Question

我有一个奇怪的问题。我有一个函数应该返回1或2个值，一个字母和一个数字。出于某种原因，它仅在我将返回指定为

时才有效

return $x, $y

但它不会像这样工作：

return $x
return $y

代码：

$ModelsDesktop = @("Dimension","Optiplex")
$ModelsLaptop = @("Latitude","Venue")

<#
Returns L or D depending on the Computer Name. Sets U if the model is uncertain.
#>
Function Get-TypeByComputerName{
    [CmdletBinding()]
    param (
        [Parameter(Mandatory = $true, ValueFromPipeline = $true, Position = 0)]
        [ValidateNotNullOrEmpty()]
        [string]$ComputerName
    )

    Process {
        if ($ComputerName -like "*-L-*" -or $ComputerName -like "*-LT-*") {
            $ModelType = "L"
        }

        elseif ($ComputerName -like "*-D-*" -or $ComputerName -like "*-WRK-*") {
            $ModelType = "D"
        }

        else {
            $ModelType = "U" #unsure
        }

        return $ModelType
    }

}


<#
Returns L or D depending on the Computer Model. Sets U if the model is uncertain.
#>
Function Get-TypeByModel{
    [CmdletBinding()]
    param (
        [Parameter(Mandatory = $true, ValueFromPipeline = $true, Position = 0)]
        [ValidateNotNullOrEmpty()]
        [string]$Model
    )

    Process {
        if (($ModelsLaptop | %{($Model) -like("*$_*")}) -contains $true) {
            $ModelType = "L"
        }

        elseif (($ModelsDesktop | %{($Model) -like("*$_*")}) -contains $true) {
            $ModelType = "D"
        }

        else {
            $ModelType = "U"
        }

        return $ModelType
    }

}


<#
Returns L or D depending on the Computer Name and Model. Sets a flag if the model is uncertain.
#>
Function Get-Type{
    [CmdletBinding()]
    param (
        [Parameter(Mandatory = $true)]
        [ValidateNotNullOrEmpty()]
        [string]$ComputerName,

        [Parameter(Mandatory = $true)]
        [ValidateNotNullOrEmpty()]
        [string]$Model
    )

    Process {
        if ((($ComputerName | Get-TypeByComputerName) -eq ($Model | Get-TypeByModel)) -and (($ComputerName | Get-TypeByComputerName) -ne "U")) {
            $ModelType = ($ComputerName | Get-TypeByComputerName)
        }

        elseif (($ComputerName | Get-TypeByComputerName) -ne "U") {
            $ModelType = ($ComputerName | Get-TypeByComputerName)
            $Flag = 1
        }

        elseif (($Model | Get-TypeByModel) -ne "U") {
            $ModelType = ($Model | Get-TypeByModel)
            $Flag = 1
        }

        else {
            $ModelType = "D"
            $Flag = 1
        }

        return $ModelType
        return $Flag
    }

}

价值：

$test = New-Object psobject -Property @{ComputerName="crd-l-02-00001";Model="opti 343"}

输出2个返回语句（如前面的代码所示）：

$test

ComputerName   Model   
------------   -----   
crd-l-02-00001 opti 343



PS C:\Users\u0096902> (Get-Type -ComputerName $test.ComputerName -Model $test.Model)
L

输出更正后的＆＃34;返回$ ModelType，$ Flag＆＃34;：

$test

ComputerName   Model   
------------   -----   
crd-l-02-00001 opti 343



PS C:\Users\u0096902> (Get-Type -ComputerName $test.ComputerName -Model $test.Model)
L
1

我错过了什么？似乎无法解决这个问题。它似乎只返回第一个＆＃34; return＆＃34;，但我不知道为什么。

此示例代码似乎完全正常：

function get-multiplereturnvalues {
  "Return value 1"
  "Return value 2"
}

$return = get-multiplereturnvalues
$return[0]  # Outputs "Return value 1"
$return[1]  # Outputs "Return value 2"

Answer 1

＆＃34;在计算机编程中，return语句导致执行离开当前子例程，并在调用子例程之后立即恢复到代码中的点，称为其返回地址。＆＃34;

一旦你致电回复，你就说你已经完成了这项功能。

Answer 2

来自Get-Help about_Return：

详细说明 Return关键字退出函数，脚本或脚本块。有可能用于退出特定点的范围，返回值或指示已达到范围的结束。

你的第一个Return强制立即退出该功能，所以第二个永远不会有机会运行。

Powershell仅返回2个值，使用＆＃34;返回x，y＆＃34;

2 个答案: