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• “Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP 28 answers

I am having trouble with form validation for checkboxes.

Here is the checkbox code:

<div class="form-group">
    <div class="checkbox checkbox-danger checkbox-inline">
        <input type="checkbox" name="defect[]" value="{$ng_id}">
    </div>
</div>

If I just do something like this then it echo's out whatever I select.

foreach($_POST['defect'] as $defects) {
echo $defects;
}

I have tried to use the empty function on its own as well as this:

if(empty($defects) || count($defects) < 1) {

but I just keep getting the error:

Undefined index: defect

</div>

Answer 1

A posted array is usually not set if nothing was ticked.

You can do the following check if you want to also check it's an array:-

if(@!is_array($_POST['defect'])){
    // Error
} else {
    foreach($_POST['defect']){
        // do stuff
    }
}

This is suppressing the warning using @ otherwise you would get the "not defined" error.

If you want to avoid that you could just do:

if(!isset($_POST['defect'])){
   // Now check it's an array and has a count > 0
}

.. but in addition you still need to ensure it's an array and has a count greater than zero. The advantage with suppressing the warning in this specific instance is that you know it's an array PLUS you know there must be a value in it otherwise no array would have been passed in from the POST. It's one of those rare moments where suppressing the warning saves you doing 2 additional IF checks to avoid other warnings.You HAVE to check it's an array otherwise you will get the "invalid foreach" warning...