C# 出现异常时显示消息框
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Show message box in case of exception
提问by Dana Yeger
I'm wondering what the correct way is to pass on an exception from one method to my form.
我想知道将异常从一种方法传递到我的表单的正确方法是什么。
public void test()
{
try
{
int num = int.Parse("gagw");
}
catch (Exception)
{
throw;
}
}
Form:
形式：
try
{
test();
}
catch (Exception ex)
{
MessageBox.Show(ex.Message);
}
in this way i cannot see my text box.
这样我就看不到我的文本框了。
采纳答案by Avi Turner
If you want just the summary of the exception use:
如果您只想要异常的摘要，请使用：
try
{
test();
}
catch (Exception ex)
{
MessageBox.Show(ex.Message);
}
If you want to see the whole stack trace (usually better for debugging) use:
如果您想查看整个堆栈跟踪（通常更适合调试），请使用：
try
{
test();
}
catch (Exception ex)
{
MessageBox.Show(ex.ToString());
}
Another method I sometime use is:
我有时使用的另一种方法是：
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:
在你的表格中，你会有类似的东西：
string ErrorMessage;
int result = DoSomthing(1, 2, out ErrorMessage);
if (!String.IsNullOrEmpty(ErrorMessage))
{
MessageBox.Show(ErrorMessage);
}
回答by User2012384
There are many ways, for example:
有很多方法，例如：
Method one:
方法一：
public string test()
{
string ErrMsg = string.Empty;
try
{
int num = int.Parse("gagw");
}
catch (Exception ex)
{
ErrMsg = ex.Message;
}
return ErrMsg
}
Method two:
方法二：
public void test(ref string ErrMsg )
{
ErrMsg = string.Empty;
try
{
int num = int.Parse("gagw");
}
catch (Exception ex)
{
ErrMsg = ex.Message;
}
}
回答by Hoby
try
{
// your code
}
catch (Exception w)
{
MessageDialog msgDialog = new MessageDialog(w.ToString());
}
