我调用myScript1。来自myScript2的ps1。ps1。

假设两个脚本都在相同的位置，首先使用以下命令获取脚本的位置:

$PSScriptRoot

然后，像这样追加你想调用的脚本名称:

& "$PSScriptRoot\myScript1.ps1"

这应该有用。

为了找到脚本的位置，使用Split-Path $MyInvocation.MyCommand。路径(确保在脚本上下文中使用该路径)。

您应该使用它而不是其他任何东西的原因可以通过这个示例脚本来说明。

## ScriptTest.ps1
Write-Host "InvocationName:" $MyInvocation.InvocationName
Write-Host "Path:" $MyInvocation.MyCommand.Path

以下是一些结果。

PS C:\Users\JasonAr> .\ScriptTest.ps1
InvocationName: .\ScriptTest.ps1
Path: C:\Users\JasonAr\ScriptTest.ps1

PS C:\Users\JasonAr> . .\ScriptTest.ps1
InvocationName: .
Path: C:\Users\JasonAr\ScriptTest.ps1

PS C:\Users\JasonAr> & ".\ScriptTest.ps1"
InvocationName: &
Path: C:\Users\JasonAr\ScriptTest.ps1

在PowerShell 3.0及以后版本中，您可以使用自动变量$PSScriptRoot:

## ScriptTest.ps1
Write-Host "Script:" $PSCommandPath
Write-Host "Path:" $PSScriptRoot

PS C:\Users\jarcher> .\ScriptTest.ps1
Script: C:\Users\jarcher\ScriptTest.ps1
Path: C:\Users\jarcher

MyScript1的当前路径。ps1和myScript2.ps1不一样。可以得到MyScript2的文件夹路径。并将其连接到MyScript1。然后执行它。两个脚本必须位于相同的位置。

## MyScript2.ps1 ##
$ScriptPath = Split-Path $MyInvocation.InvocationName
& "$ScriptPath\MyScript1.ps1"

还可以使用以下命令:

$Patch = Join-Path -Path $PSScriptRoot -ChildPath "\patch.ps1"<br>
Invoke-Expression "& `"$Patch`""

一句话解决方案:

& ((Split-Path $MyInvocation.InvocationName) + "\MyScript1.ps1")

I had a problem with this. I didn't use any clever $MyInvocation stuff to fix it though. If you open the ISE by right clicking a script file and selecting edit then open the second script from within the ISE you can invoke one from the other by just using the normal .\script.ps1 syntax. My guess is that the ISE has the notion of a current folder and opening it like this sets the current folder to the folder containing the scripts. When I invoke one script from another in normal use I just use .\script.ps1, IMO it's wrong to modify the script just to make it work in the ISE properly...