Invalid argument supplied for foreach!

Question

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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

Can you give us a `print_r` or `var_dump` of `$format`, prior to the foreach loop? — Charles, Jul 05 '10 at 16:21
$format seems to be populated after the foreach rather than before. Move the processing of the DB output to before the foreach. You should still use a check for array just in case nothing is returned from the database. — Brendan Bullen, Jul 05 '10 at 16:33

score 3 · Accepted Answer · answered Jul 05 '10 at 16:20

3

$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.

answered Jul 05 '10 at 16:20

ceejayoz

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Oh does format need to be a array? I will edit to show where $format comes from. – sark9012 Jul 05 '10 at 16:21
Yes, `foreach` only works on arrays. – ceejayoz Jul 05 '10 at 16:22
instead of doing $format = (array)$format; you can just instantly do foreach((array)$format as $form) – Not Available Jul 05 '10 at 16:24
@xeross Yeah, you can. I prefer to cast it separately for readability - it's easy to pass right over the casting if it's within the foreach call - but that's a matter of preference. – ceejayoz Jul 05 '10 at 16:25

score 0 · Answer 2 · answered Jul 05 '10 at 16:23

0

are you sure $format is an array ? put an

<?php echo gettype($format); ?>

before the foreach loop

answered Jul 05 '10 at 16:23

Not Available

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score 0 · Answer 3 · answered Jul 05 '10 at 16:28

0

$format is not array or not exists! Before foreach

if(is_array($format)){  
   foreach($format ...
}

answered Jul 05 '10 at 16:28

turbod

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Invalid argument supplied for foreach!

3 Answers3

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