Quote:
Originally Posted by giaguaro
lol as i told you i am definitely a beginner and have a poor english.
Now:
UPDATE user SET usergroupid=eliminare WHERE usergroupid=3 AND email LIKE %sina.com%
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '%sina.com%' at line 1
what happened?
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Sorry, that's my fault. You need that in quotes, like:
Code:
UPDATE user SET usergroupid=eliminare WHERE usergroupid=3 AND email LIKE '%sina.com%'
But also, you can't have 'eliminare' there, because the usergroupid must be a number.