我使用jqGrid以表格格式显示数据,使用JSP和servlet.
编辑
我想在执行插入,更新,删除等操作时显示来自服务器的错误. (数据类型：“xml”)
jqGrid的
jQuery("#list10_d").jqGrid({
height:250,
width:600,
url:'Assignment?action=Assign',
datatype: "xml",
colNames:['Sr. No.','PID', 'DATE', 'EMPID'],
colModel:[{name:'srNo',index:'srNo', width:30,sortable:false},
{name:'PID',index:'PID',width:0, sortable:true,editable:false},
{name:'DATE',index:'DATE', width:75,sortable:true,editable:true,editoptions: { dataInit: function(el) { setTimeout(function() { $(el).datepicker({dateFormat:"dd-M-yy",showButtonPanel: true,changeYear: true,changeMonth: true}).attr('readonly','readonly'); }, 200); }}},
{name:'EMPID',index:'EMPID', width:150,sortable:true,editable:true}
],
rowNum:10,
rowList:[10,20,50,100],
pager: '#pager10_d',
sortname: 'PID',
viewrecords: true,
sortorder: "asc",
},
multiselect: true,
editurl: "Assignment?action=Edit",
caption:"Assignment"
} ).navGrid('#pager10_d',{edit:false,add:true,del:false,addtext:'Assign '},
{},
{modal:true,jqModal: false,closeOnEscape:true,savekey: [true,13],closeOnEscape:true, recreateForm: true,width:500,mtype:'POST', url: 'Assignment',editData:{action: 'Assign',PID: function () {return PID;}},
afterSubmit: function (response) {
alert('After Submit \n' +'statusText: '+ response.statusText);
var myInfo = '
'
'style="float: left; margin-right: .3em;">
' +response.statusText + 'Inserted'+
'
$infoTr = $("#TblGrid_" + $.jgrid.jqID(this.id) + ">tbody>tr.tinfo"),
$infoTd = $infoTr.children("td.topinfo");
$infoTd.html(myInfo);
$infoTr.show();
// display status message to 3 sec only
setTimeout(function () {
$infoTr.slideUp("slow");
}, 5000);
return [true, "", ""]; // response should be interpreted as successful
},
errorTextFormat: function (response) {
alert('Error Text Format: \n' +'statusText: '+ response.statusText);
return '
'style="float:left; margin-right:.3em;">
' +response.statusText;},
{closeOnEscape:true, recreateForm: true,mtype: 'POST',url: 'Assignment',delData: {action: 'Delete',PID: function () {return PID;}}},
{}) ;
Servlet代码
if(request.getParameter("action").equalsIgnoreCase("Assign"))
{
PID = request.getParameter("PID");
String DATE= request.getParameter("DATE");
String EMPID= request.getParameter("EMPID");
String query = "insert into ASSIGN(PID,DATE,EMPID) values('"+ PID +"','"+ DATE +"','"+ EMPID"')";
boolean b = insert.InsertData(query);
if(b)
{
System.out.println("New record added successfully! : "+query);
response.setContentType("text/xml");
response.setCharacterEncoding("UTF-8");
//response.sendError(200, "success");
response.setStatus(200, "Inserted successfully");
}
else
{
System.out.println("Failed to add Record! : "+query);
response.setContentType("text/xml");
response.setCharacterEncoding("UTF-8");
//response.sendError(399, "not Inserted successfully");
response.setStatus(404, "Error while inserting");
}
}//INSERT
以上例子
>从jqgrid插入记录后,然后显示没有消息
网格如果成功插入记录
>错误状态：’未经授权’.错误代码：如果servlet无法在数据库中插入记录,则显示401.
我的问题是：
>从jqgrid插入记录后,如果插入了记录,那么我应该如何显示消息,向用户提供插入数据的信息.
>以及如何在插入时向用户发出错误消息(我应该使用哪个错误代码？)
提前致谢…..
