Remove All Values In A String Before 7th backslash

Question

I have a path inside a txt file that needs to be manipulated, something like:

C:\Jenkins\Automation\Blah\Foo\Bar\646433\Schema\test\473289_12321.ps1 C:\Jenkins\Automation\Blah\Foo\Bar\2112\Schema\QA\473289_123211.ps1

I want to replace everything before the 7th backslash and then replace it with C:\Question. I was doing something similar in Powershell via:

(Get-Content $FullEnvSQLFilePath) |  #get the content
Foreach-Object {$_ -replace [Regex]::Escape($StringToReplace), "$StringReplaceValue"} | #look for the string and replace

This worked fine when I knew what the exact verbiage was to look for. We now don't know that but we will want to remove everything before the 7th backslash and replace it with a value. Reverse order works fine too. I wasn't able to have much luck in Powershell via substring doing this. Thanks.

Why restricting to 7th backslash? In case you would relocate your powershell script elsewhere you will have to use 6th backslash instead. I'd say look [here](http://stackoverflow.com/questions/10972589/get-relative-path-of-files-in-sub-folders-from-the-current-directory) to get relative path to some root location, then you can use that path to form correct destination location like [here](http://stackoverflow.com/questions/30573854/powershell-copy-files-and-folders/30574964#30574964). — Vesper, Jun 02 '15 at 06:36

score 2 · Answer 1 · answered Jun 02 '15 at 14:36

2

One option:

$text = 
'C:\Jenkins\Automation\Blah\Foo\Bar\646433\Schema\test\473289_12321.ps1',
'C:\Jenkins\Automation\Blah\Foo\Bar\2112\Schema\QA\473289_123211.ps1'

$text | foreach {'C:\Question\{0}' -f $_.split('\',8)[-1]}
C:\Question\Schema\test\473289_12321.ps1
C:\Question\Schema\QA\473289_123211.ps1

answered Jun 02 '15 at 14:36

mjolinor

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does this perform faster than a regex? – Vincent De Smet Jun 02 '15 at 23:12
Haven't benchmarked it, but I'd expect it would. Typically string method operations are faster than PS operators, and among those operators, the regex operations (-match, -replace, -split) tend to be the slowest. – mjolinor Jun 03 '15 at 00:01
Thank you. I have a problem where line 1 is PRINT c:\blah\blah\blah etc. My process removes everything as soon as it finds 7 backslashes in a line. Can I have it do the replace ONLY if either the line starts with :R or if it doesn't start with PRINT and it matches 7 backslashes? – Ali Razeghi - AWS Jun 03 '15 at 16:36
That's getting outside the range of the original question. I'd post a new question for that. – mjolinor Jun 03 '15 at 17:55

Vincent De Smet · Accepted Answer · 2015-06-05T04:57:46.700

1

this ([^\\]*\\){7} regex looks 7 times for a capture group ending on a backslash and replaces it.

UPDATED: .:\\([^\\]*\\){6} regex looks for strings that look like paths starting at any root drive .:\ followed by 6 times a capture group ending on a backslash based on your comment

$text = @"
C:\Jenkins\Automation\Blah\Foo\Bar\646433\Schema\test\473289_12321.ps1
C:\Jenkins\Automation\Blah\Foo\Bar\2112\Schema\QA\473289_123211.ps1
PRINT C:\Jenkins\Automation\Blah\Foo\Baz\2112\Schema\QA\473289_123212.ps1
C:\Jenkins\Automation\Blah\Foo\quux\2112\Schema\QA\473289_123213.ps1
"@
#depending on how you get the text (single string or array)
#$text.Split("`n") | % { $_ -Replace '.:\\([^\\]*\\){6}','C:\Example\' }
$text -Replace ".:\\([^\\`n]*\\){6}","C:\Example\"

Result:

C:\Example\Schema\test\473289_12321.ps1
C:\Example\Schema\QA\473289_123211.ps1
PRINT C:\Example\Schema\QA\473289_123212.ps1
C:\Example\Schema\QA\473289_123213.ps1

edited Jun 05 '15 at 04:57

answered Jun 02 '15 at 07:56

Vincent De Smet

4,859
2
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if you use `Get-Content` you may not need the split on newline and `For-EachObject` (`%`) script block and apply regex replace direct. – Vincent De Smet Jun 02 '15 at 07:59
I excluded newline in capture group, so `%` is no longer needed – Vincent De Smet Jun 02 '15 at 08:13
Thank you. I have a problem where line 1 is PRINT c:\blah\blah\blah etc. My process removes everything as soon as it finds 7 backslashes in a line. Can I have it do the replace ONLY if either the line starts with :R or if it doesn't start with PRINT and it matches 7 backslashes? – Ali Razeghi - AWS Jun 03 '15 at 16:36
1

Easy to change the regex from `([^\\`n]*\\){7}` to `.:\\([^\\`n]*\\){7}`, which is one of the reasons why it's better to go with a clear & easy to maintain regex. Even if you might sacrifice a tiny bit of performance – Vincent De Smet Jun 05 '15 at 04:50

Remove All Values In A String Before 7th backslash

2 Answers2