SEPARATOR ','),
    count(*) FROM users GROUP BY name ORDER BY name ASC

Is it possible to only concat the job fields that do not contain ”, and if so how to? I can’t do a WHERE clause, because I still need to ones that don’t have a ‘job’.

Picture the following table:

+-------+----------+
| name  | job      |
+-------+----------+
| Bob   | doctor   |
| Bob   | teacher  |
| Frank | dentist  |
| Tom   |          |
| Tom   |          |
| Tom   |          |
| Tom   |          |
| Tom   | salesman |
+-------+----------+

The current query would result in:

+--------+---------------------------------+----------+
| name   | GROUP_CONCAT(job SEPARATOR ',') | count(*) |
+--------+---------------------------------+----------+
| Bob    | doctor, teacher                 | 2        |
| Frank  | dentist                         | 1        |
| Tom    | ,,,,salesman                    | 5        |
+--------+---------------------------------+----------+

But I want it to be:

+--------+---------------------------------+----------+
| name   | GROUP_CONCAT(job SEPARATOR ',') | count(*) |
+--------+---------------------------------+----------+
| Bob    | doctor,teacher                  | 2        |
| Frank  | dentist                         | 1        |
| Tom    | salesman                        | 5        |
+--------+---------------------------------+----------+

(Don’t mind the logic of this table. It is just a simple example to make things clear)

Reason for this is, I am going to explode() the GROUP_CONCAT(job SEPARATOR ‘,’) in php, and in the actual table, there will easily be 1000 useless ,,’s and this would make the explode function very slow.
Thanks,

lordstyx.

Answer 1

I’ve found a solution: instead of making the job ”, I’ll just make the field Null, and the query ignores it.