我想用ajax获取data-id并将其传递给同一页面中的php,但是当我单击按钮并想回显$_POST["postid"]时,什么也没发生。有什么错误吗?我想获取帖子ID。请帮助我,我是 Ajax 和 PHP 的新手。
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta http-equiv="X-UA-Compatible" content="ie=edge">
<title>Document</title>
<!-- Bootstrap css -->
<!-- Fontawesome -->
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/font-awesome/5.11.2/css/all.css">
</head>
<body>
<button title="Reply" class="replyButton btn p-0 pr-1" name="reply" data-id="11"><i class="fas fa-reply"></i></button>
<div id="mydiv">
<?php
if(isset($_POST["postid"])){
echo $_POST["postid"];
}
?>
</div>
</body>
</html>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.4.1/jquery.min.js">
</script>
<script src="http://code.jquery.com/jquery-3.3.1.min.js" integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="
crossorigin="anonymous"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.14.3/umd/popper.min.js" integrity="sha384-ZMP7rVo3mIykV+2+9J3UJ46jBk0WLaUAdn689aCwoqbBJiSnjAK/l8WvCWPIPm49"
crossorigin="anonymous"></script>
<script src="https://stackpath.bootstrapcdn.com/bootstrap/4.1.1/js/bootstrap.min.js" integrity="sha384-smHYKdLADwkXOn1EmN1qk/HfnUcbVRZyYmZ4qpPea6sjB/pTJ0euyQp0Mk8ck+5T"
crossorigin="anonymous"></script>
<script>
$(document).ready(function(){
$('.replyButton').on('click', function(){
var postid = $(this).data('id');
$post = $(this);
$.ajax({
url: 'test.php',
type: 'post',
data: {
'postid': postid
},
success: function(data){
$("#mydiv").html(data);
}
});
});
});
</script>
jQuery 中的选择器不正确。任一更改
<div class="mydiv"> <!-- selector .mydiv -->
自:
<div id="mydiv"> <!-- selector #mydiv -->
或者代替$("#mydiv")
改用$(".mydiv")
。
而且由于您发布到同一个 php 文件,因此当文件收到帖子时,您应该只输出所需的值,即 echo $_POST["postid"];
,jQuery代码将接收并放置在div
中。
注意: 考虑将 PHP 代码调整为:
<?php
if(isset($_POST["postid"])){
echo $_POST["postid"];
} else {
?>
<!-- Page content goes HERE -->
<?php } ?>