How to confirm the action from other php file?

sunnyside

I have two php files: 1.php & 2.php. Inside of 2.php I have a form and there is an action button "check out". I want to perform this function, if "check out" button in 2.php was submited, some data in my 1.php will change color ...

If these happened in one file, I just need to use

if ($_POST['CheckOut'])

but right now it associates two files. Would you tell me how I can do it?

Thank you!

devinemke

<form action="" method="POST">
choose color:<br>
<input type="submit" name="submit" value="black">
<input type="submit" name="submit" value="red">
<input type="submit" name="submit" value="blue">
<input type="submit" name="submit" value="green">
<input type="submit" name="submit" value="yellow">
</form>

<?
if (isset($_POST['submit'])) {$color = $_POST['submit'];}
else {$color = 'black';}

echo '<br><font color="' . $color . '"><b>this text ' . $color . '</b></font>';
?>

sunnyside

Hi, I understand the code you wrote. But seems this is not what I want. what you did is only happen in one file which has some data and a form contains submit button.

My case is: I have two files, one is for display information. Another one is for edit information. Once the "check out" button in edit.php has clicked, part data in my first php file will be changed.

devinemke

ok, so, put the top part of my code in one file and the bottom part in the other and change the action tag in the form to point to the second file. the end result will be the same.

sunnyside

Originally posted by devinemke
ok, so, put the top part of my code in one file and the bottom part in the other and change the action tag in the form to point to the second file. the end result will be the same.

Well Thanks anyway. My code is much more complex than what I said. The "color" thing is just an example.