CODEWITHSUNDEEP
javascriptphpmysqlsqlajax
Erik
Erik

Reputation: 2530

Trying to update a database with javascript/PHP doesn't work

I'm trying to update a database using javascript and PHP, this is my index.html code:

<!DOCTYPE html>
<html>
<head>
    <script type="text/javascript" src="https://code.jquery.com/jquery-1.11.3.min.js"></script>
</head>
<body>
    <input type="text" id="descriptioninput">
    <input type="number" id="budgetin">
    <input type="number" id="budgetout">
    <button type="button" onclick="addToDB()">Add to database</button>

    <script>
        function addToDB()
        {
            var descriptioninput = document.getElementById('descriptioninput').value;
            var budgetin = document.getElementById('budgetin').value;
            var budgetout = document.getElementById('budgetout').value;

            $.ajax ( {
                type: 'POST',
                url: 'addtodb.php',
                data:{descriptioninput:descriptioninput, budgetin:budgetin, budgetout:budgetout},
                success:function (data) {
                    // Completed successfully
                    alert('success!');
                }
            });
    </script>
</body>
</html>

Here's my addtodb.php code:

<?php

$host = "localhost";
$username = "root";
$password = "";
$dbname = "budgetdb";

$conn = new mysqli($host, $username, $password, $dbname);
if ($conn === TRUE)
{
$descriptioninput = $_GET['descriptioninput'];
$budgetin = $_GET['budgetin'];
$budgetout = $_GET['budgetout'];

$query = "INSERT INTO budget (description, budgetin, budgetout) VALUES ('$descriptioninput', '$budgetin', '$budgetout')";

$conn->query($query);

$conn->close();
}

?>

But it appears as if my PHP script doesn't run. No changes appear in my database. I've tried to do warning() in the PHP file and alert it it using.done(function(text)), but nothing is displayed.

Upvotes: 2

Views: 148

Answers (4)

Mihai Matei
Mihai Matei

Reputation: 24276

This is happening because you are doing the ajax request using POST method in js but you are trying to get the variables using the GET method in PHP. Switch it to GET and it will work.

Be aware of SQL Injection. You can prevent it either by using prepared statements or escaping the string as:

$descriptioninput = $conn->real_escape_string($_GET['descriptioninput']);

Also, the first if condition is not valid. You just need to do it like if ($conn) instead of if ($conn === TRUE)

Upvotes: 2

Abhishek Sharma
Abhishek Sharma

Reputation: 6661

Change ajax type to GET 

$.ajax ( {
type: 'GET',
url: 'addtodb.php',
data:{descriptioninput:descriptioninput, budgetin:budgetin, budgetout:budgetout},
success:function (data) {
// Completed successfully
alert('success!');
}
});

Upvotes: 1

PC3TJ
PC3TJ

Reputation: 852

I could be wrong but i believe description may be a reserved keyword in mySQL. try encapsing it 

INSERT INTO budget (`description`, `budgetin`, `budgetout`) VALUES ('$descriptioninput', '$budgetin', '$budgetout')

Upvotes: 1

PC3TJ
PC3TJ

Reputation: 852

its a little messy with your mix between ajax and JS. Try using this getHTTP function for regular JS. 

function httpGet(theUrl){
	//FETCH Data From Server
	xmlhttp=new XMLHttpRequest();
	xmlhttp.open("GET", theUrl , false );
	xmlhttp.send();   
	return xmlhttp.responseText;
}

then just build your url as +addtodb.php?param1="+param1value+"&param2="+param2value

Upvotes: 0

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