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20:07

23/06/2010

I'm building a grid based of data submitted from a form. The grid gets created as expected and everything looks right. However, when clicking the next page button or manually typing in a page number the data is submitted but it's missing the page, rows, sidx and sord parameters.
I am setting postData to what's in my form and that appears to be overwriting the default values sent by jqGrid. How can include the default values as well as my form data?
Here is how I am initializing my grid:
$("#report_table").jqGrid({
url:'ajax-get_report_data.php',
datatype: 'json',
mtype: 'POST',
colNames: data.header,
colModel: data.col_model,
pager: '#pager',
toppager: true,
rowNum:100,
rowList:[25,50,100,200,500],
sortname: data.default_sort,
sortorder: 'asc',
viewrecords: true,
height: 'auto',
width: data.width,
shrinkToFit: false,
loadui: 'block',
postData: $("#report_form").serialize()
});
16:30

Moderators
30/10/2007

Hello,
You can change the parameter names using the prmNames parameter.
Read more here:
http://www.trirand.com/jqgridw.....ki:options
Reagrds
Tony
For professional UI suites for Java Script and PHP visit us at our commercial products site - guriddo.net - by the very same guys that created jqGrid.
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