Question

如果我知道当前站点路径，如何将URI转换为URL？

请考虑以下示例：

目前的路径是：`http://www.site.com/aa/folder/page1.php

Uri：folder2 / page.php

Uri：/folder2/page.php

如果当前路径是：

`http://www.site.com/aa/folder/

或

`http://www.site.com/aa/folder

那些网址会是什么样的？

我知道这应该是简单而明显的，但我找不到完整的答案（是的，我在Google上搜索过）

Answer 1

$_SERVER超全球将拥有您正在寻找的信息，即$_SERVER['REQUEST_URI']和$_SERVER['SERVER_NAME']。 $_SERVER['QUERY_STRING']也可能有用。

请参阅：

http://php.net/manual/en/reserved.variables.server.php

Answer 2

这是一段具有您需要的功能的代码块： http://ca.php.net/manual/en/function.parse-url.php#76682

编辑：使用示例

修改上述链接功能

<?php

var_dump(resolve_url('http://www.site.com/aa/folder/page1.php','folder2/page.php?x=y&z=a'));

var_dump(resolve_url('http://www.site.com/aa/folder/page1.php','/folder2/page2.php'));

function unparse_url($components) {
    return $components['scheme'].'://'.$components['host'].$components['path'];
}

/**
 * Resolve a URL relative to a base path. This happens to work with POSIX
 * filenames as well. This is based on RFC 2396 section 5.2.
 */
function resolve_url($base, $url) {
        if (!strlen($base)) return $url;
        // Step 2
        if (!strlen($url)) return $base;
        // Step 3
        if (preg_match('!^[a-z]+:!i', $url)) return $url;
        $base = parse_url($base);
        if ($url{0} == "#") {
                // Step 2 (fragment)
                $base['fragment'] = substr($url, 1);
                return unparse_url($base);
        }
        unset($base['fragment']);
        unset($base['query']);
        if (substr($url, 0, 2) == "//") {
                // Step 4
                return unparse_url(array(
                        'scheme'=>$base['scheme'],
                        'path'=>$url,
                ));
        } else if ($url{0} == "/") {
                // Step 5
                $base['path'] = $url;
        } else {
                // Step 6
                $path = explode('/', $base['path']);
                $url_path = explode('/', $url);
                // Step 6a: drop file from base
                array_pop($path);
                // Step 6b, 6c, 6e: append url while removing "." and ".." from
                // the directory portion
                $end = array_pop($url_path);
                foreach ($url_path as $segment) {
                        if ($segment == '.') {
                                // skip
                        } else if ($segment == '..' && $path && $path[sizeof($path)-1] != '..') {
                                array_pop($path);
                        } else {
                                $path[] = $segment;
                        }
                }
                // Step 6d, 6f: remove "." and ".." from file portion
                if ($end == '.') {
                        $path[] = '';
                } else if ($end == '..' && $path && $path[sizeof($path)-1] != '..') {
                        $path[sizeof($path)-1] = '';
                } else {
                        $path[] = $end;
                }
                // Step 6h
                $base['path'] = join('/', $path);

        }
        // Step 7
        return unparse_url($base);
}

?>

Answer 3

php有pathinfo()，realpath()和parseurl()以及其他文件系统和网址路径功能。与$_SERVER超全球的信息一起使用（如andre所述），你应该能够做你需要的。

Answer 4

$uri = "http://www.site.com/aa/folder/";

$url = explode("/", $uri);
$url = $url[2];

echo $url; //www.site.com

这是你在找什么？

Answer 5

如果您安装PECL pecl_http，则可以使用http_build_url：

http_build_url("http://www.site.com/aa/folder/page1.php",
                 array("path" => "folder2/page.php"));

并将您的任何相对URI（L）传递为path。该功能将确保构建正确的。

将URI转换为URL

5 个答案: