I guess this is this closest, but apparently in this case if the "3" at the 
beginning doesn't exact, and even using LIKE '%3,%' (removing the first comma) 
would match an entry like 33, !
  I need a better solution, does mysql have something similar to explode() in 
  if it does, then how to apply it in my situation?
  of course it done using PHP but that's not feasible at all.

Tom Atkinson <[EMAIL PROTECTED]> wrote:
  If you have the literal string "3,31,11,10" in a field in your table 
then you can 'search' for numbers in the string like this:

SELECT * FROM table WHERE CONCAT(',', tablefield1, ',') LIKE '%,3,%'

I don't know if it's the best way of doing it but it works for me.

Natalie Leotta wrote:
> I'm not sure I understand your question, but my interpretation would be
> solved like this
> select * from table where variable in (select id from tablefield1)
> I hope this helps you!!
> Natalie
> On 9/16/06, Hassan wrote:
>> Hi My Best List, I have a field in a table contains "3,31,11,10" (let say
>> tablefield1) and I have a given value which I need to check if it 
>> exists in
>> that list.
>> I've tried:
>> SELECT * FROM table WHERE 3 IN ( tablefield1 );
>> Any ideas?
>> ---------------------------------
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