RE: php - mysql problem

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[snip]
> SELECT table1., *table2.* 
> FROM table1 LEFT OUTER JOIN table2 
> ON(table1.id = table2.id)
[/snip]

Oops, typo.

SELECT table1.*, table2.* 
FROM table1 LEFT OUTER JOIN table2 
ON(table1.id = table2.id)

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