Json data not saved to DB in PHP

Question

I am trying to make a functionality where I am sending Name and Age using ajax. The same is gathered by PHP and then stored to DB. I think I am doing it correctly but data not saved in database. Can someone help me out?

HTML

<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
<script>
$(document).ready(function(){
    $('button').click(function(){
        $.ajax({
            url: 'addData.php',
            type: 'POST',
            dataType: "json",
            data: {
                name: $('#name').val(),
                age: $('#age').val(),
            }
        })
    });
});
</script>
</head>
<body> 
<input id="name" type="text" />
<input id="age" type="text" />
<button>Click</button>
</body> 
</html>

PHP

<?php

//Parameters for connection
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test1";

//Reading json
$data_back = json_decode(file_get_contents('php://input'));

// set json string to php variables
$name = $data_back->{"name"};
$age = $data_back->{"age"};

// Create Connection
$conn = new mysqli($servername, $username, $password, $dbname);

//Creating Query
$sql = "INSERT INTO table1 (name, age) VALUES ($name, $age)";

//Running Query
$conn->query($sql);

//Closing Connection
$conn->close();

?>

ERROR

You'll want to parameterize your query, just inserting $name and $age into the query produces invalid SQL (and a good chance for someone to insert their own potentially malicious SQL) — Joachim Isaksson
– Joachim Isaksson, Commented Mar 5, 2016 at 20:56
you need to put quotes around strings in INSERT: ...('$name', '$age')... — Milan
– Milan, Commented Mar 5, 2016 at 21:04
and you get no error in console now? did you output you values if they are posted from ajax request.? — K. Uzair
– K. Uzair, Commented Mar 5, 2016 at 21:07

jaggedsoft · Accepted Answer · 2016-03-05 21:12:13Z

1

Even though you're sending data with jquery as json, php is still recieving it as a $_POST object. So, this should work:

$name = $_POST['name'];
$age = $_POST['age'];

answered Mar 5, 2016 at 21:12

jaggedsoft

4,0582 gold badges35 silver badges42 bronze badges

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

1 Comment

Deadpool Over a year ago

This did it for me ... ;)

Maxim Palenov · Accepted Answer · 2016-03-05 21:13:14Z

1

I think your object $data_back is empty because of errors in parsing of data by function json_decode. You should try to use var_dump($data_back); exit; after json_decode or more advanced methods such as debugging.

answered Mar 5, 2016 at 21:13

community wiki

Maxim Palenov

1 Comment

Deadpool Over a year ago

Yes, I was getting blank rows inserted +1 for this analysis.

David J Eddy · Accepted Answer · 2016-03-05 21:10:23Z

0

I believe this is the proper way to access data post json_decode

$name = $data_back->name;

I also recommend looking into prepared statements for the database query execution...Ok, I HIGHLY recommend you look into it.

Edit: Maybe try this as well: Replace file_get_contents() with $_POST;

edited Mar 5, 2016 at 21:10

answered Mar 5, 2016 at 21:01

David J Eddy

2,0471 gold badge20 silver badges37 bronze badges

1 Comment

jaggedsoft Over a year ago

You can also do json_decode($data,true) to return an associative array like this: $name = $json['name'];

Collectives™ on Stack Overflow

Json data not saved to DB in PHP

3 Answers 3

1 Comment

1 Comment

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

1 Comment

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related