Question

我想添加给定表格中的用户。我正在迭代整个表并将每个值发送到javascript文件。

<?php 
          $sql = "select * from user where user_id not in (select second_user_id from friends where first_user_id = '$user_id'\n"
    . " union\n"
    . " select first_user_id from friends where second_user_id = '$user_id') limit 20";


              $result_not_friends=mysqli_query($dbc,$sql)
              or die("error in fetching");

            // print_r($row_not_friends);

           ?>
           <table class="table table-hover table-bordered">
            <h1>Users</h1>
              <tbody>
                <?php 
                  while ( $row_not_friends = mysqli_fetch_array($result_not_friends))
                  {
                    if ( $row_not_friends['user_id'] != $user_id )
                    {
                 ?>
                  <tr>
                    <td>
                      <?php echo $row_not_friends['user_name']; ?>
                    </td>
                    <!-- here I am sending request and processing it via ajax -->
                    <td><i class="fa fa-user-plus send_request"></i></td>
                    <input type="hidden" class="send_second" value="<?php echo $row_not_friends['user_id']; ?>">
                    <input type="hidden" class="send_first" value="<?php echo $user_id; ?>">
                  </tr>

                  <?php 
                    }
                    }
                   ?>
              </tbody>
          </table>

现在我正在访问javascript文件中的每个值，如下所示： //这里发送请求

$('.send_request').on('click',
    function (e) {
        e.preventDefault();

        var first = $('.send_first').val();
        var second = $('.send_second').val();
        alert('firt id is ' + first);
        alert('second id is ' + second);
        $.ajax(
        {

            url:'send_process.php',
            type:'POST',
            dataType:"json",
            data: { first: first, second: second },
            success:function(data)
            {
                if(data.send_success)
                {
                  window.location.href = "friend.php";

                }
                else
                {
                    alert("something went wrong");
                    window.location.href = "friend.php";
                }

            },
            error : function() { console.log(arguments); }
        }
            );
    });

但此处var second = $('.send_second').val();仅提供$row_not_friends['user_id']的最高元素值。当我回显该值时，它会给出正确的结果。请帮帮我。

Answer 1

因为您正在选择页面中的所有元素，而val（）的默认行为是它返回第一个项目。它没有你想要第n项的线索。

首先，您需要修复HTML无效。您不能将输入作为tr元素的兄弟。你需要在TD内部移动它。

                <!-- here I am sending request and processing it via ajax -->
                <td><i class="fa fa-user-plus send_request"></i> <!-- removed the closing td from here  -->
                <input type="hidden" class="send_second" value="<?php echo $row_not_friends['user_id']; ?>">
                <input type="hidden" class="send_first" value="<?php echo $user_id; ?>"></td>  <!-- moved the closing td to here  -->
              </tr>

您需要在单击按钮的同一行中找到元素。由于隐藏的输入是按钮的npw兄弟，您可以使用siblings()方法。

var btn = $(this);
var first = btn.siblings('.send_first').val();
var second = btn.siblings('.send_second').val();

while循环在php中的输入字段中只设置第一个值

1 个答案: