How to use if conditions with php to evaluate false or true conditions of a submitted button?

Question

I'm using a standard form in combination with two if constructs and I am not sure if the syntax are ok. Is there a better way to refactor the following code?

if(!isset($_POST['submit']))
   {
 
   }

if(isset($_POST['submit']))
   {
 
   }


<form class="mt20" enctype="multipart/form-data" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" >

    <input type="submit" name="submit" value="speichern">

</form>

Pressing the submit button the site should be reloaded and the code within the if(!isset( ... bracket should not be executed. For hundreds of users the code is not executed. But there is one user where the code is executed.

score 0 · Answer 1 · edited May 23 '17 at 11:51

0

This should help you explain

Also if possible just do a if/else conditional instead of doing two checks which is redundant and confusing. I think that one user might be artificially manipulating the data and passing in NULL thus triggering the !isset

edited May 23 '17 at 11:51

Community

1
1

answered Feb 26 '14 at 22:13

mrpanda

104
5

score -2 · Answer 2 · edited Nov 23 '15 at 21:11

-2

You can do as:

if (!isset($_POST['submit'])) {

} else {
    if (isset($_POST['submit'])) {

    }   
}

edited Nov 23 '15 at 21:11

Tunaki

132,869
46
340
423

answered Nov 23 '15 at 21:08

Hamza

37
8

the code inside the if (!isset does not executes upon submit – Hamza Nov 23 '15 at 21:13
Use elseif{} instead – RAZERZ Nov 25 '17 at 10:54

How to use if conditions with php to evaluate false or true conditions of a submitted button?

2 Answers2