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function theme_test($form) {

  $output = '';

  foreach($form['field'] as $key => $val){

    $output .= '<div class="test">';
    $output .= drupal_render($val['uid'. $key]);
    $output .= '</div>';
    unset($key,$val);
  }

  return $output;

}

When I run this code, I get the following warning message:

Warning: Cannot use a scalar value as an array in theme_test()

How do I resolve this, and what is the reason behind this warning?

  • You should add more information, especially why you expect $val to be an array. As it is, it's more a question about PHP, and not a question specific for Drupal. The meaning of the warning doesn't change, in Drupal. – kiamlaluno Sep 23 '11 at 11:57
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There's somehting funky happening on this line:

$output .= drupal_render($val['uid'. $key]);

$val is not an array, you'll need to rethink your PHP code. Do a Google search for "Warning: Cannot use a scalar value as an array" and you'll see a bunch of cases where other users are dealing with the same issue. As an example:

PHP : Cannot use a scalar value as an array

EDIT:

Looking at your code again, this line looks suspect:

unset($key,$val);

You're in a loop that is using these values. Applying an unset could result in the issue in addition to the point I made above.

  • That is true: Using unset($key,$val); is completely useless; it would be like unsetting the local variables used in a function before it returns the control back to the caller function. – kiamlaluno Sep 23 '11 at 11:54

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