Javascript 如何在单击子li元素时保持父li打开

如果您使用php渲染并在菜单项上单击“页面刷新”，您可以将类例如“active”添加到父li，并在样式中添加

li.active{display:block}

或类似样式以显示隐藏元素

最后，我同意laravel routes的条件

---

显示您的代码和最小示例您尝试了什么？向你表示感谢。。我知道。。。但当前的问题是，当我单击子li标记时，父li将自动关闭。。我希望它在我点击儿童时仍然打开。。。我需要使用一些js吗？是的，我想，你应该在“dropdownmenu.js”文件中找到click事件，然后更改它。

<li>
    <a href="#">
        <i class="entypo-tools"></i>
        <span class="title">PHP-SDK</span>
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    </a>
    <ul>
        <li class="dropdown-submenu">
            <a href="/mypage_v2/sdk/installation">
                <i class="entypo-inbox"></i>
                <span class="title">Installation</span>
            </a>
        </li>
        <li class="dropdown-submenu">
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                <i class="entypo-docs"></i>
                <span class="title">Getting Started</span>
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                        <span class="title">sample1</span>
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                    <ul>
                        <li>
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                                <i class="entypo-newspaper"></i>
                                <span class="title">sample1</span>
                            </a>
                        </li>
                        <li>
                            <a href="/mypage_v2/sdk/newsletterEmail">
                                <i class="entypo-newspaper"></i>
                                <span class="title">sample2</span>
                            </a>
                        </li>
                        <li>
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                                <i class="entypo-newspaper"></i>
                                <span class="title">sample2</span>
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                        </li>
                        <li>
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                                <span class="title">NewslettersProcess</span>
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                        </li>
                    </ul>
                </li>

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</li>