通过谷歌浏览器查看RSS源 [英] get google chrome to view an rss feed

问题描述

我有一个.php脚本，可以动态生成RSS-Feed。

我想将.css样式表链接到此脚本中，以启用不支持的浏览器有一个内置的RSS-viewer来显示这个提要。我用

  echo<？xml version = \1.0 \encoding = \utf- 8 \？> \r\\\
; 
 echo<？xml-stylesheet type = \text / css\href = \rss.css \？>> \r\\\
; 
  

作为.php文件的第一个输出命令。

然而，像Chrome这样的浏览器仍然不会显示饲料作为网页 - 而是chrome显示我这个页面的代码。

在本教程中，我知道这是可能的...

http://www.petefreitag.com/item/208.cfm

http://www.petefreitag.com/rss/

如何做到这一点？

解决方案

您应该添加标头调用来指定内容类型，如下所示：

  header（Content-Type：application / rss + xml）; 
  

http://www.ibm.com/developerworks/library/x-phprss/ 获取更多信息。

I have an .php script that generates an RSS-Feed dynamically.

I want to link a .css stylesheet to this script to enable browsers that don't have a builtin RSS-viewer to display this feed. I do this with

echo "<?xml version=\"1.0\" encoding=\"utf-8\" ?>\r\n";
echo "<?xml-stylesheet type=\"text/css\" href=\"rss.css\" ?>\r\n";

as the first output commands of the .php-file.

However, browsers like chrome still don't display the feed as webpage - instead chrome shows me the code of this page.

From this tutorial, I know this is somehow possible...

http://www.petefreitag.com/item/208.cfm
http://www.petefreitag.com/rss/

how can this be done?

解决方案

You should add the header call to specify content type, like this:

header("Content-Type: application/rss+xml");

http://www.ibm.com/developerworks/library/x-phprss/ for more information.

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