Question on JqGrid paging.

Question

Hi,

I would like not to lose the paging and the sort parameters when I leave a page with a grid and I come back to this page.

Is it possible to get the page, rowNum, sortname, sortorder params (I think with getGridParam), put them in the URL scope, go to another page, and get those params back through the URL scope and give them in the URL hereafter in the code ? (putting Act_country.cfc?method=getAllCountries&page=url.pageNum does not give anything...)

jQuery(document).ready(function(){  
    jQuery('#list').jqGrid({  
    url:'Act_country.cfc?method=getAllCountries',
    datatype: 'json',
    mtype:'GET',
    colNames:[
    '<cfoutput>#StLabels["LBL_TAB_EDIT"]#</cfoutput>',
    '<cfoutput>#StLabels["LBL_TAB_COUNTRY_CODE"]#</cfoutput>',
        ...

Thank you in advance, Michel

Answer 1

You need a Session scope.

Check the existence of params with StructKeyExists(Session, "param"), if they are not present yet -- consider this first visit and put the variables, otherwise use values from session. If user changes the paging/sorting -- refresh the values in the session.

Answer 2

You can also save the information which you need in cookie. See http://www.intothecloud.nl/index.php/2010/04/saving-jqgrid-parameters-in-cookie/ or http://stackoverflow.com/questions/3015203/remember-persist-the-filter-sort-order-and-current-page-of-jqgrid