Question

我的想法是删除元素的内容，不包括用作Handlebarjs模板的脚本元素。

以下示例完全正常，但我想知道是否有比现在更好的方法？正如我在示例中所做的那样使用.remove（）或.empty（）不会删除TEXT节点。甚至改变孩子（）=＆gt; find（）也不起作用。

// HTML
<div id="content">

<!--This Handlebars template should remain-->
<script type="text/x-handlebars-template">
    <p>{{title}}, {{first_name}}, {{last_name}}</p>
</script>

This text node should be deleted.
<p>This element should be removed too.</p>
</div>
<!--Handlebars template example-->

// JS

// IIFE
; (function ($, undefined) {

console.log('Started');

// Cache the jQuery object
var $content = $('#content');

// Remove the contents of the div#content, but retain the Handlebars template

// $content.empty(); // <<<< Bad, as it removes everything

// These don't work, as the TEXT node isn't removed
// $content.children('*:not(script)').empty()
// $content.children('*:not(script)').remove();

// This works!
$content.contents().filter(function () {
    console.log(this.nodeName);
    // Only filter those which don't have the handlebars type and SCRIPT node name
    return this.nodeName !== 'SCRIPT' || this.type !== 'text/x-handlebars-template';

    // Remove from the DOM
}).remove();

console.log('Finished');

})(jQuery);

Answer 1

我认为这就是你所需要的：

$('#content').html($('#content script'));

删除除脚本内容以外的所有内容。

$('#content').html($('#content script'));
console.log($('#content').html());

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="content">
    
    <script type="text/x-handlebars-template">
        <p>{{title}}, {{first_name}}, {{last_name}}</p>
    </script>
    
    This text node should be deleted.
    <p>This element should be removed too.</p>

    <script type="text/x-handlebars-template">
      <p>{{more handebars stuff}}</p>
    </script>

    <p>Another element to remove.</p>    
</div>

删除元素的内容，不包括某些节点名称/类型

1 个答案: