为什么再次出现这种情况（mysqli

本文介绍了为什么再次出现这种情况（mysqli_num_rows（）需要参数）的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个名为tbl_video的表，它由两个字段组成（video_id，video_title我希望使用像社交网站一样的Ajax JQuery添加一个加载按钮。我能够获得前四个记录但是当我试图获得时然后我才得到这个例外。我做错了什么。 我尝试了什么：

I have a table named tbl_video that consist of two fields (video_id, video_title What I want to add a load more button using Ajax JQuery like social networking sites .I am able to get the first four records but when I try to get the othersthen I am getting this exception .What am i doing wrong . What I have tried:

this my index.php file I create table structure here and then fetch records. <pre lang="php"> <pre> <table class="table table-bordered" id="load_data_table"> <thead> <tr> <td width="15%">Video Id</td> <td>Video Title</td> </tr> </thead> <?php while ($row = mysqli_fetch_array($result)) { $video_id = $row['video_id']; ?> <tr> <td><?php echo $row['video_id']; ?></td> <td><?php echo $row['video_title']; ?></td> </tr> <?php } ?> <tr id="remove_row"><td><button type="button" class="more btn btn-success form-control" data-id="<?php echo $video_id; ?>">More</button></td></tr> </table> </div>

<script> function moreButton(){ $('.more').click(function(){ var id=$(this).data("id"); $.ajax({ url:"more.php", data:{id:id}, method:"post", success:function(data){ if(data!=''){ $('#load_data_table').append(data); $('#remove_row').remove(); moreButton(); } else{ $('.more:last').html("No Data"); } } }); }); } $(document).ready(function(){ moreButton(); });

and this the more.php file.I think it is self explanatory

<?php ini_set('display_errors', 1); $id = $_POST['id']; $output = ''; $db = mysqli_connect('localhost', 'root', '', 'testing'); $data = mysqli_query($db, "select * from tbl_video where video_id >{$id} limit 2"); if (mysqli_num_rows($data)) { while ($row = mysqli_fetch_array($data)) { $output .= ' <tbody> <tr> <td>'.$row['video_id'].'</td> <td>'.$row['video_title'].'</td> </tr> '; } $output .= '<tr id="remove_row"><td><button type="button" class="more btn btn-success form-control" data-id="'.$row['video_id'].'">More</button></td></tr></tbody>'; echo $output; }

任何帮助都会被评估。

any help would be appriciated.

推荐答案

row = mysqli_fetch_array（ row = mysqli_fetch_array(

result））{

video_id = video_id =