How to do global replace with -replace in Powershell?

Note that PowerShell's -replace operator is invariably global, i.e. it always looks for and replaces all matches of the given regex.

Use the following -replace operation instead:

@'
111 is different from:
111 is different from:
123 is different from:
567.
'@ -replace '(?m)^(\d{3}) .+:(\r?\n)(?!\1)(?=(\d{3})\b)', 
            '$0  *** This value $3 ***$2'

^{Note: The @'<newline>...<newline>'@ string literal used for the multiline input text is a so-called here-string.}

Output:

111 is different from:
111 is different from:
  *** This value 123 ***
123 is different from:
  *** This value 567 ***
567.

• For a detailed explanation of the regex and the ability to experiment with it, see this regex101.com page, but in short:

  • (?m) is the inline form of the Multiline .NET regex option, which makes ^ and $ match at the start and end of each line.

  • ^(\d{3}) therefore matches a 3-digit sequence only at the start of a line, in a capture group, and .+: matches a space and at least one additional character on the same line all the way to a : at the end.

  • (\r?\n) captures the specific newline sequence encountered (which may be CRLF (Windows-format) or just LF (Unix-format)) in a 2nd capture group.

    • Capturing the specific newline sequence allows you to replicate it in the substitution string via placeholder $2, to ensure that the newly inserted line is terminated with the same sequence.

    • If you don't care about potentially mixing \r\n and \n in the resulting string, you could omit the 2nd capture group and use "`n" (sic) or "`r`n" instead, using an expandable string ("...") with an escape sequence - note that, unlike in C#, \r and \n are not recognized in PowerShell string literals (it is only the .NET regex engine that recognizes them, but not in the substitution operand of -replace, which is not a regex, and where only $-prefixed placeholders are recognized).

      # Conceptually cleaner: separate the verbatim part from
      # the expandable part.
      ('$0  *** This value $2 ***' + "`n")
      
      # Alternative, using a single "..." string
      # The '$' chars. that are part of -replace *placeholders*
      # must be *escaped as `$* to prevent up-front expansion by PowerShell
      "`$0  *** This value `$2 ***`n"
      

  • (?!\1)(?=(\d{3})\b) uses both a negative ((?!...)) and positive (?=...) lookahead assertion to look for 3 digits at the start of the next line (at a word boundary, due to \b) that aren't the same as the 3 digits on the current line (\1 being a backreference to what the 1st capture group matched).

    • Note that using a capture group inside an overall by-definition non-capturing lookaround assertion is possible, and indeed used above to capture the 3-digit sequence at the start of the subsequent line, referenced via placeholder $3 in the substitution string.

  • In the substitution string, $0, $2 and $3 refer to the what the entire regex, the 2nd capture group, and the 3rd one captured, respectively ($& may be used in lieu of $0; see this answer for more info about these placeholders).

    • Note that by using a string as the substitution operand, you are limited to embedding captured text as-is, via placeholders as such as $0 (see this answer for more info about these placeholders). If you need to determine the substitution text fully dynamically, i.e. if it needs to apply transformations based on each match:

      • In PowerShell (Core) 7+, you can use a script block { ... } instead.

      • In Windows PowerShell, you'll have to call the underlying [regex]::Replace() method directly.

    • See below.

To spell out the fully dynamic substitution approach, adding 1 to the captured number in this example:

PowerShell (Core) 7+ solution, using a script block ({ ... }) as -replace's substitution operand:

@'
111 is different from:
111 is different from:
123 is different from:
567.
'@ -replace '(?m)^(\d{3}) .+:(\r?\n)(?!\1)(?=(\d{3})\b)', {
               '{0}  *** This value + 1: {1} ***{2}' -f $_.Value, ([int] $_.Groups[3].Value + 1), $_.Groups[2].Value
            }

Windows PowerShell solution, where a direct call to the underlying [regex]::Replace() method is required:

$str = @'
111 is different from:
111 is different from:
123 is different from:
567.
'@

[regex]::Replace(
  $str, 
  '(?m)^(\d{3}) .+:(\r?\n)(?!\1)(?=(\d{3})\b)', 
  {
    param($m)
    '{0}  *** This value + 1: {1} ***{2}' -f $m.Value, ([int] $m.Groups[3].Value + 1), $m.Groups[2].Value
  }
)

Output (note that 1 has been added to each captured value):

111 is different from:
111 is different from:
  *** This value + 1: 124 ***
123 is different from:
  *** This value + 1: 568 ***
567.