OpenFileDialog.ShowDialog（）抛出异常？ [英] OpenFileDialog.ShowDialog() throws an exception?

问题描述

我试图从我的WPF视图模型命令之一显示一个对话框，但是当我调用 ShowDialog（）它会抛出一个 System.ArgumentException ，我想知道有没有人可以给我一个提示，为什么？

这是我的代码：

 公共ReadOnly属性OpenParser作为ICommand 
获取
返回新的RelayCommand（Sub（param As Object）OpenParserExecute（DirectCast（param，Frame）） ）
结束Get 
结束属性
 
 Public Sub OpenParserExecute（ByVal mFrame As Frame）
 SaveParserExecute（）
 Dim mOpenDialog As OpenFileDialog = OpenDialog 
如果mOpenDialog.ShowDialog（）然后'线条抛出异常
 CurrentParser =新的ParserEditorModel（mOpenDialog.FileName）
 mFrame.Navigate（新的ParserEditor（CurrentParser））
结束如果
 End Sub 
  

StackTrace请求：

 在MS.Internal.Interop.HRESULT.ThrowIf失败（字符串消息）
在MS.Internal.AppModel.ShellUtil.GetShellItemForPath（字符串路径）
在Microsoft.Win32.FileDialog.PrepareVistaDialog（IFileDialog对话框）
在Microsoft.Win32.FileDialog。在Microsoft.Win32.CommonDialog.ShowDialog（）
在WinTransform.GUI.MainWindowModel.OpenParserExecute（框架mFrame）中的$ Microsoft $ WinCE ）在C：\Users\Alex\Desktop\MEDLI\branches\WinTransform\GUI\ViewModels\MainWindowModel.vb：行38 
  

解决方案

因为 ShowDialog（）本身返回一个 Nullable（Of Boolean）和你的如果语句期待一个非Nullable Boolean

您必须将对话框的返回值转换为布尔值，如果它是 True 通过对话框的文件名检索所选文件财产。

示例。

I am trying to show a dialog from one of my WPF view model commands however when i call ShowDialog() it throws a System.ArgumentException, I was wondering if anyone could give me a hint as to why?

Here is my code:

Public ReadOnly Property OpenParser As ICommand
    Get
        Return New RelayCommand(Sub(param As Object) OpenParserExecute(DirectCast(param, Frame)))
    End Get
End Property

Public Sub OpenParserExecute(ByVal mFrame As Frame)
    SaveParserExecute()
    Dim mOpenDialog As OpenFileDialog = OpenDialog
    If mOpenDialog.ShowDialog() Then ' Lines the throws the exception
        CurrentParser = New ParserEditorModel(mOpenDialog.FileName)
        mFrame.Navigate(New ParserEditor(CurrentParser))
    End If
End Sub

StackTrace as requested:

at MS.Internal.Interop.HRESULT.ThrowIfFailed(String message)
at MS.Internal.AppModel.ShellUtil.GetShellItemForPath(String path)
at Microsoft.Win32.FileDialog.PrepareVistaDialog(IFileDialog dialog)
at Microsoft.Win32.FileDialog.RunVistaDialog(IntPtr hwndOwner)
at Microsoft.Win32.FileDialog.RunDialog(IntPtr hwndOwner)
at Microsoft.Win32.CommonDialog.ShowDialog()
at WinTransform.GUI.MainWindowModel.OpenParserExecute(Frame mFrame) in C:\Users\Alex\Desktop\MEDLI\branches\WinTransform\GUI\ViewModels\MainWindowModel.vb:line 38

解决方案

Because ShowDialog() itself returns a Nullable(Of Boolean) and your If statement is expecting a non-Nullable Boolean.

You'll have to cast the dialog's return value to a Boolean and if it's True retreive the selected file through your dialog's Filename property.

Example.

这篇关于OpenFileDialog.ShowDialog（）抛出异常？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！