[Klug-general] Mysql subqueries in PHP
Dan Attwood
danattwood at gmail.com
Wed Jun 29 10:54:28 UTC 2011
this one works peter thanks for that
On 29 June 2011 11:28, Peter Childs <pchilds at bcs.org> wrote:
>
>
> On 29 June 2011 10:54, David Halliday <david.halliday at gmail.com> wrote:
>
>> Since I said I'd post a follow up (and I have worked it out after a
>> coffee) using a LEFT OUTER JOIN I shall.
>>
>> SELECT u.*
>> FROM usr AS u
>> LEFT OUTER JOIN mahara.group_member AS gm
>> ON u.id = gm.member
>> WHERE gm.member IS NULL
>> OR
>> (
>> gm.member IS NOT NULL
>> AND gm.group != 10
>> );
>>
>> Working on the same principal.
>> We want every row from the usr table
>> That doesn't have an entry in group_member (so no member)
>> Unless that entry has a group that isn't '10'
>>
>> So you could also use:
>> SELECT u.*
>> FROM usr AS u
>> LEFT OUTER JOIN mahara.group_member AS gm
>> ON u.id = gm.member
>> WHERE gm.group IS NULL
>> OR gm.group != 10
>> ;
>>
>> NOT IN statements are best avoided unless you have a small list of
>> predefined items for performance purposes.
>>
>> On 29 June 2011 10:07, Dan Attwood <danattwood at gmail.com> wrote:
>>
>>>
>>> Have you tested the subquery in php?
>>>>
>>>
>>> I have and that's where it wasn't working.
>>> However I have now found the reason and it was because I'm a total idiot.
>>> Next time i'll double check i'm look at the right database D'oh
>>>
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>>> Kent at mailman.lug.org.uk
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>>>
>>
>>
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>>
>
> That will not work. If there are multiple records in gm with member = id
> and one of them is 10.
>
> Works fine if you have a 1 to 1 relationship however!
>
> Sorry for pulling your query apart so quickly.
>
> Try.
>
> SELECT u.*
> FROM usr AS u
> LEFT OUTER JOIN mahara.group_member AS gm
> ON (u.id = gm.member and gm.group = 10)
> WHERE gm.group IS NULL
> ;
>
> I've not checked it however.....
>
> Peter.
>
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> Kent mailing list
> Kent at mailman.lug.org.uk
> https://mailman.lug.org.uk/mailman/listinfo/kent
>
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