PowerShell：选择匹配前的行 - 使用输入字符串变量时的Select-String -Context问题

Question

I need return a line preceeding a match on a multi-line string variable.

It seems when using a string variable for the input Select-String considers the entire string as having matched. As such the Context properties are "outside" either end of the string and are null.

Consider the below example:

$teststring = @"
line1
line2
line3
line4
line5
"@

Write-Host "Line Count:" ($teststring | Measure-Object -Line).Lines #verify PowerShell does regard input as a multi-line string (it does)

Select-String -Pattern "line3" -InputObject $teststring -AllMatches -Context 1,0 | % {
$_.Matches.Value #this prints the exact match
$_.Context #output shows all context properties to be empty 
$_.Context.PreContext[0] #this would ideally output first line before the match
$_.Context.PreContext[0] -eq $null #but instead is null
}

Am I misunderstanding something here?

What is the best way to return "line2" when matching for "line3"?

Thanks!

Edit: Additional requirements I neglected to state: Needs to provide the line above ALL matched lines for a string of indeterminate length. EG when searching the below for "line3" I need to return "line2" and "line5".

line1
line2
line3
line4
line5
line3
line6

Answer 1

Select-String operates on arrays of input, so rather than a single, multiline string you must provide an array of lines for -Context and -AllMatches to work as intended:

$teststring = @"
line1
line2
line3
line4
line5
line3
line6
"@

$teststring -split '\r?\n' | Select-String -Pattern "line3" -AllMatches -Context 1,0 | % {
  "line before:  " + $_.Context.PreContext[0]
  "matched part: " + $_.Matches.Value  # Prints the what the pattern matched
}

This yields:

line before:  line2
matched part: line3
line before:  line5
matched part: line3

• $teststring -split '\\r?\\n' splits the multi-line string into an array of lines:

  • Note: What line-break sequences your here-document uses (LF-only vs. CRLF) depends on the enclosing script file; regex \\r?\\n handles either style.

• Note that it is crucial to use the pipeline to provide Select-String 's input; if you used -InputObject , the array would be coerced back to a single string.

---

Select-String is convenient, but slow .
Especially for a single string already in memory, a solution using the .NET Framework's [Regex]::Matches() method will perform much better , though it is more complex .

Note that PowerShell's own -match and -replace operators are built on the same .NET class, but do not expose all of its functionality; -match - which does report capture groups in the automatic $Matches variable - is not an option here, because it only ever returns 1 match.

^{The following is essentially the same approach as in mjolinor's answer answer, but with several problems corrected[1].}

# Note: The sample string is defined so that it contains LF-only (\n)
#       line breaks, merely to simplify the regex below for illustration.
#       If your script file use LF-only line breaks, the 
#       `-replace '\r?\n', "`n" call isn't needed.
$teststring = @"
line1
line2
line3
line4
line5
line3
line6
"@ -replace '\r?\n', "`n" 

[Regex]::Matches($teststring, '(?:^|(.*)\n).*(line3)') | ForEach-Object { 
  "line before:  " + $_.Groups[1].Value
  "matched part: " + $_.Groups[2].Value
}

• Regex (?:^|(.*)\\n).*(line3) uses 2 capture groups ( (...) ) to capture both the (matching part of) the line to match and the line before ( (?:...) is an auxiliary non -capturing group that is needed for precedence):

  • (?:^|(.*)\\n) matches either the very start of the string ( ^ ) or ( | ) any - possibly empty - sequence of non-newline characters ( .* ) followed by a newline ( \\n ); this ensures that the line to match is also found when there is no preceding line (ie, of the line to match is the first one).
  • (line3) is the group defining the line to match; it is preceded by .* to match the behavior in the question, where pattern line3 is found even it is only part of a line.
    • If you want only full lines to match, use the following regex instead:
      (?:^|(.*)\\n)(line3)(?:\\n|$)

• [Regex]::Matches() finds all matches and returns them as a collection of System.Text.RegularExpressions.Match objects, which the ForEach-Object cmdlet call can then operate on to extract the capture-group matches ( $_.Groups[<n>].Value ).

---

<sup>[1] As of this writing:
- There is no need to match twice - the enclosing if ($teststring -match $pattern) { ... } is unnecessary.
- Inline option (?m) is not needed, because . does not match newlines by default .
- (.+?) captures only nonempty lines (and ? , the non-greedy quantifier, is not needed).
- If the line of interest is the first line - ie, if there's no line before , it won't be matched.</sup>

Answer 2

You can use a multi-line regex, with the -match operator:

$teststring = @"
line1
line2
line3
line4
line5
line3
line6
"@

$pattern = 
@'
(?m)
(.+?)
line3
'@


if ($teststring -match $pattern)
  { [Regex]::Matches($teststring,$pattern) |
    foreach {$_.groups[1].value} }