PowerShell：在字串中使用未宣告的變數

我試圖弄清楚如何在字串中使用未宣告的變數。這是一個簡化的示例來演示我的問題

[string]$sentence="My dog is $age years old"
[int]$age = 6 
$sentence | write-output

$age = 3
$sentence | write-output

尋找這個輸出：

   My dog is 6 years old

   My dog is 3 years old

得到這個輸出：

   My dog is  years old

   My dog is  years old

我嘗試了以下方法：

  "My dog is `$age years old" #My dog is $age years old
  $("My dog is `$age years old") #My dog is $age years old
  $("My dog is $age years old") #My dog is 3 years old

有沒有可能以任何方式使它更緊湊？還是這是唯一的方法？

編輯：我注意到重點是字串。我將在有問題的地方分享我的代碼。我需要將一堆從 20 行到 2000 行的 VBS 腳本（>40）遷移到 Powershell

VBS 檔案：

    Function InitializeApp

        Dim sPath
        
        ...

    End function 

Powershell腳本：

$vbs=Get-Content -path $path #get content in array, one line per index
$rules=
@(
    [pscustomobject]@{
        "name"="func_start";
        "vbs" = 'Function (?<V1>\w*)'; # The regex named group V1 will initialized when match is found
        "ps" = {"Function $1() `{`n"} # Group V1 is to be inserted in replacement string
    }
)

function Invoke-TestAndReplace($line,$ruleName){
    $r_vbs=$($rules.where({$_.name -eq $ruleName}).vbs) # Function (?<V1>\w*)
    $r_ps=$($rules.where({$_.name -eq $ruleName}).ps) # {"Function $1() `{`n"}
    if( $regex = ($line | select-string -pattern $r_vbs )) {
        $0=$regex[0].Matches.Groups.Where({$_.name -eq "0"}).value # Function InitializeApp
        $1=$regex[0].Matches.Groups.Where({$_.name -eq "V1"}).value # InitializeApp
        $2=$regex[0].Matches.Groups.Where({$_.name -eq "V2"}).value #
        $r_ps = & $r_ps # Function InitializeApp() {
        $replace = $line.replace($0,$r_ps) # Function InitializeApp -> Function InitializeApp() {
    } else {$replace = $line} #keep old value
    return $replace
}
$newVBS=@()
foreach($line in $vbs){
    $line = Invoke-TestAndReplace -line $line -rulename "func_start"
}

uj5u.com熱心網友回復：

我不知道如何使用插值來做到這一點，但我可以使用較舊的 .NetString.Format()方法來做到這一點，如下所示：

$sentence = "My dog is {0} years old."

$age = 6
[string]::Format($sentence, $age) | Write-Output
# => My dog is 6 years old

$age = 3
[string]::Format($sentence, $age) | Write-Output
# => My dog is 3 years old

[string]::Format($sentence, 12)   | Write-Output
# => My dog is 12 years old

我們可以這樣縮短（感謝評論者）：

$sentence = "My dog is {0} years old."

$age = 6
Write-Output ($sentence -f $age)
# => My dog is 6 years old

$age = 3
Write-Output ($sentence -f $age)
# => My dog is 3 years old

Write-Output ($sentence -f 12)
# => My dog is 12 years old

還要注意模板字串中的不同占位符。

uj5u.com熱心網友回復：

這個答案顯示了Joel Coehoorn的好答案的一個過于復雜的替代方案。您可以使用Script Block來存盤運算式（包括您的$age變數）然后執行它。

$sentence = { "My dog is $age years old" }
$age = 6 
& $sentence | write-output # => My dog is 6 years old

$age = 3
& $sentence | write-output # => My dog is 3 years old

如果您希望腳本塊在創建變數時“記住”變數的值，您可以使用它的.GetNewClosure()方法，例如：

$expressions = foreach($i in 0..5) {
    { "Value of `$i was $i in this iteration" }.GetNewClosure()
}
$expressions.foreach{ & $_ }

uj5u.com熱心網友回復：

您可以使用ExpandString

$sentence = 'My dog is $age years old.'

$age = 4
$ExecutionContext.InvokeCommand.ExpandString($sentence)
# My dog is 4 years old.

$age = 5
$ExecutionContext.InvokeCommand.ExpandString($sentence)
# My dog is 5 years old.

您還可以定義一個函式并呼叫該函式

function sentence([int]$age) {
    "My dog is $age years old"
}

1..5 | ForEach-Object { sentence -age $_ }

# My dog is 1 years old
# My dog is 2 years old
# My dog is 3 years old
# My dog is 4 years old
# My dog is 5 years old

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