通过yii1中记录的id打开一个特定的网格页面

Question

I work with yii1.

I have a multi-page list of records, which is displayed using CGridView, sorted in a certain way.

I have id of the record. And I want to open the page on which this record with this exact id is displayed.

Roughly speaking, link /list?too=123 opens the 12th page of the grid, and among all records on this page there is an entry with id 123.

What's the easiest way to do this?

Answer 1

My way of doing it-

• Write code to get page number from the item id. In this case controller action will calculate which page needs to be loaded, using the existing/seperate search.
• Pass this page number in the view file, where the grid is getting loaded.
• If this special page number is passed, then just load that page when passing $model->search() to the CGridView.

Hope it helps.