tahakirmani Posted December 18, 2012 Share Posted December 18, 2012 I am writing the following code in my PHP program, but mysql is not returning any row. Kindly check and help to find out the mistakes in the following code. <?php $query= "SELECT * FROM `email` where `id` =' ".$_SESSION['user_id']. "'" ; $query_run2= mysql_query($query); $query_num_rows= mysql_num_rows($query_run2); if ($query_num_rows==0) { echo "Invalid Query"; } else { $query_result = mysql_result ($query_run2, 0, 'name'); echo "Welcome" ; echo $_SESSION['user_id']; } ?> It is giving me the following Output. Log out Warning: mysql_num_rows() expects parameter 1 to be resource, string given in F:\xampp\htdocs\Email_address\welcome.php on line 20 Invalid Query Thanks, Taha. Quote Link to comment Share on other sites More sharing options...
Barand Posted December 18, 2012 Share Posted December 18, 2012 Check output from mysql_error() Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted December 18, 2012 Share Posted December 18, 2012 The code you posted cannot possibly produce the error you are getting, ... string given ... unless your actual code is something like - $query_num_rows= mysql_num_rows($query); Quote Link to comment Share on other sites More sharing options...
tahakirmani Posted December 18, 2012 Author Share Posted December 18, 2012 Thank you so much for the help..I did a stupid mistake, when i use mysql_error() function,then i come to know that i was not connected to Database. Quote Link to comment Share on other sites More sharing options...
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