php variable out of scope when using $_POST

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>

What do you expect this snippet to do, and what does it actually do? This is not entirely clear to me. — Roel Harbers
– Roel Harbers, Commented Oct 9, 2014 at 15:53
@stevo You're going to have to show us more code then. This isn't enough to go on. It's anybody's guess. Plus, make sure your submit button's named and all elements are too. — Funk Forty Niner
– Funk Forty Niner, Commented Oct 9, 2014 at 15:55
a condition (if/else) structure doesnt create a new scope) it's not possible that the first $a and the 2nd $a have different values with above code unless $a is callable and even then. — DarkMukke
– DarkMukke, Commented Oct 9, 2014 at 15:56
This is just a snippet of my code. I omitted additional code to simply the code excerpt. I bet that omitted code holds the key to this answer if, according to your update, the conditional statement is true :) — Hanky Panky
– Hanky Panky, Commented Oct 9, 2014 at 16:43

Hanky Panky · Accepted Answer · 2014-10-09 16:02:21Z

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.

edited Oct 9, 2014 at 16:02

answered Oct 9, 2014 at 15:55

Hanky Panky

46.9k9 gold badges76 silver badges95 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

DarkMukke Over a year ago

Well not really, the only way possible if there is no post with submit

Hanky Panky Over a year ago

That's the same statement in different words.

ʰᵈˑ Over a year ago

@DarkMukke Yes, if isset($_POST['submit']) doesn't return TRUE, then $a will never try and be echo'd.

ʰᵈˑ Over a year ago

@DarkMukke You're just regurgitating the answer but in different words... And I don't think you even realise you are.

Hanky Panky Over a year ago

@DarkMukke That's not the question, Question is about OP's confusion about scope and mistakenly thinking the variable is being eaten alive by the scope. Im not going to teach then which variable to check the form for.

|

Funk Forty Niner · Accepted Answer · 2014-10-09 16:52:47Z

1

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.

edited Oct 9, 2014 at 16:52

answered Oct 9, 2014 at 16:10

Funk Forty Niner

74.2k14 gold badges71 silver badges146 bronze badges

6 Comments

stevo Over a year ago

I already had the following: <form id="form" name="form" method="post" action="index.php">

Funk Forty Niner Over a year ago

@stevo My answer works. If it's still not working for you, then you have to show us full code. There's nothing more that Hanky or I can do without knowing what's in [...] or elsewhere.

Funk Forty Niner Over a year ago

@stevo Add error reporting to the top of your file(s) right after your opening <?php tag error_reporting(E_ALL); ini_set('display_errors', 1); see if it yields anything, and use var_dump(); on your variables. If you have any JS happening, let us know. You really need to post full code right up front, because it's guess work right off the bat and I don't deal with guesses, I deal with actuals. Had I known beforehand, I would not have submitted my answer.

Funk Forty Niner Over a year ago

@stevo From what I gather in your other question stackoverflow.com/q/26278179 - you most likely do have JS happening <body onload="document.FormName.reset();"> am I right on this one? Relevant?

Funk Forty Niner Over a year ago

@stevo Thanks, but there's no way I can test this. See if someone else picks up on the question.

|

stevo · Accepted Answer · 2014-10-09 18:58:40Z

0

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'];

answered Oct 9, 2014 at 18:58

stevo

1816 silver badges17 bronze badges

Collectives™ on Stack Overflow

php variable out of scope when using $_POST

3 Answers 3

6 Comments

6 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

6 Comments

6 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related