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19:58

05/09/2011

Hello, what I need to show errors from mysql but in xml, because I am loading data with this format, like this:
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
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