PHP Array Returns Wrong Value

Some comments about your code:

• You don't need to break after return.
• break does not work for if in PHP.
• You don't need to do else if you're using return already.

I made some changes to your code to make this more visible and it reduces the cyclomatic complexity as well. Probably this helps that you find your error easier:

function unnamed(/* ... unknown parameters ... */)
{
    /* ... some code ... */
    if ($num==0)
    {
        return "Do not appear to be registered. Please check your email input again.";
    }

    // there is an entry. check for consistency
    $procrastinateDB = mysql_result($result,0,'procrastinate');
    if ($procrastinate != $procrastinateDB)
    {
        return "Answer to your procrastination question is incorrect. Please try again!";
    }

    $username = mysql_result($result,0,'usrname');
    $passwd = mysql_result($result,0,'passwd');
    return array($username, $passwd, $num);
}

Your function returns a string or an array btw. If you don't check the return type and assume it's an array, you will trigger substring access (see it documented on the string manual page):

$string = 'ABC';
echo $string[0]; // A
echo $string[2]; // C

It's often better to have a function return one type, not multiple. However this depends on your design and coding style, so I can only make suggestions, like return an object that contains a status message, a status code (success/failure) and - if available - the array of data you want to return. However that depends what you want / need. You can as well check the return type with is_array as well.

an array is still returned with $usersname = 'D', $passwd = 'o', and $num = ' '.

When $num is 0, you are returning the string:

Do not appear to be registered. Please check your email input again.

In PHP, you can access strings like arrays. So, my guess is you didn't check to see if you were returned a string or an array, and did something like this:

list($username, $passwd, $num) = yourFunction();

Since strings can be accesed like arrays, $username would be D, $password would be o, and $num would be a space, as those are the 1st three characters.

I suggest using is_array before accessing the value to make sure you know what you're getting.

$ret = yourFunction();
if(is_array($ret)){
   list($username, $passwd, $num) = $ret;
}
else{
   // Something else
}