AJAX Sends data to PHP file, but PHP can't get data from $_POST

Question

I am making a page on a website in PHP where a user fills out 3 fields and hits submit. The submit button should call my AJAX function to send the data to a database connection PHP file. I can confirm the data is sent from AJAX (via an alert) and the function returns a Success. This must mean my database query file is not interpreting the data correctly. Please help me understand where I went wrong.

Code from php page where the form is:

<script type="text/javascript">
function storeInvoice() {
    //var c_name = document.getElementById('c_name');
    //var c_license = document.getElementById('c_license');                         
    //var c_licenseemail = document.getElementById('c_licenseemail');
    var data=$('#myForm').serialize();
        $.ajax({
        url: "/paydb.php",
        type: "POST",
        data: data,
        async:false,
        dataType:'html',
        success: function (value) {
            alert("Sent: "+data);
        }
    });
}
</script>

Relevant Code from Database php file:

mysqli_select_db($conn, "main_db" );

$c_license = $_POST['c_license'];
$c_name = $_POST['c_name'];
$c_licenseemail = $_POST['c_licenseemail'];

//Another method was attempted below.
//$data=$_POST['serialize'];
//$c_licenseemail = $data['c_licenseemail'];
//$c_license = $data['c_license'];
//$c_name = $data['c_name'];

$query = "INSERT INTO `invoices`(`company`, `licensenum`, `licenseemail`) VALUES ('$c_name','$c_license','$c_licenseemail');";
mysqli_query($conn, $query);

The data is sent as:

c_name=testname&c_license=3&c_licenseemail=testemail%40email.com

Any help is much appreciated!

Have you watched the AJAX request / response in the browser's developer tools? Have you included the jQuery library in the project? Are there any errors reported? Are you running this on a web-server? — Jay Blanchard
– Jay Blanchard, Commented Apr 19, 2017 at 17:58
Little Bobby says your script is at risk for SQL Injection Attacks. Learn about prepared statements for MySQLi. Even escaping the string is not safe! — Jay Blanchard
– Jay Blanchard, Commented Apr 19, 2017 at 17:58
@dan08 serialize makes the data sent in this way, he can still retrieve the data with $_POST vars — Tushar Gupta
– Tushar Gupta, Commented Apr 19, 2017 at 18:00
Please, quit using alert() for troubleshooting., use console.log() instead. — Jay Blanchard
– Jay Blanchard, Commented Apr 19, 2017 at 18:06

Agam Banga · Accepted Answer · 2017-04-19 18:44:45Z

4

Please use the

mysqli_query($conn, $query) or die(mysqli_error($conn));

For duplicate key, you need to make the primary key to auto increment in your database.

answered Apr 19, 2017 at 18:44

Agam Banga

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2 Comments

Nathan Brown Over a year ago

Very helpful, thank you. Seems to be working as intended now.

Agam Banga Over a year ago

Glad, i could help.

dokgu · Accepted Answer · 2017-04-19 18:04:52Z

1

In your success callback function replace alert(data) with alert(value) and in your database.php file echo any of the post variables to just check whether the values are correctly sent to database.php via ajax post.

edited Apr 19, 2017 at 18:04

dokgu

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answered Apr 19, 2017 at 18:03

Chaitanya Ghule

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3 Comments

Jay Blanchard Over a year ago

Please, quit using alert() for troubleshooting., use console.log() instead.

Nathan Brown Over a year ago

The echos have the correct value. For example echo "c_license" sends back 3 if 3 was typed into that field.

Nathan Brown Over a year ago

It's wierd, it sent the query to the DB after making these small changes only. But it only worked twice and now no longer works. I'm confused..

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