Reputation: 73
I have a MVC4 single page website with a form. The loading of the contents is achieve with ajax. I do not know how to get the data out from JSON in C#? Here is my code:
JavaScript:
$("#subnt").click(function (event) {
event.preventDefault();
var url = "/Home/Submit";
$.post(url, $('form[name="cnt_us-frm"]').serialize(), function (data) {
if (data.Success === true) {
$("#min-content").hide().load("/Home/PartialSubmit").fadeIn('normal'); // loads the page into 'min-content' section
}
else {
// display error message
}
})
});
});
C#:
[HttpPost]
public JsonResult Submit()
{
return Json(new { Success = true, SomeOtherData = "testing" });
}
Upvotes: 0
Views: 288
Reputation: 73
No, the other way around. How to retrieve the data from the form (json).
Upvotes: 0
Reputation: 17182
Please check below working code -
I have used exactly your working code -
[HttpPost]
public JsonResult Submit()
{
return Json(new { Success = true, SomeOtherData = "testing" });
}
Then I used following JQuery to hit the above action -
<script src="~/Scripts/jquery-1.10.2.min.js"></script>
<script>
$(function () {
$('#click').click(function (e) {
$.ajax({
url: "@Url.Action("Submit")",
type: "POST",
contentType: "application/json; charset=utf-8",
dataType: "json",
error: function (response) {
alert(response);
},
success: function (data) {
if (data.Success == true)
alert(data.SomeOtherData);
}
});
});
});
</script>
<input type="submit" value="click" id="click" />
And as the output I was able to get an alert as shown below -
Upvotes: 1
Reputation:
Your code is ok bu you can add debugger.and open developer tools check your data
.
$.post(url, $('form[name="cnt_us-frm"]').serialize(), function (data) {
debugger;
if (data.Success === true) {
$("#min-content").hide().load("/Home/PartialSubmit").fadeIn('normal'); // loads the page into 'min-content' section
}
else {
// display error message
}
Upvotes: 0
Reputation: 6738
Easiest thing to do is use the superior json.net
[HttpPost]
public string Submit()
{
var result = new { success = true, someOtherDate = "testing"};
var json = JsonConvert.SerializeObject(result);
return json;
}
Upvotes: 0