Javascript 内容可编辑更新PHP、SQL、AJAX-网络it'；他成功了吗？

我在通过PHP和Ajax将更改后的文本更新为SQL时遇到问题

我不确定哪一个是不正确的，是通过Ajax发送的数据，还是PHP文件有问题

这是我的HTML文件

<tr>
<td class="editingTab" contenteditable='true' class="texting">{{user.id}}</td>
<td contenteditable='true' >{{user.first_name}}</td>
<td contenteditable='true' >{{user.last_name}}</td>
<td contenteditable='true' >{{user.email}}</td></tr>

这是PHP文件

 <?php

    $conn = mysqli_connect('localhost','nemkeang','nemkic23','users');

    $id = $POST_['getId'];
    $name =$_POST['name'];
    $last =$_POST['last'];
    $email =$_POST['email'];

    $sql = "UPDATE user 
    SET first_name = '".$name."', 
        last_name = '".$last."', 
       email = '".$email."' 
    WHERE id = '".$id."'";

    if (!mysqli_query($conn,$sql)) {
      die('Error: ' . mysqli_error($conn));
    }
    echo  "success: ".$name." ";

    mysqli_close($con);

?>

---

我认为问题在于获取id，所以更新操作将失败。请尝试此

mysqli\u real\u escape\u string（）
函数用于转义字符串中的特殊字符，以便在SQL语句中使用。因此，请使用
mysqli\u real\u escape\u string（）


 <?php

    $conn = mysqli_connect('localhost','nemkeang','nemkic23','users');

    $id = $POST_['getId'];
    $name =$_POST['name'];
    $last =$_POST['last'];
    $email =$_POST['email'];

    $sql = "UPDATE user 
    SET first_name = '".$name."', 
        last_name = '".$last."', 
       email = '".$email."' 
    WHERE id = '".$id."'";

    if (!mysqli_query($conn,$sql)) {
      die('Error: ' . mysqli_error($conn));
    }
    echo  "success: ".$name." ";

    mysqli_close($con);

?>

<?php

    $conn = mysqli_connect('localhost','nemkeang','nemkic23','users');

    $id = mysqli_real_escape_string($conn,$_POST['id']);  //
    $name =mysqli_real_escape_string($conn,$_POST['name']);
    $last =mysqli_real_escape_string($conn,$_POST['last']);
    $email =mysqli_real_escape_string($conn,$_POST['email']);

    $sql = "UPDATE user 
    SET first_name = '$name', 
        last_name = '$last', 
       email = '$email' 
    WHERE id = '$id';

    if (!mysqli_query($conn,$sql)) {
      die('Error: ' . mysqli_error($conn));
    }
    echo  "success: ".$name." ";

    mysqli_close($con);

?>