Displaying JQGrid with default search condition

Being using JQGrid for sometime, have at one point required to display data in jqgrid but with a default search condition.

Its as simple as just passing data using the parameter serializeGridData in jqgrid

$(function () {

 var defaultsearchField = "id";
 var defaultsearchString = "5";
 var defaultsearchOper = "gt";



 jQuery("#grid").jqGrid({    
  url:'GetDetails.asmx?team=All',    
  datatype: "json",    
  colNames:['ID','Name','Country','Progress'],    
  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",
  serializeGridData: function (postData) {

  if(defaultsearchField != null)
  {
   postData.searchField = defaultsearchField; 
   postData.searchString = defaultsearchString; 
   postData.searchOper = defaultsearchOper;
  }
  else
  { 
   if (postData.searchField === undefined) postData.searchField = null;
   if (postData.searchString === undefined) postData.searchString = null;
   if (postData.searchOper === undefined) postData.searchOper = null; 
  }
  return JSON.stringify(postData);
  },
 });   

});

Now the webservice will get the default search condition and should be able to work straight away. Unless if search is not handled on web service side.

Three things are certain:
Death, taxes, and lost data.
Guess which has occurred.
~David Dixon, 1998, winning entry of the Haiku Error Messages 21st Challenge by Charlie Varon and Jim Rosenau, sponsored by Salon.com