If 声明反对动态变量 [英] If Statement Against Dynamic Variable

问题描述

我正在尝试做类似于以下的事情......

I am attempting to do something similar to the following ...

New-Variable -Name "state_$name" -Value "True"
if ("state_$name" -eq "True") {
    Write-Host "Pass"
} else {
    Write-Host "Fail"
}

我已经尝试了多种不同的方法，但它并没有完全按照我希望的方式工作.我需要编写 if 语句来说明动态变量，因为这些值将在 foreach 循环内更改.

I have attempted this a number of different ways but it is not working exactly how I would like it to work. I need to write the if statement to account for a dynamic variable as these values will change inside of a foreach loop.

我在上面提供了一个简单的例子来证明概念.

I have provided a simple example above for proof of concept.

推荐答案

替换

if ("state_$name" -eq "True") {

与:

if ((Get-Variable -ValueOnly "state_$name") -eq "True") {

也就是说，如果你的变量名只是间接，通过一个可扩展的字符串，你不能直接引用它(就像你通常使用 $ 符号一样) -您需要通过Get-Variable获取其值，如上所示.

That is, if your variable name is only known indirectly, via an expandable string, you cannot reference it directly (as you normally would with the $ sigil) - you need to obtain its value via Get-Variable, as shown above.

但是，正如 JohnLBevan 指出的那样，您可以存储变量 object在另一个(非动态)变量中，以便您可以通过 .Value 属性获取和设置动态变量的值.
在New-Variable调用中添加-PassThru直接返回变量对象，不需要后续的Get-Variable调用:

However, as JohnLBevan points out, you can store the variable object in another (non-dynamic) variable, so that you can then get and set the dynamic variable's value via the .Value property.
Adding -PassThru to the New-Variable call directly returns the variable object, without the need for a subsequent Get-Variable call:

$dynamicVarObject = New-Variable -Name "state_$name" -Value "True" -PassThru
if ($dynamicVarObject.Value -eq "True") {
    "Pass"
} else {
    "Fail"
}

<小时>

也就是说，以这种方式创建变量通常有更好的替代方法，例如使用哈希表:

$hash = @{}
$hash.$name = 'True'

if ($hash.$name -eq 'True') { 'Pass' } else { 'Fail' }

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