在 PowerShell 的替换运算符中使用格式运算符 [英] Using Format Operator within Replace Operator in PowerShell

问题描述

我正在尝试将我的文件 "Introduction _ C# _ Tutorial 1" 重命名为类似 "01.Introduction" 的名称.它需要一个 -replace 运算符以及一个 -f 运算符来填充索引号.我的代码是这样的:

I'm trying to rename my files "Introduction _ C# _ Tutorial 1" to something like "01.Introduction". It needs a -replace operator as well as a -f operator to zero pad the index number. My code is like:

$string = "Introduction _ C# _ Tutorial 1"
if ($string -match "^([^_]+)_[^\d]+(\d{1,2})$") {
    $Matches[0] -replace "^([^_]+) _[^\d]+(\d{1,2})$", ("{0:d2}. {1}" -f '$2', '$1')
    }

然而，输出就像缺少 -f 运算符:
<代码>1.简介
我怎样才能得到预期的结果?

The output, however, is like the -f operator being absent:
1. Introduction
How can I get the expected result?

顺便说一句，有没有一种简单的方法来获得 $matches 结果，而无需在 -match 语句之前加上 -match 语句或组合 -match> 对单行代码的声明?

By the way, is there a simple way to get the $matches result without a -match statement before it or combine the -match statement to a one-liner code?

推荐答案

-match 已经填充自动变量 $Matches，

The -match already fills the automatic variable $Matches,

> $Matches

Name                           Value
----                           -----
2                              1
1                              Introduction
0                              Introduction _ C# _ Tutorial 1

所以根本不需要 -replace 并重复 RegEx.

so there is no need for the -replace at all and repeat the RegEx.

但是您需要将数字转换为整数.

But you need to cast the number to an int.

$string = "Introduction _ C# _ Tutorial 1"
if ($string -match "^([^_]+)_[^\d]+(\d{1,2})$") {
    "{0:D2}. {1}" -f [int]$matches[2],$matches[1]
}

示例输出:

01. Introduction

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