Codeigniter从URL获取参数 [英] Codeigniter get parameters from url

问题描述

我有一个网址" http://localhost/codeigniter/index.php/manual_export/".我需要从假定为id的url中获取最后一段.因此，例如" http://localhost/codeigniter/index.php/manual_export/2 ".我需要获取最后一个段为"2".

I have a url "http://localhost/codeigniter/index.php/manual_export/". I need to get last segment from the url which is suppose to be a id. So for example "http://localhost/codeigniter/index.php/manual_export/2". I need to get the last segment which is "2".

我厌倦了使用以下代码:

I tired to use following code:

$id = end($this->uri->segment_array()); 

当我不在URL中添加"2"并给我"manual_export"时，此方法有效.但是，当我将id传递给url时，出现错误找不到您请求的页面.".我认为这与路由有关.我该如何解决该错误.

This works when I don't add "2" to the url and gives me "manual_export". However when I pass the id to the url I get an error "The page you requested was not found.". I think this is to do with routing. How can I fix this error.

推荐答案

另一种方法是定义路线，然后将其转换为参数

the other way to do it is by defining a route, it will then be converted to a param

例如，如果您的控制器名为manual_export且方法为getrecord

so for example if your controller is called manual_export and the method is getrecord

在文件application/routes.php中

in the file application/routes.php

$route['manual_export/(:any)'] = "manual_export/getrecord/$1";

在您的控制器中manual_export

in your controller manual_export

function getrecord($id){ // etc etc }

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