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Home/ Questions/Q 922351
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T19:00:57+00:00 2026-05-15T19:00:57+00:00

Possible Duplicate: Invalid argument supplied for foreach() I have the following code: <? foreach($format

  • 0

Possible Duplicate:
Invalid argument supplied for foreach()

I have the following code:

<?
foreach($format as $form)
{
    echo $form;
    ?>
    <ul>
        <?
        $s = $database->onlineFormatUsers($form);
        while($row=mysql_fetch_assoc($s))
        {
            $username=$row['username'];
            $id=$row['id'];?>
            <li><a href="../userprofile.php?id=<?echo $id?>"><?echo "$username";?></a></li>
        <?
        }
        ?>
    </ul>
    <?
}
?>

<? 
//the active formats
$f = $database->activeFormats();
while($row=mysql_fetch_assoc($f))
{
    $format=$row['name'];
}
?>

It is saying its an invalid argument?
Any reason why?
Thanks

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  1. Editorial Team
    Editorial Team
    2026-05-15T19:00:58+00:00Added an answer on May 15, 2026 at 7:00 pm

    $format is probably not an array.

    Wrap the foreach block in an if(is_array($format)) { } block or cast it to an array by doing $format = (array)$format.

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