3

关于如何在不展开的情况下通过 Get-PSCallStack 的任何想法。它似乎是一个 system.object[] 但从我在网上阅读的内容来看,它们在通过和“展开”时不会保持完整。我尝试在前面放置一个逗号来防止它,但这没有用。

function Pass-Callstack ([System.Object]$arg0) {
Write-Host 'Start Pass-Callstack'
$psCallStack = $arg0
$psCallStackType = $psCallStack.GetType()
$psCallStackLength = $psCallStack.Length
$psCallStackCommand0 = $psCallStack[0].command 
$psCallStackCommand1 = $psCallStack[1].command
Write-Host $psCallStackType
Write-Host $psCallStackLength
Write-Host $psCallStackCommand0
Write-Host $psCallStackCommand1
$arg0 | gm
}

function Describe-Callstack {
Write-Host 'Start Describe-Callstack'
$psCallStack = (Get-PSCallStack)
$psCallStackType = $psCallStack.GetType()
$psCallStackLength = $psCallStack.Length
$psCallStackCommand0 = $psCallStack[0].command 
$psCallStackCommand1 = $psCallStack[1].command
Write-Host $psCallStackType
Write-Host $psCallStackLength
Write-Host $psCallStackCommand0
Write-Host $psCallStackCommand1
$psCallStack | gm
}
Describe-Callstack
Pass-Callstack (,$psCallStack)
4

2 回答 2

6

当您将参数传递给函数而不使用管道时,不会展开集合,例如

function ArgShape($p)
{
    $p.GetType().Fullname
    $p.Rank
    $p.Length
    $p[0].GetType().Fullname
}

ArgShape (Get-PSCallstack)

System.Object[]
1
2
System.Management.Automation.CallStackFrame

此外,如果您期望 Pass-Callstack 参数的数组,您可以像这样指定:

function Pass-Callstack([object[]]$array)

注意使用“系统”。命名空间前缀是可选的。如果找不到类型,PowerShell 将在前面添加。此外,将参数指定为 [object] 本质上是无操作,因为这是默认类型。也就是[object]$arg0一样$arg0

您还将 $null 传递到 Pass-Callstack (尽管包装在单个元素数组中)。变量 $psCallStack 是函数私有的,并且在函数外部不可见,除非您在它前面加上$script:psCallStack. 一般来说,我不推荐这种方法。您应该像这样从 Describe-Callstack 输出 $pscallstack:

function Describe-Callstack { 
Write-Host 'Start Describe-Callstack' 
$psCallStack = (Get-PSCallStack) 
$psCallStackType = $psCallStack.GetType() 
$psCallStackLength = $psCallStack.Length 
$psCallStackCommand0 = $psCallStack[0].command  
$psCallStackCommand1 = $psCallStack[1].command 
Write-Host $psCallStackType 
Write-Host $psCallStackLength 
Write-Host $psCallStackCommand0 
Write-Host $psCallStackCommand1 
$psCallStack 
}

然后将函数调用的输出分配给一个变量:

$cs = Describe-Callstack 

并将其传递到 Pass-Callstack 例如:

Pass-Callstack $cs
于 2010-09-13T15:49:33.787 回答
0

定义一个数组并将你想要的数组放在括号内。

如果返回 null,则返回一个大小为 1 的数组,值为空。

$x = @(CallSomeMethodHere)

所以我想你会想要: $x = @($psCallStack)

于 2012-01-19T17:36:47.057 回答