powershell - Powershell Select-String:如何获取完整的路径和上下文行

Question

我有这个工作

select-string -path C:\test\*.txt -pattern "test" -allmatches –simplematch | foreach-object {
    Write-Host $_.Filename
    Write-Host $_.LineNumber
    Write-Host $_.Line
    Write-Host $_.context.postcontext
 }

但是我也想获得FullPath而不只是FileName。我也想在Line之前和之后获得上下文行，但似乎无法使其正常工作。

Answer 1

该文件路径在Path -property中可用。要获得上下文，您需要首先指定在使用-Context 2(前后2行)或-Context 1,2(之前1行，之后2行)之前和之后捕获多少行。例如

select-string -path C:\test\*.txt -pattern "test" -AllMatches -SimpleMatch -Context 1,1 | foreach-object {
    Write-Host
    Write-Host "MATCH!"
    Write-Host "-------------"
    Write-Host "Path $($_.Path)"
    Write-Host "Line number: $($_.LineNumber)"
    Write-Host "Before: $($_.Context.Precontext)"
    Write-Host "Line: $($_.Line)"
    Write-Host "After: $($_.Context.Postcontext)"
 }

输入样例(Text1.txt):

Line 1
Line 2
Line test
Line 4
Line 5

样本输出:

MATCH!
-------------
Path C:\test\Text2.txt
Line number: 4
Before: Line 3
Line: Line test
After: Line 5

MATCH!
-------------
Path C:\test\Text1.txt
Line number: 3
Before: Line 2
Line: Line test
After: Line 4