php - 将 AJAX 连接到 MySQL (PHP)

Question

我有一项与 PHP、MySQL、AJAX 和 JQuery 相关的作业到期。目前我遇到脚本问题，希望有人能帮助我。该脚本用于更新 MySQL 中的字段。该脚本似乎不起作用。这是脚本:

function updatedata(id) {
    var id = id;
    var name = $('#name' + id).val() ;
    var url = $('#url' + id).val() ;
    var imageurl = $('#imageurl' + id).val() ;
    var description = $('#description' + id).val() ;
    var datas = "name=" + name + "&url=" + url + "&imageurl=" + imageurl + "&description=" + description;

    $.ajax({
        type: "POST",
        url: "update.php?id=" + id;
        data: datas;
    }).done(function (data) {
        $('#info').html(data);
        viewdata();
    });
}

以下是“update.php”文件中包含的代码:

<?php
    include ('header.php');
    if(isset($_GET['id'])) {
        $stmt = $connection->prepare("UPDATE news_source SET name=?, url=?, description=?, imageurl=? where id=?");
        $stmt->bind_param('sssss', $full_name, $url_1, $descript_ion, $image_url, $id);
        $full_name = mysqli_real_escape_string($connection, $_POST ['name']);
        $url_1 = mysqli_real_escape_string($connection, $_POST ['url']);
        $descript_ion = mysqli_real_escape_string($connection, $_POST ['description']);
        $image_url = mysqli_real_escape_string($connection, $_POST ['imageurl']);
        $id = $_GET ['id']);

        if($stmt->execute()))
        { ?>
            <div class = "alert alert-success alert-dismissible" role="alert">
                <button type="button" class="close" data-dismiss="alert" aria-label="Close">
                    <span aria-hidden="true"> &times; </span>
                </button>
                <strong>Success!</strong> 
                Record has been added.
            </div>
        <?php } else { ?>
            <div class = "alert alert-danger alert-dismissible" role="alert">
                <button type="button" class="close" data-dismiss="alert" aria-label="Close">
                    <span aria-hidden="true"> &times; </span>
                </button>
                <strong>Error!</strong> 
                Record failed to add.
            </div>
        <?php }
    include ('footer.php');
?>

编辑:该脚本在同一个 php 文件“maintain.php”中被调用为:

<div class="modal-footer">
        <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
        <button type="button" onclick="updatedata('<?php echo $rows['id']; ?>')" class="btn btn-primary">Save changes</button>
      </div>

编辑:

问题描述:好的，问题是我无法更新MySQL数据库中的字段。主文件应该允许用户编辑 Bootstrap Modal 中存在的数据，然后运行 ​​AJAX 以动态更新数据库而不刷新页面。目前，主文件仅显示可编辑的数据，但我无法更新数据库中的数据。单击“保存更改”按钮后，没有任何反应。

注意:我已尝试使用 How do I get PHP errors to display? 中存在的信息显示 php 错误，但不幸的是它什么也没显示。

Answer 1

您正在绑定(bind)参数然后转义它们。先获取参数并转义，再绑定(bind)而且您没有清理 ID。

$full_name = mysqli_real_escape_string($connection, $_POST ['name']);
$url_1 = mysqli_real_escape_string($connection, $_POST ['url']);
$descript_ion = mysqli_real_escape_string($connection, $_POST ['description']);
$image_url = mysqli_real_escape_string($connection, $_POST ['imageurl']);
$id = intval($_GET ['id']));

$stmt->bind_param(1, $full_name);
$stmt->bind_param(2, $url_1);
$stmt->bind_param(3, $descript_ion);
$stmt->bind_param(4, $image_url);
$stmt->bind_param(5, $id);