如何使用 PowerShell Get-Content -replace 仅替换一部分字符串

米尼亚·克里斯蒂安·马林

我有一个文件，我需要在其中修改 URL，但不知道该 URL 包含什么，就像在下面的示例中一样：在文件 file.txt 中，我必须替换 URL，以便它是“https://SomeDomain /Release/SomeText”或“https://SomeDomain/Staging/SomeText”到“https://SomeDomain/Deploy/SomeText”。因此，无论 SomeDomain 和 SomeText 之间写什么，都应该用已知的字符串替换。是否有任何正则表达式可以帮助我实现这一目标？

我曾经使用以下命令来做到这一点”

((Get-Content -path "file.txt" -Raw) -replace '"https://SomeDomain/Release/SomeText");','"https://SomeDomain/Staging/SomeText");') | Set-Content -Path "file.txt"

这工作正常，但我必须在执行命令之前知道 file.txt 中的 URL 是否包含 Release 或 Staging。

谢谢！

关注

您可以使用正则表达式执行此操作，您可以-replace在其中捕获要保留的部分并使用反向引用重新创建新字符串

$fileName = 'Path\To\The\File.txt'
$newText  = 'BLAHBLAH'

# read the file as single multilined string
(Get-Content -Path $fileName -Raw) -replace '(https?://\w+/)[^/]+(/.*)', "`$1$newText`$2" | Set-Content -Path $fileName

正则表达式详细信息：

(              Match the regular expression below and capture its match into backreference number 1
   http        Match the characters “http” literally
   s           Match the character “s” literally
      ?        Between zero and one times, as many times as possible, giving back as needed (greedy)
   ://         Match the characters “://” literally
   \w          Match a single character that is a “word character” (letters, digits, etc.)
      +        Between one and unlimited times, as many times as possible, giving back as needed (greedy)
   /           Match the character “/” literally
)             
[^/]           Match any character that is NOT a “/”
   +           Between one and unlimited times, as many times as possible, giving back as needed (greedy)
(              Match the regular expression below and capture its match into backreference number 2
   /           Match the character “/” literally
   .           Match any single character that is not a line break character
      *        Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
)

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