I want to post a form via ajax call also model will be passed into the action method, but want to get Model errors via json. How can I do this?
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2 Answers
You could write a custom action filter:
public class HandleJsonErrors : ActionFilterAttribute
{
public override void OnActionExecuting(ActionExecutingContext filterContext)
{
var modelState = (filterContext.Controller as Controller).ModelState;
if (!modelState.IsValid)
{
// if the model is not valid prepare some JSON response
// including the modelstate errors and cancel the execution
// of the action.
// TODO: This JSON could be further flattened/simplified
var errors = modelState
.Where(x => x.Value.Errors.Count > 0)
.Select(x => new
{
x.Key,
x.Value.Errors
});
filterContext.Result = new JsonResult
{
Data = new { isvalid = false, errors = errors }
};
}
}
}
and then put it into action.
Model:
public class MyViewModel
{
[StringLength(10, MinimumLength = 5)]
public string Foo { get; set; }
}
Controller:
public class HomeController : Controller
{
public ActionResult Index()
{
return View();
}
[HttpPost]
[HandleJsonErrors]
public ActionResult Index(MyViewModel model)
{
// if you get this far the model was valid => process it
// and return some result
return Json(new { isvalid = true, success = "ok" });
}
}
and finally the request:
$.ajax({
url: '@Url.Action("Index")',
type: 'POST',
contentType: 'application/json; charset=utf-8',
data: JSON.stringify({
foo: 'abc'
}),
success: function (result) {
if (!result.isvalid) {
alert(result.errors[0].Errors[0].ErrorMessage);
} else {
alert(result.success);
}
}
});
2 Comments
Leniel Maccaferri
Hey Darin... great answer. I'm ready to put it into action here. Would you still recommend this approach today with ASP.NET MVC 4 and 5 out the door?
Leniel Maccaferri
I think this other answer provided by you is a better approach: stackoverflow.com/a/7287646/114029
Something like this?
$.ajax({
type: "POST",
url: $('#dialogform form').attr("action"),
data: $('#dialogform form').serialize(),
success: function (data) {
if(data.Success){
log.info("Successfully saved");
window.close();
}
else {
log.error("Save failed");
alert(data.ErrorMessage);
},
error: function(data){
alert("Error");
}
});
[HttpPost]
public JsonResult SaveServiceReport(Model model)
{
try
{
//
}
catch(Exception ex)
{
return Json(new AdminResponse { Success = false, ErrorMessage = "Failed" }, JsonRequestBehavior.AllowGet);
}
