PHP + Slim PHP框架：如何在文档头中动态添加样式表？

Question

I have created a simple slim application with a login form, dashboard, registration page and account page. 我用登录表单，仪表板，注册页面和帐户页面创建了一个简单的苗条应用程序。 Each page has its own route. 每个页面都有自己的路线。 All the pages use the same header and footer, which is included by adding <?php include __DIR__ . '/header.phtml' ?> 所有页面使用相同的页眉和页脚，通过添加<?php include __DIR__ . '/header.phtml' ?>来包含这些页眉和页脚<?php include __DIR__ . '/header.phtml' ?> <?php include __DIR__ . '/header.phtml' ?> and <?php include __DIR__ . '/footer.phtml' ?> <?php include __DIR__ . '/header.phtml' ?>和<?php include __DIR__ . '/footer.phtml' ?> <?php include __DIR__ . '/footer.phtml' ?> to the beginning and end of each page template, respectively, with the page's content placed in between. <?php include __DIR__ . '/footer.phtml' ?>到每个页面模板的开头和结尾，页面内容放置在它们之间。

I would like to dynamically include different stylesheets depending on the page viewed, but I don't want to just lazily place them in my .phtml files and have them render in the middle of the web page. 我想根据查看的页面动态地包含不同的样式表，但是我不想只是懒惰地将它们放置在我的.phtml文件中，并将它们呈现在网页的中间。

I'd like to develop some if/switch function to be placed within the header.phtml file which recognises the current route and renders a link to the appropriate stylesheet. 我想开发一些if / switch函数，将其放置在header.phtml文件中，该文件可以识别当前路由并提供指向适当样式表的链接。 Something like the following code, placed within the document head: 将以下代码放置在文档头中：

<?php
if($currRoute === "/dashboard") {
  ?>
  <link rel="stylesheet" type="text/css" href="assets/css/dashboard.css">
  <?php
} else if($currRoute === "/register") {
  ?>
  <link rel="stylesheet" type="text/css" href="assets/css/register.css">
  <?php
} else if ($currRoute === "/account") {
  ?>
  <link rel="stylesheet" type="text/css" href="assets/css/account.css">
  <?php
}
?>

Is there any solution for acquiring the value of $currRoute in the above example without some complicated regular expression, any additional languages, class libraries etc.? 在上面的示例中，是否有任何解决方案来获取$currRoute的值，而无需一些复杂的正则表达式，任何其他语言，类库等？

Answer 1

Instead of attempting to retrieve the current route, I created a string variable containing the route name within each route callback, then passed that name as an argument to the renderer (in this case I am using Slim's php-view package). 我没有尝试检索当前路由，而是在每个路由回调中创建了一个包含路由名称的字符串变量，然后将该名称作为参数传递给渲染器（在这种情况下，我使用的是Slim的php-view包）。 Here is an example, passing the string "dashboard" to the dashboard.phtml template file as an argument: 这是一个示例，将字符串"dashboard"作为参数传递给dashboard.phtml模板文件：

$app->get('/dashboard', function (Request $request, Response $response, array $args) {
  $args['pageTitle'] = "dashboard";
  return $this->renderer->render($response, 'register.phtml', $args);
});

The $args['pageTitle'] key is assigned a keyword identifier as a string, based on the name of the route. 根据路由名称，为$args['pageTitle']键分配了一个字符串形式的关键字标识符。

Then, within my header.phtml file I have created a switch statement which compares the value of $args['pageTitle'] with a set of hard-coded strings to determine the appropriate stylesheet to load: 然后，在header.phtml文件中，我创建了一个switch语句，该语句将$args['pageTitle']与一组硬编码的字符串进行比较，以确定要加载的适当样式表：

<head>
  <title>Slim 4 PHP Template</title>

  <?php
    if(isset($pageTitle)) {
      switch($pageTitle) {
        case "dashboard":
          echo "<link rel='stylesheet' type='text/css' href='assets/css/dashboard.css'>";
          break;
        case "register":
          echo "<link rel='stylesheet' type='text/css' href='assets/css/register.css'>";
          break;
      }
    }
  ?>
</head>

When visiting the route at " http://example.domain.com/dashboard " the link to the appropriate stylesheet is rendered, like so: 在“ http://example.domain.com/dashboard ”上访问路由时，将呈现到相应样式表的链接，如下所示：

<head>
  <title>Slim 4 PHP Template</title>

  <link rel="stylesheet" type="text/css" href="assets/css/dashboard.css">
</head>

PHP + Slim PHP框架：如何在文档头中动态添加样式表？

问题描述

1 个解决方案

解决方案1
0 2019-09-17 12:06:14

PHP + Slim PHP框架：如何在文档头中动态添加样式表？

问题描述

1 个解决方案

解决方案1 0 2019-09-17 12:06:14

解决方案1
0 2019-09-17 12:06:14