Why can't I skip my json object using jQuery?

Question

I am using jquery to loop through a json object ... But some of them don't seem to work ...

Here is my code

$.ajax({
            type: "POST",
            url: "Default.aspx/GetRecords",
            //      data: "{}",
            data: "{'currentPage':1,'pagesize':5}",
            contentType: "application/json; charset=utf-8",
            dataType: "json",
            success: function(jsonObj) {
            alert(jsonObj);
             for (var i = jsonObj.length - 1; i >= 0; i--) {
            var employee = jsonObj[i];
            alert(employee.Emp_Name);

When I warned that mine jsonObjreceived [object Object], but when I warned jsonObj.length, he showed undefinedany offer ....

EDIT:

using the answer below, I cannot iterate over divs,

$.each(jsonObj, function(i, employee) {
 $('<div class="resultsdiv"><br /><span class="resultName">' + employee[i].Emp_Name + '</span><span class="resultfields" style="padding-left:100px;">Category&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee[i].Desig_Name + '</span><br /><br /><span id="SalaryBasis" class="resultfields">Salary Basis&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee[i].SalaryBasis + '</span><span class="resultfields" style="padding-left:25px;">Salary&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee[i].FixedSalary + '</span><span style="font-size:110%;font-weight:bolder;padding-left:25px;">Address&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee[i].Address + '</span></div>').insertAfter('#ResultsDiv');
 });

My json object will be like this

{"Table" : [{"Row" : "1","Emp_Id" : "3","Emp_Name" : "Jerome","Address" : "Madurai","Desig_Name" : "Supervisior","SalaryBasis" : "Monthly","FixedSalary" : "25000.00"},{"Row" : "2","Emp_Id" : "4","Emp_Name" : "Mohan","Address" : "Madurai","Desig_Name" : "Acc ","SalaryBasis" : "Monthly","FixedSalary" : "200.00"},{"Row" : "3","Emp_Id" : "5","Emp_Name" : "Murugan","Address" : "Madurai","Desig_Name" : "Mason","SalaryBasis" : "Weekly","FixedSalary" : "150.00"},{"Row" : "4","Emp_Id" : "6","Emp_Name" : "Ram","Address" : "Madurai","Desig_Name" : "Mason","SalaryBasis" : "Weekly","FixedSalary" : "120.00"},{"Row" : "5","Emp_Id" : "7","Emp_Name" : "Raja","Address" : "Madurai","Desig_Name" : "Mason","SalaryBasis" : "Weekly","FixedSalary" : "135.00"}]}

My inspection through firebug on the tab jsonI got this

{"Table" : [{"Row" : "1...edSalary" : "135.00"}]}...

Response I got

{"d":"{\"Table\" : [{\"Row\" : \"1\",\"Emp_Id\" : \"3\",\"Emp_Name\" : \"Jerome\",\"Address\" : \"Madurai\",\"Desig_Name\" : \"Supervisior\",\"SalaryBasis\" : \"Monthly\",\"FixedSalary\" : \"25000.00\"},{\"Row\" : \"2\",\"Emp_Id\" : \"4\",\"Emp_Name\" : \"Mohan\",\"Address\" : \"Madurai\",\"Desig_Name\" : \"Acc \",\"SalaryBasis\" : \"Monthly\",\"FixedSalary\" : \"200.00\"},{\"Row\" : \"3\",\"Emp_Id\" : \"5\",\"Emp_Name\" : \"Murugan\",\"Address\" : \"Madurai\",\"Desig_Name\" : \"Mason\",\"SalaryBasis\" : \"Weekly\",\"FixedSalary\" : \"150.00\"},{\"Row\" : \"4\",\"Emp_Id\" : \"6\",\"Emp_Name\" : \"Ram\",\"Address\" : \"Madurai\",\"Desig_Name\" : \"Mason\",\"SalaryBasis\" : \"Weekly\",\"FixedSalary\" : \"120.00\"},{\"Row\" : \"5\",\"Emp_Id\" : \"7\",\"Emp_Name\" : \"Raja\",\"Address\" : \"Madurai\",\"Desig_Name\" : \"Mason\",\"SalaryBasis\" : \"Weekly\",\"FixedSalary\" : \"135.00\"}]}"}

Any suggestion...

Answer 1

Use jQuery method each. Documents .

success: function(jsonObj) {

         $.each(jsonObj, function(i, employee) {

            alert(employee.Emp_Name);

         }
}

, AFAIK, alert() JSON.

( ), , length.

, - , , ? ?

Update

JSON, ..

{
   Table: [
      {
         Row: '1',
         Emp_Id: '3',
         Emp_Name: 'Jerome',
         Address: 'Madurai',
         Desig_Name: 'Supervisior',
         SalaryBasis: 'Monthly',
         FixedSalary: '25000.00'
      },
      {
         Row: '2',
         Emp_Id: '4',
         Emp_Name: 'Mohan',
         Address: 'Madurai',
         Desig_Name: 'Acc ',
         SalaryBasis: 'Monthly',
         FixedSalary: '200.00'
      },
      {
         Row: '3',
         Emp_Id: '5',
         Emp_Name: 'Murugan',
         Address: 'Madurai',
         Desig_Name: 'Mason',
         SalaryBasis: 'Weekly',
         FixedSalary: '150.00'
      },
      {
         Row: '4',
         Emp_Id: '6',
         Emp_Name: 'Ram',
         Address: 'Madurai',
         Desig_Name: 'Mason',
         SalaryBasis: 'Weekly',
         FixedSalary: '120.00'
      },
      {
         Row: '5',
         Emp_Id: '7',
         Emp_Name: 'Raja',
         Address: 'Madurai',
         Desig_Name: 'Mason',
         SalaryBasis: 'Weekly',
         FixedSalary: '135.00'
      }
   ]
}

, , ,

 $.each(jsonObj.table, function(i, employee) {

                alert(employee.Emp_Name);

             }

Answer 2

.

Json, , , for :

for(aProperty in jsonObj)
{
  var employee = jsonObj[aProperty];
  alert(aProperty + " = " + employee);
}

jQuery.each:

jQuery.each(jsonObj, function(key,value))
{
   alert(key + " = " + value);
}

window.alert . Firebug.

, html $('# ResultsDiv');

:

$.each(jsonObj.Table, function(i, employee) {
 $('<div class="resultsdiv"><br /><span class="resultName">' + employee[i].Emp_Name + '</span><span class="resultfields" style="padding-left:100px;">Category&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee[i].Desig_Name + '</span><br /><br /><span id="SalaryBasis" class="resultfields">Salary Basis&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee[i].SalaryBasis + '</span><span class="resultfields" style="padding-left:25px;">Salary&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee[i].FixedSalary + '</span><span style="font-size:110%;font-weight:bolder;padding-left:25px;">Address&nbsp;:</span>&nbsp;<span class="resultfieldvalues">' + employee[i].Address + '</span></div>').insertAfter('#ResultsDiv');
 });

, , .

, $('# ResultsDiv').

ajax

$.ajax({
            type: "POST",
            url: "Default.aspx/GetRecords",
            data: "{'currentPage':1,'pagesize':5}",
            contentType: "application/json; charset=utf-8",
            dataType: "json",
            success: function(jsonObj) {
            var Employees = jsonObj.Table || jsonObj.d.Table;
            for(key in Employees) console.log(Employees);
            }
        })

Answer 3

Objects have no length. You can iterate over the properties of an object as follows:

success: function(jsonObj) {
         alert(jsonObj);
         for (i in jsonObj) {
             var employee = jsonObj[i];
             alert(employee.Emp_Name);