Question

我在powershell中运行一个脚本

MySchema:
 type: object
 required: [property1, property2, property3]
 properties:
  property1:
   type: integer
  property2:
   type: integer
  property3:
   type: integer

然后我想从输入参考值$ i

调用一个名为$ Tom的变量的值

./name -i tom

这将打印：

$tom = 29 
$andrew = 99
$bill = 5

Echo $i's age is $i

Answer 1

在powershell中，这将是这样的： name.ps1的内容：

$person = $args
$ages = @{"Tom" = 23;
  "Andrew" = 99;
  "Bill" = 5}

$age = $ages[$person]

Write-Host "$person's age is $age"

你会这样称呼它

.\name.ps1 "tom"

$args包含您发送给脚本的所有参数。因此，如果您按照以下方式调用脚本：.\name.ps1 "tom" "bill"，则结果为：tom bill's age is 23 5

Answer 2

我会使用哈希表，但如果您有全局变量，则可以使用以下命令：

#variables
$tom = 29 
$andrew = 99
$bill = 5

#your parameter
$i = "tom"

#echo
Echo "$i's age is $((Get-Variable | ? {$_.Name -eq $i}).Value)"

Answer 3

提供的替代方法。更多代码，可以说是矫枉过正，但我认为掌握PowerShell的param功能是件好事。

# PowerShell's native argument/parameter support
param(
    [string]$name
)

# Create an array with Name and Age properties as hashtable.
$people = @(
            @{ Name = "Tom"    ; Age = 29},
            @{ Name = "Andrew" ; Age = 99},
            @{ Name = "Bill"   ; Age = 5}
            )

# Find the person by comparing the argument to what is in your array
$person = $people | Where-Object {$_.Name -eq $name}

# Single name version: If the person is found, print what you would like. Otherwise let the user know name not found
if($person -ne $null){
    Write-Host "$($person.Name) is $($person.Age) years old"
}else{
    Write-Host "$name not found in list."
}

<# Multiple name version : get rid of the param section

foreach ($name in $args){
    $person = $people | Where-Object {$_.Name -eq $name}
    if($person -ne $null){
        Write-Host "$($person.Name) is $($person.Age) years old"
    }else{
        Write-Host "$name not found in list."
    }
}
#>

从变量值调用变量的值

3 个答案: