PHP &WP:根据真假条件渲染某些标记 [英] PHP & WP: Render Certain Markup Based on True False Condition

问题描述

所以，我在一个网站上工作，我想在某些页面的顶部显示静态图形，而在某些页面上我想显示滚动横幅.

到目前为止我设置的条件如下:

在我的标记中:

<?php如果($regBanner){回声 "<img src='" .$regBannerURL ."/style/images/main_site/home_page/mock_banner.jpg'/>";}别的 {echo '滑块！';}?></div><!-- 结束横幅 -->

在我的 else 语句中，我重复了滑块！"我想输出滑块的标记:

 
<img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/1.jpg" alt=""/><img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/2.jpg" alt=""/><img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/3.jpg" alt=""/>………………

我的问题是如何将 div 和所有这些图像放入我的 else echo 语句中?我在转义引号时遇到问题，并且我的滑块标记未呈现.

解决方案

<?php if($regbanner): ?><img src="<?php echo $regBannerURL; ?>/style/images/main_site/home_page/mock_banner.jpg"/><?php 其他:?><div id="滑块"><img src="<?php echo ($bannerDir = bloginfo('stylesheet_directory') .'/style/images/main_site/banners/services_banners'); ?>/1.jpg" alt=""/><img src="<?php echo $bannerDir; ?>/2.jpg" alt=""/><img src="<?php echo $bannerDir; ?>/3.jpg" alt=""/>………………
<?php endif;?></div><!-- 结束横幅 -->

So, I'm working on a site where on the top of certain pages I'd like to display a static graphic and on some pages I would like to display an scrolling banner.

So far I set up the condition as follows:

<?php 
    $regBanner = true;
    $regBannerURL =  get_bloginfo('stylesheet_directory'); //grabbing WP site URL 
?>

and in my markup:

<div id="banner">
  <?php

if ($regBanner) {
  echo "<img src='"  . $regBannerURL . "/style/images/main_site/home_page/mock_banner.jpg' />";
}

else {
  echo 'Slider!';
}

?>
</div><!-- end banner -->

In my else statement, where I'm echoing 'Slider!' I would like to output the markup for my slider:

  <div id="slider">
    <img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/1.jpg" alt="" />
    <img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/2.jpg" alt="" />
    <img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/3.jpg" alt="" />
        .............
  </div>

My question is how can I throw the div and all those images into my else echo statement? I'm having trouble escaping the quotes and my slider markup isn't rendering.

解决方案

<div id="banner">
  <?php if($regbanner): ?>
      <img src="<?php echo $regBannerURL; ?>/style/images/main_site/home_page/mock_banner.jpg" />
  <?php else: ?>
    <div id="slider">
      <img src="<?php echo ($bannerDir = bloginfo('stylesheet_directory') . '/style/images/main_site/banners/services_banners'); ?>/1.jpg" alt="" />
      <img src="<?php echo $bannerDir; ?>/2.jpg" alt="" />
      <img src="<?php echo $bannerDir; ?>/3.jpg" alt="" />
          .............
    </div>
  <?php endif; ?>
</div><!-- end banner -->

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