blocks|key|500116|text|路径通常为空。此函数更安全。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500117|function+Get-ScriptDirectory
{
++++$Invocation+=+(Get-Variable+MyInvocation+-Scope+1).Value;
++++if($Invocation.PSScriptRoot)
++++{
++++++++$Invocation.PSScriptRoot;
++++}
++++Elseif($Invocation.MyCommand.Path)
++++{
++++++++Split-Path+$Invocation.MyCommand.Path
++++}
++++else
++++{
++++++++$Invocation.InvocationName.Substring(0,$Invocation.InvocationName.LastIndexOf("\"));
++++}
}|code-block|syntax|javascript|500118|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Path is often null. This function is safer.

<pre><code>function Get-ScriptDirectory
{
 $Invocation = (Get-Variable MyInvocation -Scope 1).Value;
 if($Invocation.PSScriptRoot)
 {
 $Invocation.PSScriptRoot;
 }
 Elseif($Invocation.MyCommand.Path)
 {
 Split-Path $Invocation.MyCommand.Path
 }
 else
 {
 $Invocation.InvocationName.Substring(0,$Invocation.InvocationName.LastIndexOf("\"));
 }
}
</code></pre>

blocks|key|500174|text|您可能认为使用'.\‘作为路径意味着它是调用路径。但并不是一直都这样。例如，如果您在作业ScriptBlock中使用它。在这种情况下，它可能指向%25profile%25\Documents。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500175|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You would think that using '.\' as the path means that it's the invocation path. But not all the time. Example, if you use it inside a job ScriptBlock. In which case, it might point to %profile%\Documents.

blocks|key|500219|text|是的，这应该行得通。但是如果你需要查看绝对路径，这就是你所需要的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500220|(Get-Item+.).FullName|code-block|syntax|javascript|500221|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Yes, that should work. But if you need to see the absolute path, this is all you need:
<pre><code>(Get-Item .).FullName
</code></pre>

blocks|key|500267|text|您还可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500268|(Resolve-Path+.\).Path|code-block|syntax|javascript|500269|括号中的部分返回一个PathInfo对象。|offset|length|style|CODE|500270|(从PowerShell+2.0开始提供。)|500271|entityMap|0|LINK|mutability|MUTABLE|url|https://msdn.microsoft.com/en-us/library/system.management.automation.pathinfo.aspx^0|0|0|A|8|A|8|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@$I|11|J|12|1|13]]|A|$]]|$1|M|3|N|5|6|7|14|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

You can also use:

<pre><code>(Resolve-Path .\).Path
</code></pre>

The part in brackets returns a <a href="https://msdn.microsoft.com/en-us/library/system.management.automation.pathinfo.aspx" rel="noreferrer"><code>PathInfo</code></a> object.

(Available since PowerShell 2.0.)

blocks|key|500307|text|我喜欢the+one-line+solution+:)|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|500308|$scriptDir+=+Split-Path+-Path+$MyInvocation.MyCommand.Definition+-Parent|code-block|syntax|javascript|500309|entityMap|0|LINK|mutability|MUTABLE|url|http://techblog.dorogin.com/2011/07/powershell-current-directory-relative.html^0|3|L|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

I like <a href="http://techblog.dorogin.com/2011/07/powershell-current-directory-relative.html" rel="nofollow noreferrer">the one-line solution</a> :)

<pre><code>$scriptDir = Split-Path -Path $MyInvocation.MyCommand.Definition -Parent
</code></pre>

blocks|key|500349|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500350|(Get-Location).path|code-block|syntax|javascript|500351|或者：|500352|($pwd).path|500353|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Try :

<pre><code>(Get-Location).path
</code></pre>

or:

<pre><code>($pwd).path
</code></pre>

blocks|key|500372|text|扩展@Cradle的答案:你也可以写一个multi-purpose+function，它会根据操作员的问题得到相同的结果：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|500373|Function+Get-AbsolutePath+{

++++[CmdletBinding()]
++++Param(
++++++++[parameter(
++++++++++++Mandatory=$false,
++++++++++++ValueFromPipeline=$true
++++++++)]
++++++++[String]$relativePath=".\"
++++)

++++if+(Test-Path+-Path+$relativePath)+{
++++++++return+(Get-Item+-Path+$relativePath).FullName+-replace+"\\$",+""
++++}+else+{
++++++++Write-Error+-Message+"'$relativePath'+is+not+a+valid+path"+-ErrorId+1+-ErrorAction+Stop
++++}

}|code-block|syntax|javascript|500374|entityMap|0|LINK|mutability|MUTABLE|url|http://bit.ly/1QdZDgD^0|K|M|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

To expand on @Cradle 's answer: you could also write a <a href="http://bit.ly/1QdZDgD" rel="nofollow">multi-purpose function</a> that will get you the same result per the OP's question:

<pre><code>Function Get-AbsolutePath {

 [CmdletBinding()]
 Param(
 [parameter(
 Mandatory=$false,
 ValueFromPipeline=$true
 )]
 [String]$relativePath=".\"
 )

 if (Test-Path -Path $relativePath) {
 return (Get-Item -Path $relativePath).FullName -replace "\\$", ""
 } else {
 Write-Error -Message "'$relativePath' is not a valid path" -ErrorId 1 -ErrorAction Stop
 }

}
</code></pre>

blocks|key|715970|text|Get-Location将返回当前位置：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|715971|$Currentlocation+=+Get-Location|code-block|syntax|javascript|715972|entityMap|0|LINK|mutability|MUTABLE|url|https://msdn.microsoft.com/en-us/powershell/reference/5.1/microsoft.powershell.management/get-location^0|0|C|0|C|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

<a href="https://msdn.microsoft.com/en-us/powershell/reference/5.1/microsoft.powershell.management/get-location" rel="noreferrer"><code>Get-Location</code></a> will return the current location:

<pre><code>$Currentlocation = Get-Location
</code></pre>

blocks|key|715984|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|715985|$WorkingDir+=+Convert-Path+.|code-block|syntax|javascript|715986|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this:

<pre><code>$WorkingDir = Convert-Path .
</code></pre>

blocks|key|716030|text|在Powershell+3和更高版本中，您可以简单地使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|716031|$PSScriptRoot|offset|length|style|CODE|716032|entityMap^0|0|0|D|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

In Powershell 3 and above you can simply use

<code>$PSScriptRoot</code>

blocks|key|716069|text|值得一提的是，作为一个单行解决方案，下面是一个对我有效的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|716070|$currFolderName+=+(Get-Location).Path.Substring((Get-Location).Path.LastIndexOf("\")%2B1)|code-block|syntax|javascript|716071|末尾的1表示忽略/。|offset|length|style|CODE|716072|由于上面的帖子使用了Get-Location+cmdlet。|716073|entityMap|0|LINK|mutability|MUTABLE|url|https://msdn.microsoft.com/en-us/powershell/reference/5.1/microsoft.powershell.management/get-location^0|0|0|8|1|0|A|C|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@]|9|@$I|12|J|13|1|14]]|A|$]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

For what it's worth, to be a single-line solution, the below is a working solution for me. 

<pre><code>$currFolderName = (Get-Location).Path.Substring((Get-Location).Path.LastIndexOf("\")+1)
</code></pre>

The 1 at the end is to ignore the <code>/</code>.

Thanks to the posts above using the <a href="https://msdn.microsoft.com/en-us/powershell/reference/5.1/microsoft.powershell.management/get-location" rel="nofollow noreferrer">Get-Location</a> cmdlet.

blocks|key|500636|text|我也遇到过类似的问题，这给我带来了很多麻烦，因为我正在编写用PowerShell+(完整的最终用户图形用户界面应用程序)编写的程序，而且我有很多文件和资源需要从磁盘加载。根据我的经验，使用.来表示当前目录是不可靠的。它应该表示当前工作目录，但通常不是这样。看起来PowerShell保存了在.中调用PowerShell的位置。更准确地说，当PowerShell第一次启动时，它默认在您的主用户目录中启动。这通常是您的用户帐户的目录，如C:\USERS\YOUR+USER+NAME。之后，在显示PowerShell提示符或运行脚本之前，PowerShell会将目录更改为您从中调用它的目录，或者更改为您正在执行的脚本所在的目录。但这发生在PowerShell应用程序本身最初在您的家庭用户目录中启动之后。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|500637|.表示PowerShell启动时所在的初始目录。因此，如果从所需的目录调用PowerShell，则.仅表示当前目录。如果您稍后在PowerShell代码中更改了目录，则更改似乎不会在所有情况下都反映在.中。在某些情况下，.代表当前工作目录，而在其他情况下，PowerShell+(本身，而不是脚本)是从哪个目录被调用的，这可能会导致不一致的结果。出于这个原因，我使用调用程序脚本。内部只有一个命令的PowerShell脚本：POWERSHELL。这将确保从所需的目录调用PowerShell，从而使.表示当前目录。但只有当您稍后不在PowerShell代码中更改目录时，它才能起作用。对于脚本，我使用调用程序脚本，它类似于我提到的最后一个脚本，除了它包含一个文件选项：POWERSHELL+-FILE+DRIVE:\PATH\SCRIPT+NAME.PS1。这确保了PowerShell在当前工作目录中启动。|500638|只需单击script，无论script位于何处，都会从您的主用户目录调用PowerShell。它的结果是，当前工作目录是脚本所在的目录，而PowerShell调用目录是C:\USERS\YOUR+USER+NAME，并且.根据情况返回这两个目录中的一个，这是荒谬的。|500639|但是，为了避免所有这些麻烦并使用调用器脚本，您可以简单地使用$PWD或$PSSCRIPTROOT代替.来表示当前目录，具体取决于您希望表示当前工作目录还是从其中调用脚本的目录。如果出于某种原因，您希望检索.返回的两个目录中的其他目录，则可以使用$HOME。|500640|我个人只是在我用PowerShell开发的应用程序的根目录下有调用程序脚本，它调用我的主应用程序脚本，并且简单地记住永远不要更改我的应用程序源代码中的当前工作目录，所以我永远不必担心这个问题，我可以使用.来表示当前目录并在我的应用程序中支持相对文件寻址，而不会出现任何问题。这应该适用于较新版本的PowerShell+(比版本2更新)。|500641|entityMap^0|2M|1|41|1|61|N|0|0|1|1D|1|2S|1|32|1|5W|A|6X|1|9A|18|0|2C|N|32|1|0|U|4|Z|D|1E|1|2U|1|3E|5|0|2T|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1B|8|@$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|1R|8|@$9|1S|A|1T|B|C]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1U|8|@]|D|@]|E|$]]]|O|$]]

I had similar problems and it made me a lot of trouble since I am making programs written in PowerShell (full end user GUI applications) and I have a lot of files and resources I need to load from disk.
From my experience, using <code>.</code> to represent current directory is unreliable. It should represent current working directory, but it often does not.
It appears that PowerShell saves location from which PowerShell has been invoked inside <code>.</code>.
To be more precise, when PowerShell is first started, it starts, by default, inside your home user directory. That is usually directory of your user account, something like <code>C:\USERS\YOUR USER NAME</code>.
After that, PowerShell changes directory to either directory from which you invoked it, or to directory where script you are executing is located before either presenting you with PowerShell prompt or running the script. But that happens after PowerShell app itself originally starts inside your home user directory.

And <code>.</code> represents that initial directory inside which PowerShell started. So <code>.</code> only represents current directory in case if you invoked PowerShell from the wanted directory. If you later change directory in PowerShell code, change appears not to be reflected inside <code>.</code> in every case.
In some cases <code>.</code> represents current working directory, and in others directory from which PowerShell (itself, not the script) has been invoked, what can lead to inconsistent results.
For this reason I use invoker script. PowerShell script with single command inside:
<code>POWERSHELL</code>.
That will ensure that PowerShell is invoked from the wanted directory and thus make <code>.</code> represent current directory. But it only works if you do not change directory later in PowerShell code.
In case of a script, I use invoker script which is similar to last one I mentioned, except it contains a file option:
<code>POWERSHELL -FILE DRIVE:\PATH\SCRIPT NAME.PS1</code>.
That ensures that PowerShell is started inside current working directory.

Simply clicking on script invokes PowerShell from your home user directory no matter where script is located.
It results with current working directory being directory where script is located, but PowerShell invocation directory being <code>C:\USERS\YOUR USER NAME</code>, and with <code>.</code> returning one of these two directories depending on the situation, what is ridiculous.

But to avoid all this fuss and using invoker script, you can simply use either <code>$PWD</code> or <code>$PSSCRIPTROOT</code> instead of <code>.</code> to represent current directory depending on weather you wish to represent current working directory or directory from which script has been invoked.
And if you, for some reason, want to retrieve other of two directories which <code>.</code> returns, you can use <code>$HOME</code>.

I personally just have invoker script inside root directory of my apps I develop with PowerShell which invokes my main app script, and simply remember to never ever change current working directory inside my source code of my app, so I never have to worry about this, and I can use <code>.</code> to represent current directory and to support relative file addressing in my applications without any problems.
This should work in newer versions of PowerShell (newer than version 2).

blocks|key|500679|text|此函数将提示符位置设置为脚本路径，处理vscode、psise和pwd之间获取scriptpath的不同方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500680|function+Set-CurrentLocation
{
++++$currentPath+=+$PSScriptRoot+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++#+AzureDevOps,+Powershell
++++if+(!$currentPath)+{+$currentPath+=+Split-Path+$pseditor.GetEditorContext().CurrentFile.Path+-ErrorAction+SilentlyContinue+}+++++#+VSCode
++++if+(!$currentPath)+{+$currentPath+=+Split-Path+$psISE.CurrentFile.FullPath+-ErrorAction+SilentlyContinue+}+++++++++++++++++++++++#+PsISE

++++if+($currentPath)+{+Set-Location+$currentPath+}
}|code-block|syntax|javascript|500681|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

this function will set the prompt location to script path, dealing with the differents way to get scriptpath between vscode, psise and pwd :

<pre><code>function Set-CurrentLocation
{
 $currentPath = $PSScriptRoot # AzureDevOps, Powershell
 if (!$currentPath) { $currentPath = Split-Path $pseditor.GetEditorContext().CurrentFile.Path -ErrorAction SilentlyContinue } # VSCode
 if (!$currentPath) { $currentPath = Split-Path $psISE.CurrentFile.FullPath -ErrorAction SilentlyContinue } # PsISE

 if ($currentPath) { Set-Location $currentPath }
}
</code></pre>

blocks|key|500731|text|这就是我想出来的。它是一个数组，包含多个查找路径的方法，使用当前位置，过滤出null\空结果，并返回第一个非空值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500732|@((
++($MyInvocation.MyCommand.Module.ModuleBase),
++($PSScriptRoot),
++(Split-Path+-Parent+-Path+$MyInvocation.MyCommand.Definition+-ErrorAction+SilentlyContinue),
++(Get-Location+%7C+Select-Object+-ExpandProperty+Path)
)+%7C+Where-Object+{+$_+})[0]|code-block|syntax|javascript|500733|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is what I came up with. It's an array including multiple methods of finding a path, uses the current location, filters out null\empty results, and returns the first not-null value.
<pre class="lang-php prettyprint-override"><code>@((
 ($MyInvocation.MyCommand.Module.ModuleBase),
 ($PSScriptRoot),
 (Split-Path -Parent -Path $MyInvocation.MyCommand.Definition -ErrorAction SilentlyContinue),
 (Get-Location | Select-Object -ExpandProperty Path)
) | Where-Object { $_ })[0]
</code></pre>

blocks|key|500794|text|要仅获取当前文件夹名称，还可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|500795|(Split-Path+-Path+(Get-Location)+-Leaf)|code-block|syntax|javascript|500796|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

To only get the current folder name, you can also use:
<pre><code>(Split-Path -Path (Get-Location) -Leaf)
</code></pre>

This should be a simple task, but I have seen several attempts on how to get the path to the directory where the executed cmdlet is located with mixed success. For instance, when I execute <code>C:\temp\myscripts\mycmdlet.ps1</code> which has a settings file at <code>C:\temp\myscripts\settings.xml</code> I would like to be able to store <code>C:\temp\myscripts</code> in a variable within <code>mycmdlet.ps1</code>.
This is one solution which works (although a bit cumbersome):
<pre><code>$invocation = (Get-Variable MyInvocation).Value
$directorypath = Split-Path $invocation.MyCommand.Path
$settingspath = $directorypath + '\settings.xml'
</code></pre>
Another one suggested this solution which only works on our test environment:
<pre><code>$settingspath = '.\settings.xml'
</code></pre>
I like the latter approach a lot and prefer it to having to parse the filepath as a parameter each time, but I can't get it to work on my development environment. What should I do? Does it have something to do with how PowerShell is configured?

How to get the current directory of the cmdlet being executed

这应该是一项简单的任务，但我已经看到几次关于如何获取已执行cmdlet所在目录的路径的尝试，成功与否参差不齐。例如，当我执行在C:\temp\myscripts\settings.xml中有一个设置文件的C:\temp\myscripts\mycmdlet.ps1时，我希望能够将C:\temp\myscripts存储在...

问如何获取正在执行的cmdlet的当前目录
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问如何获取正在执行的cmdlet的当前目录
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