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Путь к файлу css в html документе при выводе текста html страницы через php file_get_contents echo


Ser8191
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Здравствуйте.

Такой вопрос.

В корневом каталоге сайта есть два каталога:

[php] -> содержит -> 1.php

[error] -> содержит -> error.html и error.css

error.html

<!DOCTYPE html>
<html>
 <head>
  <meta charset="utf-8" />
  <title>ERROR</title>
<link rel="stylesheet" href="error.css">
 </head>
 <body>
  <p>Привет, мир</p>
 </body>
</html>

1.php

<?php
        $SERVER_NAME = $_SERVER['SERVER_NAME'];
        $DOCUMENT_ROOT = $_SERVER['DOCUMENT_ROOT'];
        
        $tpl = file_get_contents("http://".$SERVER_NAME."/error/error.html");
        //$tpl = file_get_contents($DOCUMENT_ROOT."/error/error.html");
        echo $tpl;
?>

Теперь вопросы:

1. Правильнее использовать SERVER_NAME или DOCUMENT_ROOT или ?

2. При таком коде php я получу ошибку:

Refused to apply style from 'http://site.ru/php/error.css' because its MIME type ('text/html') is not a supported stylesheet MIME type, and strict MIME checking is enabled.

Как это исправить? Указывать полный путь к css файлу?

или нечто такое:

error.html

<!DOCTYPE html>
<html>
 <head>
  <meta charset="utf-8" />
  <title>ERROR</title>
<link rel="stylesheet" href="{DOMAIN}/error/error.css">
 </head>
 <body>
  <p>Привет, мир</p>
 </body>
</html>

1.php

<?php
        $SERVER_NAME = "http://" . $_SERVER['SERVER_NAME'];
        $DOCUMENT_ROOT = $_SERVER['DOCUMENT_ROOT'];
        
        
        $tpl = file_get_contents($SERVER_NAME."/error/error.html");
        //$tpl = file_get_contents($DOCUMENT_ROOT."/error/error.html");
        $tpl = str_replace("{DOMAIN}", $SERVER_NAME."", $tpl);
        
        echo $tpl;
?>

 

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