Trouble passing session variable into mysql query?

Question

Having a bit of trouble passing a session variable into a mysql query in php. Tried putting it into a variable and testing against that and no luck. Also tried various formatting already. At a loss. It's just a simple string passed in.

$result = mysqli_query($db_connection, "SELECT * FROM feedback  WHERE StudentID=" . $_SESSION['BCode'] . "");

session_start() at the top of your file?

wesside
– wesside

2013-12-12 17:42:58 +00:00
Commented Dec 12, 2013 at 17:42 — wesside
– wesside, Commented Dec 12, 2013 at 17:42
(probable) Combination of both answers below.

Funk Forty Niner
– Funk Forty Niner

2013-12-12 17:46:44 +00:00
Commented Dec 12, 2013 at 17:46 — Funk Forty Niner
– Funk Forty Niner, Commented Dec 12, 2013 at 17:46
en.wikipedia.org/wiki/SQL_injection

Dave
– Dave

2013-12-12 18:24:49 +00:00
Commented Dec 12, 2013 at 18:24 — Dave
– Dave, Commented Dec 12, 2013 at 18:24

MikeF · Accepted Answer · 2013-12-12 18:05:32Z

3

Are you setting:

session_start();

at the header of your .php file?

<?php 
session_start(); 

$studentid = $_SESSION['BCode'];

$result = mysqli_query($db_connection, "SELECT * FROM feedback WHERE StudentID='" . $studentid . "'"); 

$finalResult = array(); 

while ($row = $result->fetch_assoc())
{ 
    $finalResult[] = $row; 
} 

$jsonoutput = json_encode($finalResult); // Encodes the query result into JSON 
?>

edited Dec 12, 2013 at 18:05

answered Dec 12, 2013 at 17:43

MikeF

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Comments

Funk Forty Niner · Accepted Answer · 2013-12-12 17:47:54Z

1

Check for session_start(); and try

mysqli_query($db_connection, "SELECT * FROM feedback WHERE StudentID='" . $_SESSION['BCode'] . "'");

edited Dec 12, 2013 at 17:47

Funk Forty Niner

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answered Dec 12, 2013 at 17:46

zavg

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3 Comments

user3078580 Over a year ago

<?php session_start(); mysqli_query($db_connection, "SELECT * FROM feedback WHERE StudentID='" . $_SESSION['BCode'] . "'"); $finalResult = array(); while ($row = $result->fetch_assoc()){ $finalResult[] = $row; } $jsonoutput = json_encode($finalResult); // Encodes the query result into JSON ?>

user3078580 Over a year ago

Now it's jumping down to the while loop. Error

Funk Forty Niner Over a year ago

You really should post full code right away when it comes to issues like this. @user3078580 This is the classic "Can of Worms" scenario.

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