Question

我试图用jQuery $ .post方法从php文件中获取数据。在ajax调用之后，我将附加一个来自php文件的按钮，它将通过删除自身来加载更多帖子，但是在ajax调用之后jQuery不再有效。

var result = function (str,end)
{
    $.get("query.php",{ start:str, end: end },function(ajaxresult){
    $(ajaxresult).appendTo(".focus");  
    })
}
result(4,0);

$('#loadmore').on('click', function(e){
    result(8,4);
    alert('delete');
    e.preventDefault();
    e.remove();
});

Here is my Php file code.

<?php
$start = $_GET['start'];
$end = $_GET['end'];


$con = mysqli_connect('localhost','root','binarystar','test');
if (!$con)
{
    die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,"users");
$sql="SELECT * FROM post ORDER BY id DESC LIMIT $end,$start";

$result = mysqli_query($con,$sql);


while($row = mysqli_fetch_array($result))
{
    echo "<section>";
    echo "<span class='feed_username'>" . $row['username'] ."</span>   ";
    echo "<span class='date'>" . $row['time'] ."</span><br>";
    echo "<span class='postx'>" . $row['post'] ."</span><br>";
    echo "</section>";

}
echo "<button id='loadmore'>Load More Posts</button>";

mysqli_close($con);

Answer 1

由于id是唯一的，因此您需要使用类：

echo "<button class='loadmore'>Load More Posts</button>";

然后，您可以将 event delegation 用于动态加载的元素：

$('body').on('click', '.loadmore', function() {
    // Your code here
});

$ ajax调用后jQuery无法正常工作

1 个答案: