PHP Query subtraction not working

Question

Hello I am trying to subtract a users credit's after a transaction but something keeps going wrong when updating.while testing accepted_bidder = 15 and credit row in customer is 100. for some reason when I update it is -15 rather than 75 anyone know why this may be?

output: Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, array given in C:\xampp\htdocs\payment2.php on line 20

Notice: Undefined variable: credit in C:\xampp\htdocs\payment2.php on line 26

<?php
session_start();
require 'config.php';


$id = $_SESSION['login_user'];
$jobid    = $_POST['job_id'];
$poster_id    = $_POST['poster_id'];
$accepted_bidder    = $_POST['accepted_bidder'];
$accepted_bid    = (int) $_POST['accepted_bid'];
$poster_id = $_POST['poster_id'];

$query = "SELECT credit FROM `customer` WHERE email_adress = '$id'";

$success = $conn->query($query);


$result = mysqli_fetch_array($success);

while($row = mysqli_fetch_array($result)):
$credit = (int)$row['credit'];
endwhile;

//var_dump($result);
//var_dump($accepted_bid); 
$updated_credit = $credit - $accepted_bid;
//echo $updated_credit;


$query2   = "UPDATE job SET start_escrow = '1' WHERE job_id = '$jobid'";
$success2 = $conn->query($query2);

$query3   = " UPDATE customer SET credit = '$updated_credit' WHERE email_adress = '$id'";
$success3 = $conn->query($query3);



if (!$success) {
    die("Couldn't enter data: ".$conn->error);

}

echo "Thank You For Contacting Us <br>";
 //header("location: myjobs.php");



$conn->close();




?>

Note: The object-oriented interface to mysqli is significantly less verbose, making code easier to read and audit, and is not easily confused with the obsolete mysql_query interface. Before you get too invested in the procedural style it’s worth switching over. Example: $db = new mysqli(…) and $db->prepare("…") The procedural interface is an artifact from the PHP 4 era when mysqli API was introduced and should not be used in new code. — tadman
– tadman, Commented Apr 19, 2018 at 17:06
WARNING: When using mysqli you should be using parameterized queries and bind_param to add user data to your query. DO NOT use string interpolation or concatenation to accomplish this because you have created a severe SQL injection bug. NEVER put $_POST, $_GET or any user data directly into a query, it can be very harmful if someone seeks to exploit your mistake. — tadman
– tadman, Commented Apr 19, 2018 at 17:06
Note: Try and get out of the habit of declaring SQL statements in throw-away variables that are used only once. It's a lot easier to follow code where the query is supplied directly to the function, and there's no longer a chance of messing up and sending in $sql3 instead of the visually similar $sql8. — tadman
– tadman, Commented Apr 19, 2018 at 17:06
This code also has race conditions. When adjusting columns try and do the operation atomically, like UPDATE customer SET credit=credit-?, instead of as two queries which can run out of sync. — tadman
– tadman, Commented Apr 19, 2018 at 17:07

Barmar · Accepted Answer · 2018-04-19 17:08:52Z

3

$result is not the result of the query, it's the first row you read using:

$result = mysqli_fetch_array($success);

So when you then do:

while ($row = mysqli_fetch_array($result))

you're trying to use that row as the query result.

Replace these two lines:

$success = $conn->query($query);
$result = mysqli_fetch_array($success);

with

$result = $conn->query($query);

answered Apr 19, 2018 at 17:08

Barmar

789k57 gold badges554 silver badges669 bronze badges

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

Collectives™ on Stack Overflow

PHP Query subtraction not working

1 Answer 1

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related