PHP form checkbox and undefined index

Question

I am getting an \"Undefined index\" error when submitting a form with an un-checked checkbox. Is there any other way besides running an \"isset\" or \"empty\" check on each

Answer 1

If you want a on/off checkbox you can write a hidden value before you write the checkbox.

<input type="hidden" name="checkbox1" value="no" />
<input type="checkbox" name="checkbox1" value="yes" />

This will always return a value, either no (default unless checkbox is checked by default) or yes.

You can validate input with the filter functions with FILTER_VALIDATE_BOOLEAN.

Its easier if you write a function for this, like formCheckbox($name), with options for values (value 'on' means checkbox is checked by default), attributes, etc.

Answer 2

try the following

<?php
//first part of the query
$query = "UPDATE table SET ";

$howManyCheckboxes = 5;
//for every checkbox , see if it exist , if it is, add it to the query
for($i=1;$i<=$howManyCheckboxes;$i++)
{
   if(isset($_POST['checkbox'.$i]))
   {
      $query = $query . " checkbox".$i."='".escape($_POST['checkbox'.$i])."',";
   }
}
//lets remove the last coma
$query = substr($query,0,-1);
//add the rest of the query
$query = $query . " LIMIT 1";

$result = mysql_query( $query );
if( ! $result ) echo "um, not everything went as expected.";
?>

Answer 3

You could write a function that checks whether a checkbox was checked:

function checkbox_value($name) {
    return (isset($_POST[$name]) ? 1 : 0);
}

Now call that function in your query like this:

$sql =  'UPDATE table SET '.
        'checkbox1 = '. checkbox_value('checkbox1') .','.
        'checkbox2 = '. checkbox_value('checkbox2') .','.
        'checkbox3 = '. checkbox_value('checkbox3') .','.
        'checkbox4 = '. checkbox_value('checkbox4') .','.
        'checkbox5 = '. checkbox_value('checkbox5') .','. "LIMIT 1";