只要文件名,不要路径和后缀
下面的代码得到文件名是带路径的,我怎么处理才能得到不带路径和文件后缀的文件名呢?比如"c:\aa.xls" 我只要"aa"
OpenFileDialog1.InitialDirectory = "c:\"
OpenFileDialog1.Filter = "excel文件(*.xls)|*.xls"
OpenFileDialog1.Title = "选择数据表"

OpenFileDialog1.FileName = ""
Dim fname As String
If OpenFileDialog1.ShowDialog() <> Windows.Forms.DialogResult.Cancel Then
fname = Trim(OpenFileDialog1.FileName)
Else
fname = ""
End If

[解决办法]
Path.GetFileNameWithoutExtension("d:\temp\c.xls")
[解决办法]
支持楼上

Path.GetFileNameWithoutExtension 方法
返回不具有扩展名的指定路径字符串的文件名。


下面的代码示例演示 GetFileNameWithoutExtension 方法的用法。

Dim fileName As String = "C:\mydir\myfile.ext"
Dim pathname As String = "C:\mydir\"
Dim result As String

result = Path.GetFileNameWithoutExtension(fileName)
Console.WriteLine("GetFileNameWithoutExtension('{0}') returns '{1}'", fileName, result)

result = Path.GetFileName(pathname)
Console.WriteLine("GetFileName('{0}') returns '{1}'", pathname, result)

' This code produces output similar to the following:
'
' GetFileNameWithoutExtension('C:\mydir\myfile.ext') returns 'myfile'
' GetFileName('C:\mydir\') returns ''
[解决办法]
修正代码:
Dim fileName As String = "C:\mydir\myfile.ext"
Dim pathname As String = "C:\mydir\"
Dim result As String

result = System.IO.Path.GetFileNameWithoutExtension(fileName)
Console.WriteLine("GetFileNameWithoutExtension('{0}') returns '{1}'", fileName, result)

result = System.IO.Path.GetFileName(pathname)
Console.WriteLine("GetFileName('{0}') returns '{1}'", pathname, result)
[解决办法]
System.IO.Path