PowerShell 函数不会返回对象 [英] PowerShell function won't return object

问题描述

我有一个创建通用列表的简单函数:

I have a simple function that creates a generic List:

function test()
{
    $genericType = [Type] "System.Collections.Generic.List``1"
    [type[]] $typedParameters = ,"System.String"
    $closedType = $genericType.MakeGenericType($typedParameters)
    [Activator]::CreateInstance($closedType)
}

$a = test

问题是无论我尝试什么， $a 始终为空.如果我在函数之外执行相同的代码，它可以正常工作.

The problem is that $a is always null no matter what I try. If I execute the same code outside of the function it works properly.

想法?

推荐答案

恕我直言，这是陷阱 #1.如果您从以某种方式可枚举的函数中返回一个对象(我不确切知道是否实现 IEnumerable 是唯一的情况)，PowerShell 会展开该对象并返回其中的项目.

IMHO that's pitfall #1. If you return an object from the function that is somehow enumerable (I don't know exactly if implementing IEnumerable is the only case), PowerShell unrolls the object and returns the items in that.

您新创建的列表是空的，因此没有返回任何内容.要使其工作，只需使用此:

Your newly created list was empty, so nothing was returned. To make it work just use this:

,[Activator]::CreateInstance($closedType)

这将创建一个展开的单项数组，并将该项(通用列表)分配给 $a.

That will make an one item array that gets unrolled and the item (the generic list) is assigned to $a.

更多信息

以下是类似问题的列表，可帮助您了解正在发生的事情:

Here is list of similar question that will help you to understand what's going on:

• Powershell 陷阱
• 在 Powershell 中避免不可知的锯齿状数组展平
• PowerShell 函数返回数据集/数据表的奇怪行为
• 什么决定了 Powershell 管道是否会展开收藏?

注意:不需要用括号声明函数头.如果需要添加参数，函数如下:

Note: you dont need to declare the function header with parenthesis. If you need to add parameters, the function will look like this:

function test {
 param($myParameter, $myParameter2)
}

或

function  {
param(
  [Parameter(Mandatory=true, Position=0)]$myParameter,
  ... again $myParameter2)
...

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