关于将 null null 有什么办法可以解决这个问题吗？ 我做错了什么？
为了便于访问，我在元素周围添加了一个容器，并将其内容替换为函数的输出 null标记转换为和 标记的问题，我试图遵循@ishtiaq-Ahmed和@yob的答案。 但是，输出并不实际显示函数的功能，而是将 类保留为
类。 代码如下:
var div2table = $('.tr').map(function() {
var issue = $(this);
var tdline = issue.find('.td').map(function() {
return '<td>' + $(this).text();
}).get().join('</td>');
return '<tr>' + tdline + '</td>';
}).get().join('</tr>');
div2table = '<table>' + div2table + '</tr></table>';
console.log(div2table);<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class='tr'>
<div class='td'>this will be 1st TD</div>
<div class='td'>this will be 2nd TD</div>
<div class='td'>this will be 3rd TD</div>
</div>
<div class='tr'>
<div class='td'>this will be 1st TD</div>
<div class='td'>this will be 2nd TD</div>
<div class='td'>this will be 3rd TD</div>
</div>
<div class='tr'>
<div class='td'>this will be 1st TD</div>
<div class='td'>this will be 2nd TD</div>
<div class='td'>this will be 3rd TD</div>
</div>
共1个答案
匿名用户
var div2table = $('.container .tr').map(function() {
var issue = $(this);
var tdline = issue.find('.td').map(function() {
return '<td>' + $(this).text();
}).get().join('</td>');
return '<tr>' + tdline + '</td>';
}).get().join('</tr>');
div2table = '<table>' + div2table + '</tr></table>';
$('.container').html(div2table);<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="container">
<div class='tr'>
<div class='td'>this will be 1st TD</div>
<div class='td'>this will be 2nd TD</div>
<div class='td'>this will be 3rd TD</div>
</div>
<div class='tr'>
<div class='td'>this will be 1st TD</div>
<div class='td'>this will be 2nd TD</div>
<div class='td'>this will be 3rd TD</div>
</div>
<div class='tr'>
<div class='td'>this will be 1st TD</div>
<div class='td'>this will be 2nd TD</div>
<div class='td'>this will be 3rd TD</div>
</div>
</div>
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