On 11/22/2012 04:10 PM, Ben Mildren wrote:
SELECT id FROM mytable WHERE type IN(x,y,z) GROUP BY id;
Ben you were almost there ;)
SELECT id FROM mytable WHERE type IN(x,y,z) GROUP BY id HAVING COUNT(id)=<num of
params>
The only bad is the hardcoded parameter in the HAVING, may be it might be
improved.
Anyway if the query is handwritten then you just hand-modify that too,
if it is built from code I can't imagine counting the parameters in the code
being so hard.
Cheers
Claudio
On 22 November 2012 15:01, Neil Tompkins <neil.tompk...@googlemail.com> wrote:
Michael,
Thanks this kind of works if I'm checking two types. But what about if I
have 5 types ?
On Thu, Nov 22, 2012 at 2:53 PM, Michael Dykman <mdyk...@gmail.com> wrote:
response did not go to the list..
I assume that you mean the id must be associated with both type=5 AND
type=2 as opposed to type=5 OR type=2;
in some dialect of SQL (not mysql) you can do this:
select distinct id from 'table' where type=5
intersect
select distinct id from 'table' where type=2
As INTERSECT is not avilable under mysql, we will have to go the JOIN route
select distinct a.id from mytable a
inner join mytable b on (a.id=b.id)
where a.type= 2 and b.type = 5;
- michael dykman
On Thu, Nov 22, 2012 at 9:30 AM, Neil Tompkins
<neil.tompk...@googlemail.com> wrote:
Hi,
I'm struggling with what I think is a basic select but can't think how to
do it : My data is
id,type
1000,5
1001,5
1002,2
1001,2
1003,2
1005,2
1006,1
From this I what to get a distinct list of id where the type equals 2
and 5
Any ideas ?
Neil
--
- michael dykman
- mdyk...@gmail.com
May the Source be with you.
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- michael dykman
- mdyk...@gmail.com
May the Source be with you.
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Claudio
