Question

我使用这些类型的网址：

http://test.com/rest/api.php?action=test

但我需要这些类型的网址：

http://test.com/rest/api/action/test

我尝试.htaccess在这里提交这些代码

RewriteEngine On
RewriteCond %{REQUEST_FILENAME} !-d
RewriteCond %{REQUEST_FILENAME} !-s
RewriteRule ^(.*)$ api.php?rquest=$1 [QSA,NC,L]
RewriteCond %{REQUEST_FILENAME} -d
RewriteRule ^(.*)$ api.php [QSA,NC,L]
RewriteCond %{REQUEST_FILENAME} -s
RewriteRule ^(.*)$ api.php [QSA,NC,L]

Answer 1

通过将您的代码与.htaccess目录下的DOCUMENT_ROOT/rest相关联来放置此代码：

Options +FollowSymLinks -MultiViews
# Turn mod_rewrite on
RewriteEngine On
RewriteBase /rest/

RewriteCond %{REQUEST_FILENAME} !-f
RewriteCond %{REQUEST_FILENAME} !-d
RewriteRule ^api/([^/]+)/?$ api.php?action=$1 [QSA,NC,L]

Answer 2

这应该解决它：

RewriteEngine On
RewriteRule ^rest/api.php?action=test$ /rest/api/action/test [L]

Answer 3

您的目标可以通过使用语法如下的正则表达式来实现：

RewriteRule ^rest/api/action/(\d+)*$ ./rest/api.php?action=$1

尝试阅读这些文章：

http://net.tutsplus.com/tutorials/other/using-htaccess-files-for-pretty-urls/和 http://net.tutsplus.com/tutorials/other/the-ultimate-guide-to-htaccess-files/

如何使用PHP在我的目录中创建友好的URL

3 个答案: