Powershell正则表达式用可变长度的字符串替换数字 [英] Powershell regular expression to replace a digit with a string of variable length

问题描述

我有以下字符串

  $ FileNamePattern ='blah_ {4} _ {5} _blah_ {4}  -  { 2} .CSV'
  

我想用大括号中的数字替换一个问题的字符串标记，n个字符长

作为一个例子，我希望它返回'blah _ ???? _ ????? _ blah_？ ?? - ??。CSV'

我到目前为止，但似乎没有得到扩展的替换工作

  [regex] ::替换（$ FileNamePattern，'{（\d +）}'，'？ $ 1'）
  

任何帮助将不胜感激！

Matthew

解决方案

您需要在回调方法内进行匹配处理：

  $ callback = {param（$ match）？ * [int] $ match.Groups [1] .Value} 
 $ FileNamePattern ='blah_ {4} _ {5} _blah_ {4}  -  {2} .CSV'
 $ rex = [regex ]'{（\d +）}'
 $ rex.Replace（$ FileNamePattern，$ callback）
  

正则表达式 {（\d +）} 匹配 {和} 并在其间捕获1位数字。该子对象在回调中被解析为一个整数（参见 [int] $ match.Groups [1] .Value ）然后？重复使用？ * [int] $ match.Groups [1] .Value 。

I have the following string

$FileNamePattern =  'blah_{4}_{5}_blah_{4}-{2}.CSV'

and I want to replace the numbers in the curly braces with a string of question marks, n characters long

As an example I would like it to return 'blah_????_?????_blah_????-??.CSV'

I have this so far, but can't seem to get the 'expansion' in the replace working

[regex]::Replace($FileNamePattern,'{(\d+)}','"?"*$1')

Any help would be greatly appreciated!

Matthew

解决方案

You need to do the processing of the match inside a callback method:

$callback = {  param($match) "?" * [int]$match.Groups[1].Value }
$FileNamePattern =  'blah_{4}_{5}_blah_{4}-{2}.CSV'
$rex = [regex]'{(\d+)}'
$rex.Replace($FileNamePattern, $callback)

The regex {(\d+)} matches { and } and captures 1+ digits in between. The submatch is parsed as an integer inside the callback (see [int]$match.Groups[1].Value) and then the ? is repeated that amount of times with "?" * [int]$match.Groups[1].Value.

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