我希望按钮出现在弹出窗口内。这里的问题是 html 代码而不是弹出窗口中的按钮出现。我认为这段代码有问题： $('[data-bs-toggle="popover"]').popover('show');
如何在弹出窗口内添加带有按钮标签的按钮而不将 data-bs-toggle 更改为 data-toggle？
<!doctype html>
<html lang="en">
<head>
<!-- Required meta tags -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<!-- Bootstrap CSS -->
<link href="https://cdn.jsdelivr.net/npm/bootstrap@5.0.2/dist/css/bootstrap.min.css" rel="stylesheet" integrity="sha384-EVSTQN3/azprG1Anm3QDgpJLIm9Nao0Yz1ztcQTwFspd3yD65VohhpuuCOmLASjC" crossorigin="anonymous">
</head>
<body>
<div class="container">
<button type="button" class="btn btn-secondary" data-bs-container="body" data-bs-toggle="popover" data-bs-placement="bottom" data-bs-content="Bottom popover" id="example"> Popover on bottom
</button>
</div>
<script src="https://code.jquery.com/jquery-3.6.0.js"></script>
<script>
var popString = "";
popString = popString + "<a href='#'><button class='btn btn-primary' id ='checkout'>Checkout</button></a> <button class='btn btn-primary' id ='clearCart'>Clear Cart</button> "
document.getElementById('example').setAttribute('data-bs-content', popString);
$(document).ready(function(){
$('[data-bs-toggle="popover"]').popover('show'); …