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Home/ Questions/Q 8327213
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T01:02:00+00:00 2026-06-09T01:02:00+00:00

$retrieve = mysql_query(SELECT * FROM `users`) or die(mysql_error()); $object -> userDetails = array(); while($retrieveArray

  • 0
$retrieve = mysql_query("SELECT * FROM `users`") or die(mysql_error());

$object -> userDetails = array();

while($retrieveArray = mysql_fetch_array($retrieve))
{
    $item -> userId =  $retrieveArray['id'];
    $item -> userEmail = $retrieveArray['email'];
    $item -> userLocation = $retrieveArray['location'];
    $item -> userFirstName = $retrieveArray['firstname'];
    $item -> userLastName = $retrieveArray['lastname'];
    $object ->userDetails[] = $item;
}   
$json = json_encode($object);
echo $json;

Is there something wrong with this code? My output only displays the first row of my database.

{"userDetails":[{"userId":"1","userEmail":"EmailAddress@gmail.com","userLocation":"HomeAddress","userFirstName":"Allan","userLastName":"Knocks"}]}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T01:02:01+00:00Added an answer on June 9, 2026 at 1:02 am

    Try this:

    $retrieve = mysql_query("SELECT id, email, location, firstname, lastname FROM `users`") or die(mysql_error());
    $userDetails= array();
    while($row = mysql_fetch_assoc($retrieve)) {
      $userDetails[] = $row;
    }   
    $json = json_encode(array('userDetails' => $userDetails));
    echo $json;
    
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