PHP output buffering variable

Question

Is it possible to use a POST variable on the buffered script? The code is the following

<?php
header('Content-type: application/json');

$someVar= $_POST["var"];

ob_start();
require "someScript.php?var=" . $someVar; // instead of passing through GET, is there a way to use the $someVar directly?
$output = ob_get_clean();
echo $output;
?>

Note: I've already tried to access $someVar directly from someScript but without success. For that is why I'm asking. Thanks

EDIT: Possible someScript.php

<?php
    header('Content-type: application/json');
    require_once("BD.php");
    require_once("json/json_header.php");
    require_once("tools.php");
    require_once('class.phpmailer.php');

    $BD = new BDInfo(); 


    function encodeIDs($id1,$id2){

        // doesn't matter
    }

    $response = array("errorCode" => 0, "errorDesc" => "No Error");


    if(isset($someVar)){
        try{
            $link = mysqli_connect($BD->BDServer, $BD->BDUsername, $BD->BDPassword);

            if( !$link )
                throw new Exception( "..." );

            if( !mysqli_select_db($link, $BD->BDDatabase) )
                throw new Exception( "..." );

            $SQL="SELECT (...) us.VAR='" . $someVar . "';";

            $RS = mysqli_query($link,$SQL);
            if($RS && $row = mysqli_fetch_array($RS)){
                // process query result
            }
            else
            {
                $response["errorCode"]=4;    
                $response["errorDesc"]="Error descr";   
            }
            mysqli_close($link);

        }catch (Exception $e) {
            $response["errorCode"]=1;
            $response["errorDesc"]="Database Error: " . $e->getMessage();
        }
    }else{
        $response["errorCode"]=2;
        $response["errorDesc"]="Invalid Parameters";
    }

    echo json_encode($response);
?>

I get Invalid Parameters, showing that isset($var) failed

possible duplicate of PHP - include() file not working when variables are put in url? — DaveRandom
– DaveRandom, Commented Mar 14, 2012 at 11:26
@sandeep, what i'm trying to do is to use a variable on an imported script. I thought I made it clear — jose
– jose, Commented Mar 14, 2012 at 11:36
I think @Yotaware guessed best what you are trying to do, so give that answer a try... — Martin Hennings
– Martin Hennings, Commented Mar 14, 2012 at 11:52

diolemo · Accepted Answer · 2012-03-14 11:25:14Z

3

The variable $someVar is already available to you in someScript.php. You should not use a query string to do transfer it.

If you want to avoid modifying someScript.php for some reason then you can do this:

$_GET['var'] = $someVar;

Or just this:

$_GET['var'] = $_POST['var'];

answered Mar 14, 2012 at 11:25

diolemo

2,6712 gold badges21 silver badges28 bronze badges

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

2 Comments

jose Over a year ago

I can't access $someVar on someScript.php. I've tried just now.

diolemo Over a year ago

It should work. Try to setup a much simpler example and see if it works then. See here for an example.

Jon · Accepted Answer · 2012-03-14 11:25:13Z

1

When you include a script with include or require, that script has access to all variables in scope at the inclusion site. In your example, you can access $someVar directly from someScript.php (you could also directly access $_POST or anything else).

answered Mar 14, 2012 at 11:25

Jon

439k85 gold badges756 silver badges820 bronze badges

1 Comment

m1crdy Over a year ago

But not with that output buffer he uses.

Martin Hennings · Accepted Answer · 2012-03-14 11:25:58Z

1

Yes, you can.

Since require() works similar to include(), the included file inherits the variable scope.

From the manual:

"When a file is included, the code it contains inherits the variable scope of the 
line on which the include occurs. Any variables available at that line in the 
calling file will be available within the called file, from that point forward. 
However, all functions and classes defined in the included file have the global 
scope."

Imagine an include (or require) as that the included code replaces the line include ....

answered Mar 14, 2012 at 11:25

Martin Hennings

17k9 gold badges51 silver badges70 bronze badges

Collectives™ on Stack Overflow

PHP output buffering variable

3 Answers 3

2 Comments

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related