The save button is not functioning properly

Currently, I am in the process of creating an order form. One of the key functionalities I have been working on is the "save changes function". This feature allows users to make changes to the table and then update the database by clicking a specific button.

In addition, the "Bestelling toevoegen" section serves as the form for adding new entries. When entering the required information and selecting "Toevoegen", a new record will be displayed in the table above.

Despite these developments, I have encountered an issue with saving changes to the "Status" field.

Below is the comprehensive code snippet:


$dbname = "localhost";
$dblogin = "root";
$dbpass = "";
$dbtable = "bestelformulier";

$con=mysqli_connect("$dbname","$dblogin","$dbpass","$dbtable");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM overzicht");

echo "<form name='wijzigen' method='post'>";
echo "<table align='center' width='700px' border='2'>
<tr>
<th>Ordernr</th>
...
phpmysql

Answer №1

It seems like the issue you are facing is due to outputting multiple select elements with the same name, status. To resolve this problem, consider updating the following line of code:

echo "<td><select name='status'>

to:

echo "<td><select name='status[$ordernr]'>

This adjustment will create an array of status elements with order numbers as keys. Next, modify the code snippet below:

$status = $_POST['status'];

if(isset($_POST['wijzigen'])) {
$query = mysqli_query("UPDATE overzicht SET status='$status' WHERE ordernr='$ordernr'");
}

to:

$statuses = $_POST['status'];

if(isset($_POST['wijzigen'])) {
    foreach($statuses as $ordernr => $status)
    {
        if($status != "")
            $query = mysqli_query($con, "UPDATE overzicht SET status='$status' WHERE ordernr='$ordernr'");
    }
}

Answer №2

Here is some advice for improving your code:

$query = mysqli_query("UPDATE overzicht SET status='$status' WHERE ordernr='$ordernr'");

Make sure that your $ordernr is not empty to successfully update existing records. Pass the ordernr via GET or POST in the form.

If you want to update multiple order statuses simultaneously, consider passing all ordernrs and their statuses as an array or use AJAX for more efficient processing.

Lastly, ensure you protect against SQL injection attacks. If this code goes live, it will be vulnerable to hacking attempts on your database.

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