php - 如何注释掉其中包含 HTML 和 PHP 代码的多行？

Question

我想注释掉下面的整个代码。在不使用多个评论的情况下，最简单的方法是什么？提前致谢。

<nav id="site-navigation" class="main-navigation" role="navigation">
    <h3 class="menu-toggle"><?php _e( 'Menu', 'twentytwelve' ); ?></h3>
    <a class="assistive-text" href="#content" title="<?php esc_attr_e( 'Skip to content', 'twentytwelve' ); ?>"><?php _e( 'Skip to content', 'twentytwelve' ); ?></a>
    <?php wp_nav_menu( array( 'theme_location' => 'primary', 'menu_class' => 'nav-menu' ) ); ?>
</nav><!-- #site-navigation -->

Answer 1

用这样的 php 注释将整个块包围起来：

<?php /*
<nav id="site-navigation" class="main-navigation" role="navigation">
    <h3 class="menu-toggle"><?php _e( 'Menu', 'twentytwelve' ); ?></h3>
    <a class="assistive-text" href="#content" title="<?php esc_attr_e( 'Skip to content', 'twentytwelve' ); ?>"><?php _e( 'Skip to content', 'twentytwelve' ); ?></a>
    <?php wp_nav_menu( array( 'theme_location' => 'primary', 'menu_class' => 'nav-menu' ) ); ?>
</nav><!-- #site-navigation -->
*/?>

Answer 2

<!-- <nav id="site-navigation" class="main-navigation" role="navigation">
<h3 class="menu-toggle"><?php _e( 'Menu', 'twentytwelve' ); ?></h3>
<a class="assistive-text" href="#content" title="<?php esc_attr_e( 'Skip to content', 'twentytwelve' ); ?>"><?php _e( 'Skip to content', 'twentytwelve' ); ?></a>
<?php wp_nav_menu( array( 'theme_location' => 'primary', 'menu_class' => 'nav-menu' ) ); ?>
</nav> -->

应该够了。它们是多行注释。

这应该会有所帮助：https ://stackoverflow.com/questions/5781375/comment-out-html-and-php-together