Question

因此，我正在从事一个项目，目的是允许我的用户从那里的个人文件中选择一个mp3 / wav文件，并将其保存在我的程序目录中，以便以后可以访问。

Public Class Uploader_Control


Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button_Upload.Click 'Button click triggers open file dialog
    OpenFileDialog1.InitialDirectory = "C:\" 
    OpenFileDialog1.Title = "Upload A Song" 
    OpenFileDialog1.Filter = "Audio Files|*.mp3; *.wav"
    OpenFileDialog1.Multiselect = False
    OpenFileDialog1.ShowDialog()
    Dim New_Upload As String = OpenFileDialog1.FileName 'Sets 'New_Upload' to the location of the selected file.

    Dim Upload_Path = Get_Upload_Path() 'Triggers function to find the destination path that i want the file to be copied too



End Sub
Function Get_Upload_Path() 'Tries to find the location for the file to saved too
    Dim Upload_Path As String = Application.ExecutablePath
    Upload_Path = Upload_Path.Substring(0, Upload_Path.LastIndexOf("/")) & "\Resources\" 

    Return Upload_Path
End Function

我尝试了几种方法尝试将其复制，但是我似乎找不到找到将文件复制到目录中的方法。如果任何人都可以对我可以实现的方法提供任何见解，将不胜感激。

Answer 1

首先要解决的是您的GetUploadPath。

Function Get_Upload_Path() as String
    Dim newPath As String = Application.ExecutablePath
    newPath = Path.Combine(Path.GetDirectoryName(newPath), "Resources")
    Return newPath
End Function

在这里，我正在使用安全的Path类方法来提取您的文件夹并追加Resource文件夹。您的代码改为使用字符“ /”，这不是Windows中的默认路径分隔符。可能您在那里会遇到异常

最后，您使用File.Copy复制文件，但请记住，应始终为File.Copy提供文件名。复制源和目标参数，而不仅仅是目录名

  if OpenFileDialog1.ShowDialog() = DialogResult.OK Then
     Dim newPath = Get_Upload_Path()
     Dim file = Path.GetFileName(OpenFileDialog1.FileName)
     Dim newUploadName = Path.Combine(newPath, file)
     File.Copy(OpenFileDialog1.FileName, newUploadName)
  End If

Answer 2

最简单的方法是使用System.IO.FileInfo：

Dim file as New FileInfo(path)
file.CopyTo(newPath)

Answer 3

此解决方案将起作用，尝试使内容与原始内容保持接近：

    'Set up and open file dialog
    OpenFileDialog1.Title = "Upload A Song"
    OpenFileDialog1.Filter = "Audio Files|*.mp3; *.wav"
    OpenFileDialog1.Multiselect = False
    OpenFileDialog1.InitialDirectory = "C:\"
    OpenFileDialog1.FileName = String.Empty

    'Only proceed if OK pressed
    If OpenFileDialog1.ShowDialog = DialogResult.OK Then
        'Get Destination Path
        Dim Upload_Path As String = IO.Path.Combine(Application.StartupPath, "Resources")
        If IO.Directory.Exists(Upload_Path) Then
            'Copy file
            IO.File.Copy(OpenFileDialog1.FileName, IO.Path.Combine(Upload_Path, IO.Path.GetFileName(OpenFileDialog1.FileName)))
        End If
    End If

在您尝试复制之前，我添加了一个文件夹是否存在的检查。

您需要考虑的其他事项：

如果该文件夹不存在，请考虑创建它或发出消息。

您正在将文件复制到您的apps目录。根据您应用的位置，您可能会遇到“权限”问题。

是否会引用您在工作中加载的原始文件？音频文件的引用列表，将其加载而不进行复制，然后在需要时进行处理？

Answer 4

此外，如果您愿意，也可以使用以下方法移动它：

My.Computer.FileSystem.MoveFile(c:\\source\my.mp3, d:\\destination\my.mp3)

如果没有，请改用CopyFile。

VB.net如何将文件复制到新位置

4 个答案: