MyFM | oriented social agregator platform written in Ruby | Application Framework library

Several problems needed to be fixed:

1. You're using settype() incorrectly. It changes the variable in place, it doesn't return the new value. You can use intval() for that. And you shouldn't convert screen1 to an integer; the first argument to base_convert must be a string (if you convert it to an integer, it will be parsed as decimal, but the whole point of this is to parse it in a specific base).
2. You should use if, not do-while, to test if the form was submitted.
3. You can't use the PHP variable in JavaScript the way you are. JavaScript runs on the client after the PHP script has finish. So your onclick will show the value from the previous form submission.
4. Instead, echo the result variable into the value of the output field. This has to be after the code that does the base conversion.
5. The button needs to be a submit button so it submits the form. If you don't want to submit the form and reload the page, you'll need to learn to use AJAX, which I'm not going to show how to do in this answer.
6. You misspelled $screen3 as $creen3 (missing s) in your call to base_convert().
7. You need to echo $_SERVER['PHP_SELF'] in the action attribute.

The answer is just add display:none, make change of the value to 0 and checked. EXAMPLE: echo "";

Your PHP should only get the difference in time. That means that PHP will always output a format of HH:mm:ss, and no other text or values, which you then get in JavaScript. Ensuring that the output is always the same, unless you're using encoded arrays, means that you can design your code to expect the values you always send.

@arkascha and @Mehdi thanks for help. my problem solved. there is confusion in two word in query $tid and $testid I replaced both with $testid and my code working properly