用这两种方式组合路径有什么区别？ [英] What is the difference between combining paths in those 2 ways?

问题描述

您能解释一下

$attachment = [String]::Concat($workingDir,"\", $fileName)

和

$attachment = [IO.Path]::Combine($workingDir, $fileName)

要在Powershell中合并路径吗？

when it comes to combining paths in Powershell?

推荐答案

请考虑以下情况： $ workingDir 带有反斜杠，而 $ fileName 带有反斜杠，例如：

Consider a situation where $workingDir has a trailing backslash and $fileName has a leading one, e.g.:

$workingDir = "C:\foo\"
$fileName   = "\bar.txt"

这两个命令将产生以下结果：

The 2 commands will produce the following results:

PS C:\> [String]::Concat($workingDir, "\", $fileName)
C:\foo\\\bar.txt
PS C:\> [IO.Path]::Combine($workingDir, $fileName)
\bar.txt

在PowerShell中，最好使用 Join-Path ：

In PowerShell it's better to use Join-Path:

PS C:\> Join-Path $workingDir $fileName
C:\foo\bar.txt

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