Exception management in ASP.NET MVC using jquery

Question

I have 2 question, in the first one, I get a list, I'd like if excetpion in C# code (controller) has the possibility to a View (an error view) and display it a specific div. How can I get the view in the .error. ? HTML

<div><a href="#" class="MnuCustomerList">List</a></div>

jQuery

$(".MnuCustomerList").click(function () {
    var jqxhr = $.post("/Customer/List", function (data) {
        $('#rightcolumn').html(data);
    })
    .success(function () { alert("success"); })
    .error(function () { alert("error"); })
    .complete(function () { alert("complete"); });
})

;

Controller:

public PartialViewResult List()
{
    throw new Exception("myexception");
    return PartialView("List");
}

Second question : I have a form

@model MyModel
@using (Html.BeginForm("Save", "Customer", FormMethod.Post))
{
    <table style="width:100%;">
    <tr>
        <td>Code</td>
        <td>@Html.TextBoxFor(m => m.Customer.Code, new { id = "tbCode" })</td>
    </tr>
    <tr>
        <td>LastName</td>
        <td>@Html.TextBoxFor(m => m.Customer.LastName, new { id = "tb", maxlength = 50, style = "width:40%;" })</td>
    </tr>
    <tr>
        <td colspan="2"><input type="submit" value="A submit button"/></td>
    </tr>
    </table>
}

In the Controller, I check if the code already exist, if yes CustomerException("Code exist").

My questions, is it possible to post this form using jQuery BUT still using this format that's mean with the model and NOT get one value by one value like the sample below

$.ajax({
    type: "POST",
    url: "/Customer/Save",
    data: {
        id: $('#Id').val(),
        firstName: $('#FirstName').val(),
        lastName: $('#LastName').val(),
        isEnable: $('#IsEnable').attr('checked')        
    },
    success: function (html) {
        $("#ContentDataSection").html(html);
    },
    error: function (XMLHttpRequest, textStatus, errorThrown) { }
});

Thanks,

Craig M · Accepted Answer · 2011-04-07 20:58:07Z

6

The error callback is only going to fire if there is an error in making the ajax request. If an error occurs server side, you'll need to pass data back (in Json format would be a good choice) to the client indicating there was a failure server side and handle it in the success callback.

Edit to add code showing how to set response code to 500 so it can be handled in the error call back per Ryan's comment:

Controller action:

public ActionResult FiveHundred()
{
    Response.StatusCode = 500;
    return Json(new {Error = "Uh oh!"});
}

Javascript:

$.ajax({
    url: "/home/fivehundred",
    type: "POST",
    success: function(data) {
        // handle normally
    },
    error: function(data) {
        alert("error " + data.Error);
    }
});

edited Apr 7, 2011 at 20:58

answered Apr 7, 2011 at 19:59

Craig M

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7 Comments

TheBoubou Over a year ago

When I post my form, I check if code exist, ik exist I'd like : 1. show message a client side 2. Stop the from posting for this second point is it possible ?

Ryan Over a year ago

You could always set the status code to 500 (or any other error code) to trigger the error callback to handle the json response.

TheBoubou Over a year ago

@Ryan do you have a sample how implement this ?

Craig M Over a year ago

Set Response.Status = 500; public ActionResult FiveHundred() { Response.StatusCode = 500; return Json(new {Error = "Uh oh!"}); }

TheBoubou Over a year ago

@Craig but with this the form is gone, I'd like keep my form to repost and the form is not posted by jQuery

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