FD_F Posted April 7, 2009 Share Posted April 7, 2009 thay both geting same value FDF@gmail.com when i echo but when i using SELECT id FROM users WHERE user_email = '$temp' only $temp = "FDF@gmail.com" works when i using $temp = $table[7]; i get Incorrect table name '' thanks for helping Quote Link to comment https://forums.phpfreaks.com/topic/152940-solved-weird-problem-difrent-between-temp-table7-and-temp-fdfgmailcom/ Share on other sites More sharing options...
Zane Posted April 7, 2009 Share Posted April 7, 2009 what is $table[7]? echo it Quote Link to comment https://forums.phpfreaks.com/topic/152940-solved-weird-problem-difrent-between-temp-table7-and-temp-fdfgmailcom/#findComment-803204 Share on other sites More sharing options...
FD_F Posted April 7, 2009 Author Share Posted April 7, 2009 when i echo i get same result in both echo $table[7]; gets -> FDF@gmail.com and also in $temp = "FDF@gmail.com" echo $temp; geting FDF@gmail.com Quote Link to comment https://forums.phpfreaks.com/topic/152940-solved-weird-problem-difrent-between-temp-table7-and-temp-fdfgmailcom/#findComment-803207 Share on other sites More sharing options...
Zane Posted April 7, 2009 Share Posted April 7, 2009 well, we've established that table[7] and temp are the same thing. to be blunt....what are doing? what's your question Quote Link to comment https://forums.phpfreaks.com/topic/152940-solved-weird-problem-difrent-between-temp-table7-and-temp-fdfgmailcom/#findComment-803222 Share on other sites More sharing options...
FD_F Posted April 7, 2009 Author Share Posted April 7, 2009 solved i used $name = trim($temp); $sql = ("SELECT id FROM users WHERE user_email = '$name'");. it was spaces problem thanks Quote Link to comment https://forums.phpfreaks.com/topic/152940-solved-weird-problem-difrent-between-temp-table7-and-temp-fdfgmailcom/#findComment-803273 Share on other sites More sharing options...
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