using variable on pageload and modify if changed - php

Question

i got this variable:
$machine=isset($_POST['machine'])?$_POST['machine']:'';
for my selectbox:
<select id="machine" name="machine">
<option value=""></option>
<option value="1">A</option>
<option value="2">B</option>
<option value="3">C</option>
<option value="4">D</option>
</select>
now i want to made some modifications. i have saved in my settings $favmachine=4.
for example i load the page with
if($favmachine!=0)
{
$machine=$favmachine;
}
my form looks like this (exactly what i want):
<select id="machine" name="machine">
<option value=""></option>
<option value="1">A</option>
<option value="2">B</option>
<option value="3">C</option>
<option value="4" selected="selected">D</option>
</select>
the problem now is when i want to change for example from D to A it goes to D again.. right exactly what i wrote... but i dont know how to handle this problem
i want my $favmachine selected on pageload (works) BUT want to be able to change it to another machine and set this to $machine by $_POST
hope you understand my problem :D

Answer 1

I think you want to overwrite $machine with $favmachine only if the check for a POST value resulted in “nothing” before:
$machine=isset($_POST['machine'])?$_POST['machine']:'';
if($machine==='' && $favmachine!=0)
{
$machine=$favmachine;
}
or in shorter form, using the ? operator once more:
$machine = isset($_POST['machine']) ?
$_POST['machine']
:
( $favmachine != 0 ? $favmachine : '');
(Multiple lines and indentation for readability only.)