.net powershell - 提取文件名和扩展名
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powershell - extract file name and extension
提问by culter
I need to extract file name and extension from e.g. my.file.xlsx. I don't know the name of file or extension and there may be more dots in the name, so I need to search the string from the right and when I find first dot (or last from the left), extract the part on the right side and the part on the left side from that dot.
我需要从例如 my.file.xlsx 中提取文件名和扩展名。我不知道文件或扩展名的名称，名称中可能有更多的点，所以我需要从右边搜索字符串，当我找到第一个点（或从左边开始）时，提取部分右侧和该点左侧的部分。
Maybe there is better solution, but I did'n find anything here or anywhere else. Thank you
也许有更好的解决方案，但我在这里或其他任何地方都没有找到任何东西。谢谢
回答by Goyuix
If the file is coming off the disk and as others have stated, use the BaseNameand Extensionproperties:
如果文件从磁盘上脱落并且如其他人所述，请使用BaseName和Extension属性：
PS C:\> dir *.xlsx | select BaseName,Extension
BaseName Extension
-------- ---------
StackOverflow.com Test Config .xlsx
If you are given the file name as part of string (say coming from a text file), I would use the GetFileNameWithoutExtensionand GetExtensionstatic methods from the System.IO.Pathclass:
如果您将文件名作为字符串的一部分（比如来自文本文件），我将使用System.IO.Path类中的GetFileNameWithoutExtension和GetExtension静态方法：
PS C:\> [System.IO.Path]::GetFileNameWithoutExtension("Test Config.xlsx")
Test Config
PS H:\> [System.IO.Path]::GetExtension("Test Config.xlsx")
.xlsx
回答by Straff
PS C:\Windows\System32\WindowsPowerShell\v1.0>split-path "H:\Documents\devops\tp-mkt-SPD-38.4.10.msi" -leaf
tp-mkt-SPD-38.4.10.msi
PS C:\Windows\System32\WindowsPowerShell\v1.0> $psversiontable
Name Value
---- -----
CLRVersion 2.0.50727.5477
BuildVersion 6.1.7601.17514
PSVersion 2.0
WSManStackVersion 2.0
PSCompatibleVersions {1.0, 2.0}
SerializationVersion 1.1.0.1
PSRemotingProtocolVersion 2.1
回答by CB.
If is from a text file and and presuming name file are surrounded by white spaces this is a way:
如果来自文本文件并且假设名称文件被空格包围，这是一种方法：
$a = get-content c:\myfile.txt
$b = $a | select-string -pattern "\s.+\..{3,4}\s" | select -ExpandProperty matches | select -ExpandProperty value
$b | % {"File name:{0} - Extension:{1}" -f $_.substring(0, $_.lastindexof('.')) , $_.substring($_.lastindexof('.'), ($_.length - $_.lastindexof('.'))) }
If is a file you can use something like this based on your needs:
如果是一个文件，您可以根据需要使用类似的东西：
$a = dir .\my.file.xlsx # or $a = get-item c:\my.file.xlsx
$a
Directory: Microsoft.PowerShell.Core\FileSystem::C:\ps
Mode LastWriteTime Length Name
---- ------------- ------ ----
-a--- 25/01/10 11.51 624 my.file.xlsx
$a.BaseName
my.file
$a.Extension
.xlsx
回答by Jaqueline Vanek
PS C:\Users\joshua> $file = New-Object System.IO.FileInfo('file.type')
PS C:\Users\joshua> $file.BaseName, $file.Extension
file
.type
回答by Shay Levy
Check the BaseName and Extension properties of the FileInfo object.
检查 FileInfo 对象的 BaseName 和 Extension 属性。
回答by Ernest Correale
Use Split-Path
使用分割路径
$filePath = "C:\PS\Test.Documents\myTestFile.txt";
$fileName = (Split-Path -Path $filePath -Leaf).Split(".")[0];
$extension = (Split-Path -Path $filePath -Leaf).Split(".")[1];
回答by Esperento57
just do it:
去做就对了：
$file=Get-Item "C:\temp\file.htm"
$file.Basename
$file.Extension
回答by hans57sauc
This is an adaptation, if anyone is curious. I needed to test whether RoboCopy successfully copied one file to multiple servers for its integrity:
这是一种改编，如果有人好奇的话。我需要测试 RoboCopy 是否成功将一个文件复制到多个服务器以确保其完整性：
$Comp = get-content c:\myfile.txt
ForEach ($PC in $Comp) {
dir "\$PC\Folder\Share\*.*" | Select-Object $_.BaseName
}
Nice and simple, and it shows the directory and the file inside it. If you want to specify one file name or extension, just replace the *'s with whatever you want.
很好很简单，它显示了目录和里面的文件。如果您想指定一个文件名或扩展名，只需将 * 替换为您想要的任何内容。
Directory: \SERVER\Folder\Share
Mode LastWriteTime Length Name
---- ------------- ------ ----
-a--- 2/27/2015 5:33 PM 1458935 Test.pptx
