How to use the pagination with view parser? |
Hello there!
I am building a view with pagination with view parser, the problem is how I spend the value of pager to build my links? since it is an object and in view parser it is not possible to use them Code: $data = [
At the bottom of the UG page (https://codeigniter4.github.io/userguide...ation.html), a pagination objet will have `links()` that you can inject into your view at the proper place.
I thought there was something inside the Model class as well, but it isn't jumping out at me.
Not having idea how to do it... If you find the solution please tell me too.
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07-02-2021, 02:59 AM
(This post was last modified: 07-02-2021, 06:24 AM by eternity6. Edit Reason: beautiful formatting )
Well...You can simple put HTML result of pager inside an array with a specific label in your control and then parser it with a { code } in the template.
For example Controller: PHP Code: $this->data['page_pagination'] = $data['pager']->links(); In your template view use code with ! for don't print HTML code Code: <div class="pagination">{! page_pagination !}</div> |
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