Consider the following code that makes an ajax call:
jQuery("#grid").jqGrid({ url:'GetDetails.asmx?team=All', datatype: "json", colNames:['ID','Team', 'Name', 'Designation'], colModel:[ {name:'id',index:'id', width:55}, {name:'Icon',index:'Icon', width:90}, {name:'name',index:'name asc, invdate', width:100}, {name:'designation',index:'designation', width:80}], rowNum:10, rowList:[10,20,30], pager: '#pager', sortname: 'id', viewrecords: true, sortorder: "desc", caption:"JSON Data" });
The return data from the GetDetails.asmx should be in json format and the data should look like :-
{ "page":"1", "total":1, "records":"5", "rows":[ {"id":"1","cell":["1","","Scott","Manager"]}, {"id":"2","cell":["2","","Charles","Accountant"]}, {"id":"3","cell":["3","","Dan","Sales"]}, {"id":"4","cell":["4","","Chris","Administrator"]}, {"id":"5","cell":["5","","Paul","Developer"]} ]}
The data cell value should contain an image location so that JQGrid will render and display an image in the grid.
“Tell me and I forget. Teach me and I remember. Involve me and I learn.”
– Benjamin Franklin
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