javascript - 如何使用 Ajax 将对象数组发送到 PHP 并更新 MySQL 服务器

Question

我有这个对象数组:

var desks = [];

function Desk(id, name, x, y) {
    var desk = {};
    desk.id = id;
    desk.name = name;
    desk.x = x;
    desk.y = y;
    desks.push(desk);
}

我有一个包含 desk_id、desk_name、desk_x 和 desk_y 列的 MySQL 表。 当我单击一个按钮时，我想用 Desk 对象更新表中的值。

这基本上是我正在努力实现的目标:

JS

$("#btn").click(function() {
    $.ajax({
        method: "POST",
        url: "phpfile.php",
        data: desks
    })
});

PHP

<?php
if ($_SERVER['REQUEST_METHOD'] === 'POST') {
    require "../db.php";

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

    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 

    $desks = $_POST("desks");
    for($i = 0; $i < sizeof($desks); $i++) {
        $sql = "UPDATE desks SET desk_x=$desks[i]->x WHERE desk_id=$desks[i]->id";
        $conn->query($sql);
        $sql = "UPDATE desks SET desk_y=$desks[i]->y WHERE desk_id=$desks[i]->id";
        $conn->query($sql);
    }
    $conn->close();
}
?> 

不知道有什么问题。感谢您的帮助。

Answer 1

这样做
JS

$("#btn").click(function() {
    $.ajax({
        method: "POST",
        url: "phpfile.php",
        data: {desks:desks}
    })
});

PHP

$desks = $_POST("desks");
$sql = "UPDATE desks SET desk_x=$desks->x,desk_y=$desks->y WHERE desk_id=$desks->id";
$conn->query($sql);