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Home/ Questions/Q 8372077
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T14:19:36+00:00 2026-06-09T14:19:36+00:00

Probably a simple solution here, but here is my code: <?php include(connect.php); $data =

  • 0

Probably a simple solution here, but here is my code:

<?php
include("connect.php");
$data = mysql_query("SELECT * FROM table ORDER BY name");
$info = mysql_fetch_array($data);
?>

and then printing it with:

<select>
 <?php
  while($info = mysql_fetch_array($data)) {
     echo '<option value="'.$info['value'].'" rel="'.$info['del'].'">'.$info['name'].'</option>';
  }
 ?>
</select>

For some reason it’s returning all results except for the very first one? Ex. I’m ordering by name and the first one should be Apples (id = 9), but it’s skipping that and returning Blueberries (id = 5)

Thanks for any help!

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  1. Editorial Team
    Editorial Team
    2026-06-09T14:19:37+00:00Added an answer on June 9, 2026 at 2:19 pm

    You are calling mysql_fetch_array() immediately after mysql_query() before your loop, which retrieves one row into $info and advances the record pointer to the second row. When you then enter your fetch loop, the rowset is already pointing to the second row.

    $data = mysql_query("SELECT * FROM table ORDER BY name");
    // Don't do this!
    // $info = mysql_fetch_array($data);
    // Instead just proceed with your while loop to fetch rows...
    
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