jQuery("#list1").jqGrid({

url:'server.php?q=1', datatype: "xml",

colNames:['Inv No','Date', 'Client', 'Amount','Tax','Total','Notes'],

              colModel:[ {name:'id',index:'id', width:75},

                   {name:'invdate',index:'invdate', width:90},

                  {name:'name',index:'name', width:100},

                  {name:'amount',index:'amount', width:80, align:"right"},

                  {name:'tax',ndex:'tax', width:80, align:"right"},

                 {name:'total',index:'total', width:80,align:"right"},

                  {name:'note',index:'note', width:150, sortable:false} ],

rowNum:10, autowidth:

true, rowList:[10,20,30],pager: jQuery('#pager1'), sortname: 'id', viewrecords: true, sortorder: "desc", caption:"XML Example" }).

I have done this with JSON like this:

var json;

var editMessage = function(response,postdata){

 json   = response.responseText; // response text is returned from server.

var result = JSON.parse(json); // convert json object into javascript object

return [result.status,result.message,null];

}

jQuery.jgrid.edit = {

afterSubmit:editMessagemsg:{required:"El campo es obligatorio"},processData: "Processing...",reloadAfterSubmit:true,closeAfterEdit:true,bCancel: "Cerrar",bSubmit: "Guardar",editCaption: "Editar",

};

in the PHP file this to show the errors from mysql

$message="duplicated row";

$output['status']  = false;

$output['message'] = $message;

echo json_encode($output);

I need this but en XML

If someone needs the source code this is my e-mail: alxx28@hotmail.com