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Home/ Questions/Q 6534291
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T10:13:59+00:00 2026-05-25T10:13:59+00:00

Hello and Good Morning, I am still learning PHP and for some reason my

  • 0

Hello and Good Morning,

I am still learning PHP and for some reason my script will not post any data in my foreach loop. Any Idea why? The emailRow Echos out fine but I am going to remove My code is below:

<?php 

include 'includes/header.php';


$accountUser = array();
$upgradeEmail = $_GET['currEmail'];
$emailQuery = "SELECT fbID, firstName, lastName FROM users WHERE emailOne='".$upgradeEmail."' AND authLevel=0";

<?php echo $emailRow['fbID']; ?>
<?php echo $emailRow['firstName']; ?> 
<?php echo $emailRow['lastName']; ?>

 while($emailRow = mysql_fetch_assoc($emailQuery, $conn))
 {
    $accountUser[]=$emailRow;
 }
?> 
 <table>
<?php foreach($accountUser as $emailData) { ?>


<tr><td> <?php emailData['fbID']; ?> </td><td><?php emailData['firstName']; ?></td><td><?php emailData['lastName']; ?></td></tr>

<?php } ?>
</table>
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  1. Editorial Team
    Editorial Team
    2026-05-25T10:14:00+00:00Added an answer on May 25, 2026 at 10:14 am

    You have constructed your SQL query in $emailQuery, but never executed it. Call mysql_query(), and pass its result resource to mysql_fetch_assoc().

    $emailQuery = "SELECT fbID, firstName, lastName FROM users WHERE emailOne='".$upgradeEmail."' AND authLevel=0";
    $result = mysql_query($emailQuery);
    if ($result)
    {
      while($emailRow = mysql_fetch_assoc($result, $conn))
      {
         $accountUser[]=$emailRow;
      }
    }
    else // your query failed
    {
       // handle the failure
    }
    

    Please also be sure to protect your database from SQL injection by calling mysql_real_escape_string() on $upgradeEmail since you’re receiving it from $_GET.

    $upgradeEmail = mysql_real_escape_string($_GET['currEmail']);
    
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