请求不带GET参数的字符串

问题描述

是否有一种简单的方法来获取没有GET参数的请求的文件或目录?例如，如果URL是http://example.com/directory/file.php?paramater=value，我只想返回http://example.com/directory/file.php.我很惊讶$_SERVER[]中没有简单的索引.我想念一个吗?

Is there a simple way to get the requested file or directory without the GET arguments? For example, if the URL is http://example.com/directory/file.php?paramater=value I would like to return just http://example.com/directory/file.php. I was surprised that there is not a simple index in $_SERVER[]. Did I miss one?

推荐答案

您可以使用$_SERVER['REQUEST_URI']获取请求的路径.然后，您需要删除参数...

You can use $_SERVER['REQUEST_URI'] to get requested path. Then, you'll need to remove the parameters...

$uri_parts = explode('?', $_SERVER['REQUEST_URI'], 2);

然后，添加主机名和协议.

Then, add in the hostname and protocol.

echo 'http://' . $_SERVER['HTTP_HOST'] . $uri_parts[0];

如果混合使用http:和https://，则还必须检测协议. $_SERVER['REQUEST_SCHEME']返回协议.

You'll have to detect protocol as well, if you mix http: and https://. $_SERVER['REQUEST_SCHEME'] returns the protocol.

echo $_SERVER['REQUEST_SCHEME'] .'://'. $_SERVER['HTTP_HOST']
     . explode('?', $_SERVER['REQUEST_URI'], 2)[0];

...返回，例如:

http://example.com/directory/file.php

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php.com文档:

• $_SERVER —服务器和执行环境信息
• explode -用字符串将字符串分隔
• parse_url —解析URL并返回其URL组件(可能是更好的解决方案)

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php.com Documentation:

• $_SERVER — Server and execution environment information
• explode — Split a string by a string
• parse_url — Parse a URL and return its components (possibly a better solution)

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