post values with ajax, php put it in database - php

Question

I want to accept user data from a form, use AJAX to post the values to a PHP script, which will insert them into the database.
JQuery:
$(".add").click(function(){
var order = {
Name: $name.val(),
Drink: $drink.val(),
}
$.ajax({
type: 'POST',
url: '/ajax/api/orders.php',
data: order,
success: function(newOrder){
$orders.append('<li>Name: ' + newOrder.Name +', Drink: '+ newOrder.Drink +'</li>')
}
})
});
Php:
//open connection to mysql db
$connection = mysqli_connect("localhost","root","xxxxxxxxx","test") or die("Error " . mysqli_error($connection));
//fetch table rows from mysql db
$sql = "select * from orders";
$result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));
$name = $_POST["Name"];
$drink = $_POST["Drink"];
$sql2 = "INSERT INTO orders (Id, Name, Drink) VALUES (NULL, '$name', '$drink')";
//create an array
$emparray = array();
while($row =mysqli_fetch_assoc($result))
{
$emparray[] = $row;
}
echo json_encode($emparray);
//close the db connection
mysqli_close($connection);
can somebody help me with it?

Answer 1

i think you should insert before you get all orders
$name = $_POST["Name"];
$drink = $_POST["Drink"];
$sql2 = "INSERT INTO orders (Id, Name, Drink) VALUES (NULL, '$name', '$drink')";
mysqli_query($connection, $sql) or
die("Error in Inserting" . mysqli_error($connection))
$id = mysqli_insert_id ($connection);
$sql = "select * from orders where Id = $id";
$result = mysqli_query($connection, $sql) or
die("Error in Selecting " . mysqli_error($connection));
//create an array
$emparray = array();
while($row =mysqli_fetch_assoc($result))
{
$emparray[] = $row;
}
echo json_encode($emparray);
The javascript will be right this:
$(".add").click(function(){
var order = {
Name: $name.val(),
Drink: $drink.val(),
}
$.ajax({
type: 'POST',
url: '/ajax/api/orders.php',
data: order,
success: function(newOrder) {
$.each(newOrder, function(order) {
//$order is a ul that have id = orders
$orders.append('<li>Name: ' + order.Name +', Drink: '+ order.Drink +'</li>');
});
}
});
});
and if you need anything ask :]
UPDATE
please update the success function you have to this
success: function(data) {
data.forEach(function(order) {
//$orders is a ul that have id = orders
$orders.append('<li>Name: ' + order.Name +', Drink: '+ order.Drink +'</li>');
});
}
UPDATE 2
success: function(data) {
var obj = JSON.parse(data);
var length = obj.length;
for (var i = 0; i < length; i++) {
$orders.append('<li>Name: ' + obj[i].Name +', Drink: '+ obj[i].Drink +'</li>');
}
}
UPDATE 3
$name = $_POST["Name"];
$drink = $_POST["Drink"];
$sql2 = "INSERT INTO orders (Id, Name, Drink) VALUES (NULL, '$name', '$drink')";
mysqli_query($connection, $sql2) or
die("Error in Inserting" . mysqli_error($connection))
$id = mysqli_insert_id($connection);
$sql = "select * from orders where Id = $id";
$result = mysqli_query($connection, $sql) or
die("Error in Selecting " . mysqli_error($connection));
//create an array
$emparray = array();
while($row =mysqli_fetch_assoc($result))
{
$emparray[] = $row;
}
echo json_encode($emparray);