以 [ 作为参数的 Yii 链接 [英] Yii link with [ as a parameter

问题描述

我使用 GII 创建了一个记录列表.我使用管理视图，因此它们位于表视图中.在表格的顶部是带有记录状态的搜索.当状态下拉列表更改时，我提交表单并搜索表格.我希望管理员的默认视图只显示活动记录，所以我想在菜单中创建一个链接:中/管理员/?中[状态]=活动实际链接当然是medium/admin/?Medium%5Bstatus%5D=active

With GII I have created a list of records. I use the admin view so they are in a table view. On top of the table it is the search with a status for the records. When the status dropdown is changed I submit the form and the table gets searched. I want the default view of the admin to show only the active records so I want to create a link in the menu to this: medium/admin/?Medium[status]=active The actual link of course is medium/admin/?Medium%5Bstatus%5D=active

我尝试这样做:

CHtml::link('Mediums', array("medium/admin", array('Medium[status]' => 'active')))
CHtml::link('Mediums', array("medium/admin", array('Medium%5Bstatus%5D' => 'active'))) 
CHtml::link('Mediums', array("medium/admin", array('Medium' => array('status' => 'active')))) 

但是所有的链接都不正确，所以表格的默认视图是显示所有记录.

But all of the links are incorrect so the default view of the table is with all the records shown.

创建此类链接的正确方法是什么?

What is the correct way to create such a link?

谢谢.

推荐答案

http://www.yiiframework.com/doc/api/1.1/CHtml#link-detail 和 http://www.yiiframework.com/wiki/48/ 对你很有用.

CHtml::link(CHtml::encode('Mediums'),array("medium/admin", "status"=>"active"));

然后确保在你的控制器中有这样的东西:

Then ensure that in your controller you have something like this:

public function actionAdmin($status)

现在您可以在操作中使用 'status'.

Now you ca use 'status' in your action.

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