Verbat Posted January 23, 2008 Share Posted January 23, 2008 $query45 = "UPDATE test SET $ecode = '$cjkiop' WHERE username='$nick'"; $fdsftrt = dbquery($query45 ,$db_connect); For some reason it doesnt read the $ecode info to insert into the test table. The $ecode info is just few latters info. Quote Link to comment Share on other sites More sharing options...
revraz Posted January 23, 2008 Share Posted January 23, 2008 Where do you define it? Quote Link to comment Share on other sites More sharing options...
Verbat Posted January 23, 2008 Author Share Posted January 23, 2008 Where do you define it? Does it matter? It takes the info from another page - $ecode = $_POST['ecode']; Quote Link to comment Share on other sites More sharing options...
kts Posted January 23, 2008 Share Posted January 23, 2008 I would suggest echoing the variables to make sure things are being passed properly. Quote Link to comment Share on other sites More sharing options...
revraz Posted January 23, 2008 Share Posted January 23, 2008 Sure it matters if you want help and don't provide all the code to help you with. You say it doesn't read the variable, yet you don't state where it's set. We're not mind readers. Where do you define it? Does it matter? It takes the info from another page - $ecode = $_POST['ecode']; Quote Link to comment Share on other sites More sharing options...
Verbat Posted January 23, 2008 Author Share Posted January 23, 2008 I got it guy's, thanks. Just had some problem in the code. Quote Link to comment Share on other sites More sharing options...
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