C＃Webmethod + jQuery发布错误消息

Question

I am trying to throw a handled exception through JQuery Post and a Webmethod so that I can get an error message back to the alert function. So this is just a test script.

I have this set up, and am catching the error, now I need to return the server error back to the alert function!

[WebMethod]
[ScriptMethod(ResponseFormat = ResponseFormat.Json)]
public static string Test()
{
    try
    {
        string value = null;
        if (value.Length == 0) // <-- Causes exception
        {
            Console.WriteLine(value); // <-- Never reached
        }

    }

    catch (Exception E)
    {            

        StreamWriter sw = new StreamWriter(@"C:\inetpub\wwwroot\jQuery_AJAX_Database\error.txt", true);
        sw.WriteLine(E.Message);

        sw.Close();
        sw.Dispose();

        throw new Exception("error");

    }

    return "bla";

And this my Post script:

 $("#btnTest").click(function () {
            $.ajax({
                type: 'POST',
                url: "my_test2.aspx/Test",
                contentType: "application/json; charset=utf-8",
                success: function (data) {
                    alert('Success!')
                },
                error: function (xhr, status, error) {
                    var exception = JSON.parse(xhr.responseText);
                    alert(exception.Message);
                    // Here I need to return server error
                }
            });
        });

Thanks P

Answer 1

You need an http status code to do that.

Modify your code into this

catch (Exception E)
    {            

        StreamWriter sw = new StreamWriter(@"C:\inetpub\wwwroot\jQuery_AJAX_Database\error.txt", true);
        sw.WriteLine(E.Message);

        sw.Close();
        sw.Dispose();

        return new HttpStatusCodeResult(500, E.Message);
    }

500 stands for internal server error . have a look for another if it fits better.

I think 400 - Bad Request might be more suited for your situation.

Answer 2

Thanks @ted you got me on the right track. Solved as below!

throw new HttpException(E.Message);