php - 如何从 json (php) 获取正确的值？

Question

更新模式( Bootstrap )

<div class="modal fade" id="updateModal" tabindex="-1" role="dialog" aria-labelledby="exampleModalLabel" aria-hidden="true">
    <div class="modal-dialog" role="document">
        <div class="modal-content">
            <div class="modal-header">
                <h5 class="modal-title" id="exampleModalLabel">Edit：</h5>
                <button class="close" type="button" data-dismiss="modal" aria-label="Close">
                    <span aria-hidden="true">×</span>
                </button>
            </div>
            <div class="modal-body" id="body">
                <input type="text" class="form-control" placeholder="product_title" id="title">
            </div>
            <div class="modal-footer">
                <button class="btn btn-secondary" type="button" data-dismiss="modal">Cancel</button>
                <a class="btn btn-primary" href="login.php">Update</a>
            </div>
        </div>
    </div>
</div>

触发 update_modal(bootstrap)

    <td>
       <a data-toggle='modal' data-target='#updateModal'>
       <button class='btn btn-success' id='edit' data-id='$product_no'>Edit</button>
       </a>
    </td>

js

  <script>
      $(document).ready(function() {
          $(document).on('click','#edit',function () {
              let id=$(this).attr('data-id');
              // console.log(id);
              $.ajax({
                  url:'get_record.php',
                  method:'post',
                  data:{
                      p_no:id
                  },
                  dataType:'JSON',
                  success:function (data) {
                        // $('#body').html(data);
                      console.log(data[0]);
                      console.log(data[1]);
                      console.log(data[2]);
                  }

              });
          });
      })

  </script>

get_record.php

<?php
include ("db.php");
 global $conn;
if (isset($_POST['p_no'])){
    $p_no=$_POST['p_no'];
    $query="select * from products where p_no='$p_no'";
    $result=mysqli_query($conn,$query);
    while($row=mysqli_fetch_array($result)){
        $user_data=' ';
        $user_data [0]=$row['product_title'];
        $user_data [1]=$row['date'];
        $user_data [2]=$row['product_price'];
//        echo $pro_title;
    }
    echo json_encode($user_data);
}

My problem is when I click the edit button, the update_modal displays successfully, but it doesn't get correct values from my database. I click F12 on my chrome and it tested like this photo

Answer 1

您正在将 get_record.php 中的 $user_data 初始化为字符串:$user_data = ' ';

尝试将该变量初始化为数组:$user_data = [];

因为您将 $user_data 初始化为字符串，所以设置 $user_data[0] 仅将字符串的第一个字符设置为值 的第一个字符$row['product_title']。

<小时/>

编辑:正如评论中所指出的，通过将发布的数据直接插入到您的 SQL 中，您也会面临 SQL 注入(inject)攻击:

$p_no=$_POST['p_no'];
$query="select * from products where p_no='$p_no'"

在查询中使用所有用户输入之前，您应该转义它们。在 PHP 中对 MySQL 执行此操作的一种方法是:

$p_no=mysqli_real_escape_string($_POST['p_no']);
$query="select * from products where p_no='$p_no'"

使用prepared statements将是理想的。