kigroy Posted December 16, 2008 Share Posted December 16, 2008 I get the resource ID #3 error when I run this code: In my header tags: <?php $conn = mysql_connect('dbname','username','password') or trigger_error("SQL", E_USER_ERROR); $db = mysql_select_db('tablename',$conn) or trigger_error("SQL", E_USER_ERROR); $name = mysql_real_escape_string($_GET['name']); //**from form submitted. the form code works i doubled checked**// $query = "SELECT column_name1 FROM column_name2 WHERE column_name2 = '$name'"; //** query works too, i tested in phpMyAdmin**// $result = mysql_query($query, $conn); ?> In my body tags: <?PHP echo $result ?> My server is WAMPSERVER 2.0 and my php running is 5.2.7. Thoughts? Thanks Quote Link to comment Share on other sites More sharing options...
rhodesa Posted December 16, 2008 Share Posted December 16, 2008 you need to fetch a row from the result... <?php $conn = mysql_connect('dbname','username','password') or trigger_error("SQL", E_USER_ERROR); $db = mysql_select_db('tablename',$conn) or trigger_error("SQL", E_USER_ERROR); $name = mysql_real_escape_string($_GET['name']); //**from form submitted. the form code works i doubled checked**// $query = "SELECT column_name1 FROM column_name2 WHERE column_name2 = '$name'"; //** query works too, i tested in phpMyAdmin**// $result = mysql_query($query, $conn); $row = mysql_fetch_array($result); ?> <?php echo $row[0]; ?> Quote Link to comment Share on other sites More sharing options...
kigroy Posted December 16, 2008 Author Share Posted December 16, 2008 Thanks! Worked like a charm. BTW, I'm a noob if you haven't figured it out. Quote Link to comment Share on other sites More sharing options...
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