display the values of jquery ajax success result
in jquery ajax, i get the values from the database and need to display in drop down. first i pass the id and fetch the level. with tat level id and name, again i fetch the values related to the selected level and display in the dropdown.
function Level(Id) {
$.ajax({
url: 'GetLevel' + '/?Id=' + Id,
type: "POST",
dataType: "json",
contentType: 'application/json',
success: function (result) {
testing(value.Value);
},
complete: function () { },
error: ServiceFailed// When Service call fails
});
}
function testing(LevelId) {
result = getDropdownValues();
$('#drpe >option').remove();
for (var i = result.length; i--; ) {
$.each(result, function (key, value) {
$("#drp").append($("<option><开发者_如何学编程;/option>").val(value.Key).html(value.Value));
// it does not display the values in drop down
//it is empty
});
}
function getDropdownValues (LevelId, selectedLevel) {
var passVal = null;
$.ajax({
url: 'GetValues' + '/?selectedLevel=' + selectedLevel + '&LevelId=' + LevelId,
type: "POST",
dataType: "json",
contentType: 'application/json',
async: false,
success: function (result) {
passVal = result;
},
complete: function () { },
error: ServiceFailed// When Service call fails
});
return passVal;
}
and i am using this c# class
public class Level
{
public int Id { get; set; }
public string Name { get; set; }
public List<ListEntity> Value { get; set; }
}
If I undertand correctly your question you are asking how to populate the drop down right? If so and assuming your website return a json result and you create a list like that
{ Text = "Valuename", Value = ValueId }
you can do this:
$('#MyDropDown').empty();
$.each(result, function (index, optionData) {
$('#MyDropDown').append("<option value='" + optionData.Value + "'>" + optionData.Text + "</option>");
});
精彩评论