PHP form variables, error for mysql_fetch_array()

Question

Not sure what I am doing wrong here.. it is giving me the error: *Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource*

What am I missing? Just edited code.. still has a problem with mysql_fetch_array()

<?
    //Extract data from form
    if(isset($_POST["editUserName"])){
 $myUserName = mysqli_real_escape_string($myConn, $_POST["editUserName"]);
    }

 if(isset($_POST["updateSubmit"])){ 
  $mySubmit = mysqli_real_escape_string($myConn, $_POST["updateSubmit"]);
    }


    //Verify form was submitted before beginning database interaction
    if ($mySubmit != "")
    {

    //Create an SQL delete statement to select the desired record
    $mySQLselect = "SELECT * FROM tblUsers WHERE userName = '.$myUserName.'";
    $myRS = mysqli_query($myConn, $mySQLselect) or die('Error: ' .mysqli_error($myConn));
    $myData = mysql_fetch_array($myRS);

    //Create form output for editing
    echo("<form name='frmEdit' id='frmEdit' action='doEdit.php' method='post'>");
    echo("<input type='hidden' name='hidUserName' id='hidUserName' value='.$myUserName.'/>");       
    echo("<p>User Name: <input type='text' name='BuserName' id='BuserName' value='$myData[userName]'/></p>");
    echo("<p>Password: <input type='text' name='BuserPass' id='BuserPass' value='$myData[userPass]'/></p>");
    echo("<p>First Name: <input type='text' name='BfirstName' id='BfirstName' value='$myData[userFirst]'/></p>");
    echo("<p>Last Name: <input type='text' name='BlastName' id='BlastName' value='$myData[userLast]'/></p>");
    echo("<p>Address: <input type='text' name='Baddress' id='Baddress' value='$myData[address]'/></p>");
    echo("<p>City: <input type='text' name='Bcity' id='Bcity'value='$myData[city]'/></p>");
    echo("<p>State: <input type='text' name='Bstate' id='Bstate' value='$myData[state]'/></p>");
    echo("<p>Zip: <input type='text' name='Bzip' id='Bzip' value='$myData[zip]'/></p>");
    echo("<p>Email: <input type='text' name='Bemail' id='Bemail'$myData[email]'/></p>");
    echo("<p>Phone: <input type='text' name='Bphone' id='Bphone'$myData[phone]'/></p>");
    echo("<p><input type='submit' name='btnDoEdit' id='btnDoEdit' value='Make Changes'/></p>");
    echo("</form>");
            }   

            ?>

Mujahid · Accepted Answer · 2011-06-09 08:22:18Z

1

All the way in your code you've used mysqli but for fetching data you're using mysql_fetch_array, isn't is mysqli_fetch_array()?

answered Jun 9, 2011 at 8:22

Mujahid

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AR. · Accepted Answer · 2011-06-08 22:59:50Z

0

'isset' will give you a boolean value, not a value in the $_POST array. It just checks whether that value exists (is set).

     if(isset($_POST["editUserName"])){
     $myUserName = mysqli_real_escape_string($myConn, $_POST["editUserName"]);
        }

     if(isset($_POST["updateSubmit"])){ 
      $mySubmit = mysqli_real_escape_string($myConn, $_POST["updateSubmit"]);
        }

answered Jun 8, 2011 at 22:59

AR.

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