I have the following string
我有下面的字符串。
$FileNamePattern = 'blah_{4}_{5}_blah_{4}-{2}.CSV'
and I want to replace the numbers in the curly braces with a string of question marks, n characters long
我想把花括号里的数字换成一串问号，n个字符长
As an example I would like it to return 'blah_????_?????_blah_????-??.CSV'
作为一个例子我想它返回blah_ ? ? ? ? _ ? ? ? ? ? _blah_ ? ? ? ? ? ? . csv”
I have this so far, but can't seem to get the 'expansion' in the replace working
到目前为止，我已经有了这个功能，但似乎无法在替换过程中实现“扩展”
[regex]::Replace($FileNamePattern,'{(\d+)}','"?"*$1')
Any help would be greatly appreciated!
如有任何帮助，我们将不胜感激!
Matthew
马太福音
1 个解决方案
#1
2
You need to do the processing of the match inside a callback method:
您需要在回调方法中对匹配进行处理:
$callback = { param($match) "?" * [int]$match.Groups[1].Value }
$FileNamePattern = 'blah_{4}_{5}_blah_{4}-{2}.CSV'
$rex = [regex]'{(\d+)}'
$rex.Replace($FileNamePattern, $callback)
The regex {(\d+)} matches { and } and captures 1+ digits in between. The submatch is parsed as an integer inside the callback (see [int]$match.Groups[1].Value) and then the ? is repeated that amount of times with "?" * [int]$match.Groups[1].Value.
regex {(\d+)}匹配{和}并捕获中间的1+位数。子匹配被解析为回调内部的一个整数(参见[int]$match.Groups[1].Value)，然后是?用“?”来重复这个次数?*(int)$ match.Groups[1]value。
#1
2
You need to do the processing of the match inside a callback method:
您需要在回调方法中对匹配进行处理:
$callback = { param($match) "?" * [int]$match.Groups[1].Value }
$FileNamePattern = 'blah_{4}_{5}_blah_{4}-{2}.CSV'
$rex = [regex]'{(\d+)}'
$rex.Replace($FileNamePattern, $callback)
The regex {(\d+)} matches { and } and captures 1+ digits in between. The submatch is parsed as an integer inside the callback (see [int]$match.Groups[1].Value) and then the ? is repeated that amount of times with "?" * [int]$match.Groups[1].Value.
regex {(\d+)}匹配{和}并捕获中间的1+位数。子匹配被解析为回调内部的一个整数(参见[int]$match.Groups[1].Value)，然后是?用“?”来重复这个次数?*(int)$ match.Groups[1]value。