从列表框打开文件 [英] Open file from a listbox

问题描述

我试图从列表框打开文件，这些文件可能是 Word ， PDF ，< Excel> 等。是否需要为每个文件类型单独的代码，或者有什么方法可以打开文件时双击？

列表框通过使用更新按钮来填充。

 公共类frmMain 
 
 Private Sub ButtonUpdate_Click（sender As Object，e As EventArgs）处理ButtonUpdate.Click 
 Dim folderInfo As新的IO.DirectoryInfo（我的目录在这里）
 Dim arrFilesInFolder（）作为IO.FileInfo 
昏暗的fileInFolder作为IO.FileInfo 
 arrFilesInFolder = folderInfo.GetFiles（*。*） 
 For each fileInFolder in arrFilesInFolder 
 ListBox1.Items.Add（fileInFolder.Name）
 Next 
 End Sub 
 
 Private Sub ButtonExit_Click（sender As Object， e作为EventArgs）处理ButtonExit.Click 
 Me.Cl ose（）
 End Sub 
 
 Private Sub ListBox1_DoubleClick（sender As Object，e As EventArgs）Handles ListBox1.DoubleClick 
 
 End Sub 
 
 End Class 
  

解决方案

将文件名传递给Process.Start方法
$ b $ pre $ Private $ ListBox1_DoubleClick（sender As Object，e As EventArgs）Handles ListBox1.DoubleClick
Dim fullPath = Path.Combine（YourDirectoryHere，ListBox1.SelectedItem.ToString（））
System.Diagnostics.Process.Start（fullPath）
End Sub

另一个问题是您尝试打开的文件类型（扩展名）。填充列表框的方法使用 *。* 加载FileInfo。所以每一种文件都被添加到列表框中，如果没有与该扩展相关的程序，这可能是一个问题。

更多关于 Process.Start（string）here

I'm trying to open files from a listbox, the files could be Word, PDF, Excel, etc. Does there need to be separate code for each file type, or is there some way to just open the file when its double clicked?

The listbox populates fine through the use of the update button I have.

Public Class frmMain

Private Sub ButtonUpdate_Click(sender As Object, e As EventArgs) Handles ButtonUpdate.Click
    Dim folderInfo As New IO.DirectoryInfo("my directory is here")
    Dim arrFilesInFolder() As IO.FileInfo
    Dim fileInFolder As IO.FileInfo
    arrFilesInFolder = folderInfo.GetFiles("*.*")
    For Each fileInFolder In arrFilesInFolder
        ListBox1.Items.Add(fileInFolder.Name)
    Next
End Sub

Private Sub ButtonExit_Click(sender As Object, e As EventArgs) Handles ButtonExit.Click
    Me.Close()
End Sub

Private Sub ListBox1_DoubleClick(sender As Object, e As EventArgs) Handles ListBox1.DoubleClick

End Sub

End Class

解决方案

In its simplest form you just need to pass the filename to the Process.Start method

Private Sub ListBox1_DoubleClick(sender As Object, e As EventArgs) Handles ListBox1.DoubleClick
   Dim fullPath = Path.Combine("YourDirectoryHere", ListBox1.SelectedItem.ToString())
   System.Diagnostics.Process.Start(fullPath)
End Sub

However, this requires that you have saved the directory and recombine it with your file name.

Another problem is the file type (extension) that you try to open. The method that fills the listbox use *.* to load the FileInfo. So every kind of file is added to the listbox and this could be a problem if there is no program associated with that extension.

See more info on Process.Start(string) here

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