我使用jqGrid以表格格式顯示數據,使用JSP和servlet.

編輯

我想在執行插入,更新,刪除等操作時顯示來自服務器的錯誤. (數據類型：“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代碼

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

>從jqgrid插入記錄后,然后顯示沒有消息

網格如果成功插入記錄

>錯誤狀態：’未經授權’.錯誤代碼：如果servlet無法在數據庫中插入記錄,則顯示401.

我的問題是：

>從jqgrid插入記錄后,如果插入了記錄,那么我應該如何顯示消息,向用戶提供插入數據的信息.

>以及如何在插入時向用戶發出錯誤消息(我應該使用哪個錯誤代碼？)

提前致謝…..

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