2015-09-02 121 views
0

我试图解压缩.zip文件。这一切都很好,但它总是创建一个名为.zip文件的文件夹。解压缩文件而不创建文件夹

如何在不创建新文件夹的情况下获取文件?

Imports System.IO.Compression 

Private Sub XYform_Load(sender As Object, e As EventArgs) Handles Me.Load 
    Try 
     If (Not System.IO.Directory.Exists(System.IO.Path.GetTempPath & "\XML")) Then 
      System.IO.Directory.CreateDirectory(System.IO.Path.GetTempPath & "\XML") 
     End If 
    Catch 
    End Try 

    Try 
     ZipFile.ExtractToDirectory("D:\Test\Test data.zip", System.IO.Path.GetTempPath & "\XML") 
    Catch 
     'Allready Exists 
    End Try 
End Sub 

如果我这样做,它将始终在“XML”文件夹中创建一个“测试数据”文件夹。

回答

1

如果您想操纵文件,您需要单独使用ZipArchiveEntries。

下面是一个例子:

Dim zipPath As String = "c:\example\start.zip" 
Dim extractPath As String = "c:\example\extract" 

Using archive As ZipArchive = ZipFile.OpenRead(zipPath) 
    For Each entry As ZipArchiveEntry In archive.Entries 
     entry.ExtractToFile(Path.Combine(extractPath, entry.FullName)) 
    Next 
End Using 

如果你只是想解压到不同的目录,虽然,你可以使用:

Dim zipPath As String = "c:\users\exampleuser\end.zip" 
Dim extractPath As String = "c:\users\exampleuser\extract" 

Using archive As ZipArchive = ZipFile.Open(zipPath, ZipArchiveMode.Update) 
    archive.ExtractToDirectory(extractPath) 
End Using 

Source


编辑:

如果你不想保持结构,试试这个:

Dim zipPath As String = "c:\example\start.zip" 
Dim extractPath As String = "c:\example\extract" 

Dim zipPath As String = "c:\example\start.zip" 
Dim extractPath As String = "c:\example\extract" 

Using archive As ZipArchive = ZipFile.OpenRead(zipPath) 
    For Each entry As ZipArchiveEntry In archive.Entries.Where(Function(a) Not String.IsNullOrEmpty(a.Name)) 
     entry.ExtractToFile(entry.Name) 
    Next 
End Using 
+0

THX,但机器人的解决方案仍然创建该文件夹 – Moosli

+0

那么它可能是因为zip文件的目录。看看我的编辑,看看它是否有效。如果是这种情况,那么相应地处理它。你需要自己做一些调查。我们无法为您完成每个场景。 – Cory