Hi Peter,

Thanks a lot - that does the trick!

Incredibly quick response :)

Ian

> -----Original Message-----
> From: Peter Normann [mailto:[EMAIL PROTECTED]]
> Sent: 16 June 2002 17:06
> To: 'mySQL list'; [EMAIL PROTECTED]
> Subject: RE: A LEFT OUTER JOIN question.
> 
> 
> Try
> 
> SELECT Items.ItemID, Stuff.info FROM Items LEFT JOIN Stuff ON
> (Items.ItemID = Stuff.ItemID && (type=0 || type IS NULL));
> 
> 
> Peter Normann
> 
> -----Original Message-----
> From: mySQL list [mailto:[EMAIL PROTECTED]] 
> Sent: 16. juni 2002 17:59
> To: [EMAIL PROTECTED]
> Subject: A LEFT OUTER JOIN question.
> 
> 
> I have two tables, 'items' and 'stuff'.
> 
> Items has a primary key of itemid. and looksmlike this
> 
> ItemID  other fields...
> 
> 1       ...
> 2
> 3
> 4
> 5
> 6
> 
> Stuff contains something like this:
> 
> ItemID  type  info
> 1       0    blah blah
> 2       0    something
> 2       1    ...
> 3       1    ...
> 
> I want a query which returns all the items, and if it exists, the info
> field from related 'stuff' of type 0.
> 
> So far I've got:
> 
> SELECT Items.ItemID, Stuff.info FROM Items LEFT OUT JOIN Stuff ON
> Items.ItemID = Stuff.ItemID WHERE (type = 0 OR type = NULL);
> 
> This works if there are no Stuff records with a type other than 0, eg,
> the last one in the example. The Query returns items 1,2,4,5, & 6, no 3.
> 
> How can I make a query which returns all items and type 0's info (if it
> exists)?
> 
> ian
> 
> 
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