Question

从搜索结果更新用户的名字和姓氏数据库记录时出现问题。查询运行正常并更新数据库记录，但它会为错误的选定用户执行此操作。

search_user.php

        <?php require_once("includes/db.php"); ?>
        <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"> </script>


        <script type="text/javascript">
               $(document).ready(function(){
$('.edit').on('click', function (e) {

    var firstname = $('#firstname').val();
    var lastname = $('#lastname').val();
   var user_id = $('#user_id').val();

    var dataString = 'firstname='+firstname + '&lastname='+lastname+ '&user_id='+user_id ;



                  if(firstname=='' || lastname=='')
                    {
                    alert("Please fill all fields");
                    }

                      else
                    {
                    $.ajax({
                    type: "POST",
                    url: "update_users.php",
                    data: dataString,
                    success: function(){

                        alert("User details have been updated.")
                    }
                    });
                    }
     return false;
});

               });
  </script>

         <?php

            $output = "";


            $query = "SELECT * FROM users";

            $result = mysqli_query($connection, $query);

            if(mysqli_num_rows($result)> 0){

                     $output .= '<h4 align="center"> Search Result </h4>';

                 $output .= '


                    <table class="table table-bordered">

                    <thead>

                    <th>First Name </th>
                    <th>Last Name </th>
                    <th colspan="2">Options </th>

                    </thead>

            ';

                while($row = mysqli_fetch_array($result)){

                    $_SESSION['user_id'] = $row['user_id'];

                    $output .= '

                    <tbody>
                        <tr>

                        <td><input type="text" id="firstname" name="firstname" value="'.$row["first_name"].'"></td>
                        <td> <input type="text" id="lastname" name="lastname" value="'.$row["last_name"].'"></td>
                        <td> <a href="search_user.php?edit='.$row["user_id"].'" class="edit">Edit </a> </td>
                         <td> <a href="search_user.php?delete='.$row["user_id"].'" class="delete_confirmation">Delete </a> </td>
                        </tr>

<input type="hidden" name="user_id" id="user_id" value="'.$row["user_id"].'">

                        </tbody>

                    ';

                }

            echo $output;

            } else {

                echo "User not found!";
            }



            ?>

update_user.php

 <?php require_once("includes/db.php"); ?>


     <?php

     if (isset($_POST['firstname'], $_POST['lastname'], $_POST['user_id'])) {


     $firstname = mysqli_real_escape_string($_POST['firstname']);
     $lastname = mysqli_real_escape_string($_POST['lastname']);


     $query = "UPDATE users
        SET first_name  ='" . $_POST['firstname'] . "',
         last_name ='" . $_POST['lastname'] . "'

                  WHERE
        user_id = '" .  $_POST['user_id']  . "'";


                  $edit_user_query = mysqli_query($connection, $query);
                    } 
                else {
            echo "invalid response";
                }

              ?>

Answer 1

正如我在评论中提到的，你需要失去对会话的依赖。而是在实际的编辑链接上放置一个id。见下面的修改代码：

<?php require_once("includes/db.php"); ?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"> </script>
<script type="text/javascript">
    $(document).ready(function(){
    $('.edit').on('click', function (e) {

        var firstname = $('#firstname').val();
        var lastname = $('#lastname').val();
        var id=$(this).attr('id');

         var dataString = 'firstname='+firstname + '&lastname='+lastname + '&userid='+id;
          if(firstname=='' || lastname=='')
            {
            alert("Please fill all fields");
            }

              else
            {
            $.ajax({
            type: "POST",
            url: "update_users.php",
            data: dataString,
            success: function(){

                alert("User details have been updated.")
            }
            });
            }
         return false;
    });
});

     <?php

        $output = "";


        $query = "SELECT * FROM users";

        $result = mysqli_query($connection, $query);

        if(mysqli_num_rows($result)> 0){

                 $output .= '<h4 align="center"> Search Result </h4>';

             $output .= '


                <table class="table table-bordered">

                <thead>

                <th>First Name </th>
                <th>Last Name </th>
                <th colspan="2">Options </th>

                </thead>

        ';

            while($row = mysqli_fetch_array($result)){

                $_SESSION['user_id'] = $row['user_id'];

                $output .= '

                <tbody>
                    <tr>

                    <td><input type="text" id="firstname" name="firstname" value="'.$row["first_name"].'"></td>
                    <td> <input type="text" id="lastname" name="lastname" value="'.$row["last_name"].'"></td>
                    <td> <a href="search_user.php?edit='.$row["user_id"].'" class="edit"  id="'.$row["user_id"].'">Edit </a> </td>
                     <td> <a href="search_user.php?delete='.$row["user_id"].'" class="delete_confirmation">Delete </a> </td>
                    </tr>

                    </tbody>

                ';

            }

        echo $output;

        } else {

            echo "User not found!";
        }



        ?>

您可以通过帖子访问ID：

$userid = mysqli_real_escape_string($_POST['userid']);

使用Ajax更新数据库记录

1 个答案: