php variable out of scope when using $_POST - php

Question

I have variables $a and $location. However when I press the 'Submit' button the variable has gone out of scope. I can pass in in the variable from input box T2 but I can't manage to manage to do it to 'a' which is a regular variable. I added echo $b to debug it and I do get that output which verifies the conditional statement is true.
<?php
[...]
$a = '5';
$location = 'home';
if(isset($_POST['submit'])) {
$location=$_POST['T2'];
echo $location;
echo $a;
$b = '6';
echo $b;
[...]
}
?>
<input type="text" name="T2" value="<?php echo $location; ?>">
<button type="submit" name="submit" value="create">Submit</button>

Answer 1

$a="Testing";
echo $a;
if(isset($_POST['submit'])) { // you're not getting here
$location=$_POST['T2'];
echo $a;
echo "Testing one two three"; // even this won't show :)
}
There is only one plausible reason for that second echo not to work, i.e. Your if condition doesnt evaluate to true. There is no other reason that can cause that. Now check your field or button named submit on the form :)
A simple print_r($_POST); will tell you all that was posted, you can check there if submit was also posted.

Answer 2

Far as I can tell, you're missing the form tags along with the method which should be post, least from what you posted for code.
Sidenote: If what you posted isn't full code, do. The following works which printed home56 on submit.
I do believe that's what the expected result should be.
<?php
$a = '5';
$location = 'home';
if(isset($_POST['submit'])) {
$location=$_POST['T2'];
echo $location;
echo $a;
$b = '6';
echo $b;
}
?>
<form action="" method = "post">
<input type="text" name="T2" value="<?php echo $location; ?>">
<input type="submit" name="submit" value="create">
</form>
Sidenote: You can keep your present button if you wish instead of the input I tested with:
<button type="submit" name="submit" value="create">Submit</button>
Footnotes:
Both [...] - any relevance? I'll bet there is.

Answer 3

Got it to work by echoing it in a hidden input
<input type="hidden" name="myvar" value="<?php echo $a; ?>">
and then using $_POST to extract it
$a=$_POST['myvar'];