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
3 answers
solution1
1 ACCPTED 2016-03-05 21:12:13
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'];
solution2
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.
solution3
0 2016-03-05 21:01:04
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 ;
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