使用ajax发布值，php将其放入数据库

我想从表单中接受用户数据，使用AJAX将值发布到PHP脚本中，PHP脚本将把它们插入数据库

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>')
    }
    })
});

---

有人能帮我吗？

我想你应该在收到所有订单之前插入

$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);

javascript在以下方面是正确的：

$(".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>');
            });
        }
    });
});

---

您是否收到任何错误？除了其他问题外，您的代码对SQL注入非常开放。您没有在此处查询/连接到db$sql2=INSERT INTO orders Id、名称、饮料值NULL、$Name、$DREAND；所以你的插入永远不会发生。php将其放入我的数据库中不会的@KylianWesternow谁会在这件事上指名道姓？^谁知道你的HTML是否正确。你的代码有什么问题？我在控制台newOrder中得到一个错误。每个都不是一个错误function@KylianWester请让我看看console.lognewOrder的结果；我得到的新订单没有定义，我正在学习所有这一切，所以我认为这是一件愚蠢的事情xUncaught类型错误：不能使用'in'运算符在[{Id:1，Name:kylian，Drink:Coffee}，{Id:2，Name:Chris，Drink:Tea}，{Id:3，Name:Jelle，Drink:Coffee}，{Id:4，Name:rud，Drink:IceTea}]中搜索'length'我得到这个控制台错误捕获类型错误：data.forEach不是一个函数

$(".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>');
            });
        }
    });
});

        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>');
            });
        }

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>');
    }
}

$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);