TEDSON Posted September 7, 2010 Share Posted September 7, 2010 I was expecting a return string, but got Resource id #2 instead. How do I have a string returned instead of that? heres my table user code Bob One Ted Two I dont get it :-\ <html> <body> <?php $con = mysql_connect("localhost","user","PassWord"); if (!$con) { echo 'Could not connect to MySQL server. <br />Error # ', mysql_errno(), ' Error msg: ', mysql_error(); exit; } $db = mysql_select_db("userdb") or die("Unable to select database"); if (!$db) { echo 'Could not select db. <br />Error # ', mysql_errno(), ' Error msg: ', mysql_error(); exit; } $query = "SELECT code from usertbl WHERE user = 'Ted' LIMIT 0 , 30"; $result = mysql_query($query, $con); if (!$result) { echo 'Could not query server. <br />Error # ', mysql_errno(), ' Error msg: ', mysql_error(); exit; } echo $result; ?> </body> </html> When I use that query in phpmyadmin it works Any pointers much appreciated Link to comment https://forums.phpfreaks.com/topic/212732-resource-id-2/ Share on other sites More sharing options...
Pikachu2000 Posted September 7, 2010 Share Posted September 7, 2010 You have to actually do something with the result resource with mysql_fetch_array(), mysql_fetch_assoc(), $mysql_fetch_row(), etc. $query = "SELECT code from usertbl WHERE user = 'Ted' LIMIT 0 , 30"; $result = mysql_query($query, $con); if (!$result) { echo 'Could not query server. <br />Error # ', mysql_errno(), ' Error msg: ', mysql_error(); exit; } while( $array = mysql_fetch_assoc($result) ) { echo $array['code'] . '<br />'; } Link to comment https://forums.phpfreaks.com/topic/212732-resource-id-2/#findComment-1108131 Share on other sites More sharing options...
TEDSON Posted September 7, 2010 Author Share Posted September 7, 2010 Crikey! I was not expecting such a quick response, I'm not complaining and I appreciate it, thanks. Perfect. Panic over. Link to comment https://forums.phpfreaks.com/topic/212732-resource-id-2/#findComment-1108134 Share on other sites More sharing options...
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