PHP form checkbox and undefined index - php

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 individual posted value?
I have looked at this Question and am having trouble believing that this is the only solution.
Below is some example code:
EDIT: please not that these are not the actual names of the tables columns; they are named uniquely (like "postAddress, displayPhone, student, etc.)

Answer 1

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";

Answer 2

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 3

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.";
?>