Hello Pawel,
cool that you used ZendX_JQuery as basis :-)
The problem is described very simple. Zend_Json cannot handle javascript
expressions. Its escaping everything like it were a string. Sadly the
Zend_Json_Expr proposal didn't make it into the 1.7 release.
To fix this you have to hack your Zend_Json output with preg_replace somehow to
strip the " around the pager: value.
greetings,
Benjamin
On Friday 14 November 2008 16:35:46 Paweł Chuchmała wrote:
> Hi.
>
> I want to use jqGrid with ZF, so I wrote simple helper based on
> ZendX_JQuery helpers.
> I have a problem with some params.
>
> As result I must have:
> jQuery("#list2").jqGrid({
> url: '/artist/list/format/json/',
> datatype: 'json',
> viewrecords: true,
> * pager: jQuery('#pager2'),*
> rowNum:20,
> rowList:[10,20,30],
> imgpath: "/css/themes/sand/images",
> sortname: 'id',
> colNames: ['Id', 'Nazwa'],
> colModel: [
> {name: 'artist_id', index:'artist_id', width:55},
> {name: 'artist_name', index:'artist_name', width:300}
>
> ],
> });
>
>
> so I make array with this options.
> $params = array(
> ...
> 'pager' => "jquery('#pager2')",
> ...
> );
> After Zend_Json::encode($myarray) i have:
> ...
> "pager": "jquery('#pager2')"
> ...
>
> from jqGridDocumentation: pager defines that we want to use a pager bar to
> navigate through the records. This must be a valid html element;
> In my example it's a string.
>
> What is the best solution for this?
>
> regards,
--
Benjamin Eberlei
http://www.beberlei.de
