codeigniter如何根据uri段的no设置路由到默认控制器 [英] codeigniter how to set routes to default controller according to the no of uri segments

问题描述

我在项目中使用codeigniter作为模块来显示类别及其项目的信息。

I am using codeigniter as a module in my project to display information as categories and its items.

我想要的URL如下所示

I want the url's like the following

1. sitename.com/module/将以随机顺序显示所有项目。


2. sitename.com/moduel/categoryname/显示该特定类别中的所有项目


3. sitename.com/module/categoryname/title-of-the-item/这将显示该特定项目的详细信息

这里sitename.com/module/是codeigniter的根目录，默认控制器是home。

here sitename.com/module/ is the root of the codeigniter and the default controller is home.

我的问题是如何为一个段或uri和两个uri段设置路线

my question is how to set routes for one segment or uri and two segments of uri like

1. sitename.com/module/应加载默认


2. sitename.com/module/categoryname/应载入首页/类别/类别名称


3. sitename.com/module/ categoryname / item-of-the-item应加载index / title / title-of-the-item

函数如下

1. 函数索引（）...


2. 函数类别（$ categoryname）。 ...


3. function title（$ title）....

我想根据段数创建一个路由，或者如果有任何其他替代来完成这个与路由，那是很好。

so, either i want to create a route according to the num of segments or if there is any other alternate to accomplish this with routing then that is fine.

我想使用

所以，

推荐答案

$route['page/(:any)'] = "home/page/$1";
$route['page'] = "home/index";
$route['(:any)/page/(:any)'] = "home/category/$1/page/$2";
$route['(:any)/page'] = "home/category/$1";
$route['page'] = "home/index";

$route['(:any)/(:any)'] = "home/title/$2";
$route['(:any)'] = "home/category/$1";

最后两行（7和8）是我需要的... （我的目的）

最后两行之前的行（1,2,3,4,5）是如何对最后两行使用分页7和8）

the lines before the last two(1,2,3,4,5) is how i use pagination for the last two lines(7 and 8)

只有首页/类别和首页/有分页，而不是首页/标题，即结束页。

only home/category and home/ has pagination and not home/title which is the end page.

同时，如果任何人发现代码可能是buggy或需要完美，那么请添加一个答案，以便我可以更新。

meanwhile, if anybody found the code could be buggy or need perfection then please add an answer so that i can update.

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