4

我有一个“保存”按钮，所以当用户单击时，它将保存 xml 文件（xml 序列化）。此处使用了保存文件对话框，当我在未选择任何文件的情况下按取消时，会出现“参数异常”并显示“空路径名不合法”。我该如何处理这个异常？即使在保存文件对话框中没有选择任何路径，我也希望表单保持不变。非常感谢。

我的保存文件对话框片段：

private void SaveButton_Click(object sender, RoutedEventArgs e)
{
        string savepath;
        SaveFileDialog DialogSave = new SaveFileDialog();
        // Default file extension
        DialogSave.DefaultExt = "txt";
        // Available file extensions
        DialogSave.Filter = "XML file (*.xml)|*.xml|All files (*.*)|*.*";
        // Adds a extension if the user does not
        DialogSave.AddExtension = true;
        // Restores the selected directory, next time
        DialogSave.RestoreDirectory = true;
        // Dialog title
        DialogSave.Title = "Where do you want to save the file?";
        // Startup directory
        DialogSave.InitialDirectory = @"C:/";
        DialogSave.ShowDialog();
        savepath = DialogSave.FileName;
        DialogSave.Dispose();
        DialogSave = null;
        ...
        using (Stream savestream = new FileStream(savepath, FileMode.Create))
        {
                XmlSerializer serializer = new XmlSerializer(typeof(FormSaving));
                serializer.Serialize(savestream, formsaving);
        }

}

我的论点异常发生在这一行：

using (Stream savestream = new FileStream(savepath, FileMode.Create))
{
        XmlSerializer serializer = new XmlSerializer(typeof(FormSaving));
        serializer.Serialize(savestream, formsaving);
}
4

2 回答 2

5

这里的问题是您不关心保存对话框的结果，即使用户单击了取消，您也会尝试保存。您应该将代码更改为如下所示：

...
DialogSave.InitialDirectory = @"C:/";
if( DialogSave.ShowDialog() == DialogResult.OK )
{
  savepath = DialogSave.FileName;
  DialogSave = null;
  ...
  using (Stream savestream = new FileStream(savepath, FileMode.Create))
  {
     XmlSerializer serializer = new XmlSerializer(typeof(FormSaving));
     serializer.Serialize(savestream, formsaving);
  }
}
DialogSave.Dispose();
于 2011-02-28T08:30:09.580 回答
4

如果用户取消对话框，您可能不想保存？检查结果ShowDialog并采取相应措施：

if (DialogSave.ShowDialog() == true)
{
    savepath = DialogSave.FileName;
            ...
    using (Stream savestream = new FileStream(savepath, FileMode.Create))
    {
        XmlSerializer serializer = new XmlSerializer(typeof(FormSaving));
        serializer.Serialize(savestream, formsaving);
    }
}
于 2011-02-28T08:30:02.933 回答