如何在PowerShell中将字符串转换为整数 [英] How to convert string to integer in PowerShell

问题描述

我有一个带数字的目录列表.我必须找到最大的数字并将其增加 1 并创建一个具有该增量值的新目录.我能够对下面的数组进行排序，但我无法增加最后一个元素，因为它是一个字符串.

I have a list of directories with numbers. I have to find the highest number and and increment it by 1 and create a new directory with that increment value. I am able to sort the below array, but I am not able to increment the last element as it is a string.

如何将下面的数组元素转换为整数?

How do I convert this below array element to an integer?

PS C:\Users\Suman\Desktop> $FileList

Name
----
11
2
1

推荐答案

您可以在变量之前指定类型以强制其类型.这称为(动态)转换(更多信息在此处):

You can specify the type of a variable before it to force its type. It's called (dynamic) casting (more information is here):

$string = "1654"
$integer = [int]$string

$string + 1
# Outputs 16541

$integer + 1
# Outputs 1655

<小时>

作为示例，以下代码段向 $fileList 中的每个对象添加了一个 IntVal 属性，其中包含 Name 的整数值属性，然后在这个新属性上对 $fileList 进行排序(默认为升序)，取最后一个(最高的 IntVal)对象的 IntVal 值，递增最后创建一个以它命名的文件夹:

---

As an example, the following snippet adds, to each object in $fileList, an IntVal property with the integer value of the Name property, then sorts $fileList on this new property (the default is ascending), takes the last (highest IntVal) object's IntVal value, increments it and finally creates a folder named after it:

# For testing purposes
#$fileList = @([PSCustomObject]@{ Name = "11" }, [PSCustomObject]@{ Name = "2" }, [PSCustomObject]@{ Name = "1" })
# OR
#$fileList = New-Object -TypeName System.Collections.ArrayList
#$fileList.AddRange(@([PSCustomObject]@{ Name = "11" }, [PSCustomObject]@{ Name = "2" }, [PSCustomObject]@{ Name = "1" })) | Out-Null

$highest = $fileList |
    Select-Object *, @{ n = "IntVal"; e = { [int]($_.Name) } } |
    Sort-Object IntVal |
    Select-Object -Last 1

$newName = $highest.IntVal + 1

New-Item $newName -ItemType Directory

Sort-Object IntVal 不是必需的，因此您可以删除它如果您愿意.

Sort-Object IntVal is not needed so you can remove it if you prefer.

[int]::MaxValue = 2147483647 所以你需要使用超出这个值的 [long] 类型 ([long]::MaxValue = 9223372036854775807).

[int]::MaxValue = 2147483647 so you need to use the [long] type beyond this value ([long]::MaxValue = 9223372036854775807).

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