PHP: Retrieving value of checkboxes when name doesn't have array

Question

A form I don't have any control over is POSTing data to my PHP script. The form contains checkboxes along these lines:

<input type="checkbox" value="val1" name="option"/>
<input type="checkbox" value="val2" name="option"/>

If I were to write the code for the form, I'd write name="option[]" instead of name="option". But this is not a change I can do. Now, if both checkboxes are checked, $_POST["option"] returns just one of the values. How can I, in PHP retrieve all the values selected?

Answer 1

You can read the raw post data. For example:

<fieldset>
    <legend>Data</legend>
    <?php
    $data = file_get_contents("php://input");
    echo $data."<br />";
    ?>
</fieldset>

<fieldset>
    <legend>Form</legend>
    <form method="post" action="formtest.php">
        <input type="checkbox" value="val1" name="option"/><br />
        <input type="checkbox" value="val2" name="option"/><br />
        <input type="submit" />
    </form>
</fieldset>

Check both boxes and the output will be:

option=val1&option=val2

Here's a live demo. All you have to do then is to parse the string yourself, into a suitable format. Here's an example of a function that does something like that:

function parse($data)
{
    $pairs = explode("&", $data);
    $keys = array();
    foreach ($pairs as $pair) {
        list($k,$v) = explode("=", $pair);
        if (array_key_exists($k, $keys)) {
            $keys[$k]++;
        } else {
            $keys[$k] = 1;
        }
    }

    $output = array();
    foreach ($pairs as $pair) {
        list($k,$v) = explode("=", $pair);
        if ($keys[$k] > 1) {
            if (!array_key_exists($k, $output)) {
                $output[$k] = array($v);
            } else {
                $output[$k][] = $v;
            }
        } else {
            $output[$k] = $v;
        }
    }

    return $output;
}

$data = "foo=bar&option=val1&option=val2";

print_r(parse($data));

Outputs:

Array
(
    [foo] => bar
    [option] => Array
        (
            [0] => val1
            [1] => val2
        )

)

There might be a few cases where this function doesn't work as expected though, so be careful.

Answer 2

Thaks for the great explanation :) You people are sooo good..