dropdown return value in html and php - php

Question

I have dropdown list (menu) and a button created with this code:
<form name="period" action="all.php" method="POST">
<div align="center">
<select name=period_dropdown>
<option value="nill"></option>
<option value="48">48</option>
<option value="72">72</option>
<option value="96">96</option>
<option value="120">120</option>
</select>
<input type="submit" value= "OK" >
</div></form>
When option of dropdown is selected, must be on button instead of OK and when button is pressed to be assigned to variable $period1. So, when I select 72, button must have 72 instead of OK and when I click button, variable $period1 to get value 72 (integer). Please, no javascript. Just html and php.
Thanks

Answer 1

You have to use javascript for what you are trying to do. If you want the button to change on the fly without submitting the form, you have to do client-side scripting (javascript).
Either:
Change the value of button after the dropdown is selected and form is submitted
Use two lines of javascript to change the value of the button when the select box is changed

Answer 2

In all.php:
if (isset($_POST['period_dropdown'])) {
$period1 = $_POST['period_dropdown'];
//do something with $period1
}
?>

Answer 3

<form name="period" action="all.php" method="POST">
<div align="center">
<select id="period_dropdown" name="period_dropdown" onchange="updateButton()">
<option value="nill"></option>
<option value="48">48</option>
<option value="72">72</option>
<option value="96">96</option>
<option value="120">120</option>
</select>
<input id="period_button" type="submit" value= "OK" >
</div></form>
<script type="text/javascript">
function updateButton() {
document.getElementById("period_button").value = document.getElementById("period_dropdown").value;
}
</script>