VB.net文件夹路径不起作用

时间:2016-12-15 22:02:27

标签: vb.net directory explorer

在我的代码中,我必须使用通配符" *"搜索simliar的文件夹名称。此代码在VBA中运行良好(Process.Start ...行除外),但它现在所做的只是打开我的文档。

应该打开一个网络文件夹。

字符串pathStr生成最终的文件夹路径,这是正确的

我已经对文件夹路径进行了三重检查(我复制了此代码在Windows资源管理器中生成和粘贴的内容并且路径检出)

为什么我的代码只打开我的文档?

Private Sub OpenJobToolStripMenuItem_Click(sender As Object, e As EventArgs) Handles OpenJobToolStripMenuItem.Click
    Dim jobnum As String = ListView1.SelectedItems(0).Text

    'Define the master path to all job numbers
    Dim masterpath As String = "\\ussatf02\Production\00A Job Folders\"

    'Get the child folder path
    Dim fullFolder As String = masterpath & jobnum.Substring(0, 5) & "xxx\"

    'Get first 8 characters of job number
    Dim jobFolder As String = jobnum.Substring(0, 8)

    'Define the full path to the jobFolder
    Dim xPath As String = fullFolder & jobFolder

    'Check the full path with a wildcard to see if there is a folder named something simliar to what we have
    Dim foundFolder As String = Dir(xPath & "*", vbDirectory)

    'If the folder is not found (length < 2) then throw an error, else open the folder
    If (foundFolder.Length < 2) Then
        Dim msgRes As MsgBoxResult = MsgBox("The job folder for: " & jobnum & " could not be found.", vbCritical, "Error Finding Folder")
    Else

        'Define the final path of the folder
        Dim pathStr As String = xPath & "\" & foundFolder

        'Open the folder
        System.Diagnostics.Process.Start("explorer.exe", pathStr)
    End If

End Sub

任何帮助将不胜感激!!

修改

我现在替换了

 Dim pathStr As String = xPath & "\" & foundFolder

有关

 Dim pathStr As String = System.IO.Path.GetDirectoryName(xPath & "\" & foundFolder)

我仍然得到相同的结果

1 个答案:

答案 0 :(得分:1)

在这种情况下,

资源管理器将转到默认文件夹&#39;我的文档&#39;,如果您尝试打开的文件夹不在那里。确保pathStr存在。

您的文件夹可能包含Unicode字符,请参阅此网址C#: System.Diagnostics.Process.Start("Explorer.exe", @"/select" + FilePath). Can not open file when file's name is unicode character

中的更多内容

System.Diagnostics.Process.Start(&#34; Explorer.exe&#34;,&#34; / select,&#34;&#34;&#34;&amp; pathStr&amp;&#34; &#34;&#34;&#34)