Question

我已经尝试了几个小时才能使其正常工作。我有一个<div id=""data_friends>标签和一个hidden input field我想使用AJAX更新。 div标签如下：

<div class="friends-tab-list" id="data_friends">

    <?php 
    //Default query limits results to 8

    $sql = ("SELECT * FROM users WHERE FIND_IN_SET(?, friend_array) > 0 LIMIT 0,8");
    $query = mysqli_prepare($con, $sql);
    $query->bind_param('s', $username);
    $query->execute();

    $result = $query->get_result();

    $num_rows = $result->num_rows;

        while ($row = $result->fetch_assoc()) {
        $row['profile_pic']; 
        $row['username'];

        echo "<a class='profile-img-item' href='" . $row['username'] . "'>
              <img src='" . $row['profile_pic'] . "' title='" . $row['username'] . "'>
              </a>";
                }
        $query->close();

          ?>
</div>

隐藏的输入如下：<input id='max' type='hidden' value='<?php echo $num_rows; ?>'>

我单击“查看更多朋友”按钮，并使用以下命令将数据发送到includes/handlers/ajax_load_profile_friends.php：

$.ajax({

    url:'includes/handlers/ajax_load_profile_friends.php',
    type:'POST',
    dataType: 'json',
    data:{'username':username, 'num_friends':num_friends, 'counter':counter},

        success: function(data) {
            $('#data_friends').html(data.html);
            $('#max').val(data.num_rows);
                }
        });

来自ajax_load_profile_friends.php的数据如下：

$query = mysqli_prepare($con,"SELECT * FROM users WHERE FIND_IN_SET(?, friend_array) LIMIT $counter"); 
$query->bind_param('s', $username);
$query->execute();

$result = $query->get_result();

$num_rows = $result->num_rows;
}       

while ($row = $result->fetch_assoc()) {
$row['profile_pic']; 
$row['username'];

$html = "<a class='profile-img-item' href='" . $row['username'] . "'>
          <img src='" . $row['profile_pic'] . "' title='" . $row['username'] . "'>
         </a>";

}   

echo json_encode(array('num_rows' => $num_rows, 'html' => $html));

运行此命令时，如果我想通过在成功函数$('#data_friends').html(data.html);中执行此操作，我认为每次点击都会获得16条记录的回报，那么我得到的回报只有一个

我的隐藏输入字段<input id='max' type='hidden' value='<?php echo $num_rows; ?>'>中的值未使用此$('#max').val(data.num_rows);

更新

ajax_load_profile_friends.php中是否缺少引起这些行为的东西？

**请记住，当我不使用json_encode时可以使用此功能，并编写类似$('#data_friends').html(data.html);的成功函数，然后从AJAX中删除dataType: 'json',。这里的问题在于，无论哪种方式，我都无法更新隐藏的输入值。我认为我会尝试并纠正此方法，因为大多数示例都将json_encode()指定为返回数据的方法。

Answer 1

header( "Content-Type: application/json", TRUE );
$query = mysqli_prepare($con,"SELECT * FROM users WHERE FIND_IN_SET(?, friend_array) LIMIT $counter"); 
$query->bind_param('s', $username);
$query->execute();

$result = $query->get_result();

$num_rows = $result->num_rows;

$html='';

while ($row = $result->fetch_assoc()) {


$html .= "<a class='profile-img-item' href='" . $row['username'] . "'>
          <img src='" . $row['profile_pic'] . "' title='" . $row['username'] . "'>
         </a>";

}   

echo json_encode(array('num_rows' => $num_rows, 'html' => $html));

Answer 2

您没有在while循环之前声明$ html变量。试试这个

<?php

$query = mysqli_prepare($con,"SELECT * FROM users WHERE FIND_IN_SET(?, friend_array) LIMIT $counter"); 
$query->bind_param('s', $username);
$query->execute();

$result = $query->get_result();

$num_rows = $result->num_rows;
}       
$html = '';
while ($row = $result->fetch_assoc()) {
// $row['profile_pic']; 
// $row['username'];

$html .= "<a class='profile-img-item' href='" . $row['username'] . "'>
          <img src='" . $row['profile_pic'] . "' title='" . $row['username'] . "'>
         </a>";

}   

echo json_encode(array('num_rows' => $num_rows, 'html' => $html));

PHP变量无法使用json_encode（）正确返回成功的AJAX / jQuery POST

2 个答案: