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 Quote Link to comment 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 />'; } Quote Link to comment 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. Quote Link to comment Share on other sites More sharing options...
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