Show message box in case of exception

问题:

I'm wondering what the correct way is to pass on an exception from one method to my form.

public void test()
{
    try
    {
        int num = int.Parse("gagw");
    }
    catch (Exception)
    {
        throw;
    }
}

Form:

try
{
    test();
}
catch (Exception ex)
{
    MessageBox.Show(ex.Message);
}

in this way i cannot see my text box.

回答1:

If you want just the summary of the exception use:

    try
    {
        test();
    }
    catch (Exception ex)
    {
        MessageBox.Show(ex.Message);
    }

If you want to see the whole stack trace (usually better for debugging) use:

    try
    {
        test();
    }
    catch (Exception ex)
    {
        MessageBox.Show(ex.ToString());
    }

---

Another method I sometime use is:

    private DoSomthing(int arg1, int arg2, out string errorMessage)
    {
         int result ;
        errorMessage = String.Empty;
        try 
        {           
            //do stuff
            int result = 42;
        }
        catch (Exception ex)
        {

            errorMessage = ex.Message;//OR ex.ToString(); OR Free text OR an custom object
            result = -1;
        }
        return result;
    }

And In your form you will have something like:

    string ErrorMessage;
    int result = DoSomthing(1, 2, out ErrorMessage);
    if (!String.IsNullOrEmpty(ErrorMessage))
    {
        MessageBox.Show(ErrorMessage);
    }

回答2:

There are many ways, for example:

Method one:

public string test()
{
string ErrMsg = string.Empty;
 try
    {
        int num = int.Parse("gagw");
    }
    catch (Exception ex)
    {
        ErrMsg = ex.Message;
    }
return ErrMsg
}

Method two:

public void test(ref string ErrMsg )
{

    ErrMsg = string.Empty;
     try
        {
            int num = int.Parse("gagw");
        }
        catch (Exception ex)
        {
            ErrMsg = ex.Message;
        }
}

回答3:

        try
        {
           // your code
        }
        catch (Exception w)
        {
            MessageDialog msgDialog = new MessageDialog(w.ToString());
        }