How to remove comments from text file - windows

Question

My text file contains one line comments that all being with "// ". Two forward slashes and a space. These may either take up the whole line or just the last part of a line. Each comment does not extend beyond the line that it's on. So no /* */ type comments crossing multiple lines.
In simple terms, all comments start with "//space" anywhere on the line. Anything starting with "//space" should be removed and trailing spaces on that line should also be removed. Leading spaces should stay. Any blank lines should be removed.
Sample file:
// This is a comment
x = 1 // This is also a comment after the double slash
x = 2
x = 3 // The above is a blank line
// Comment on this record but nothing precedes it, so should be deleted.
y = 4 // A line with leading spaces that should be kept.
z = "//path"; // The first double slashes are not a comment since the space is missing after the "//"
// Last comment line.
Result file (no trailing spaces, but keep leading spaces.:
x = 1
x = 2
x = 3
y = 4
z = "//path";
I can remove the blank lines using gc file.txt | Where-Object { $_ -ne ''} > result.txt. However I'm having trouble with reading just the beginning part of a line up to the "//" comment part.
I also tried findstr but haven't found how to read each line up to the "//" and then trim spaces out.
I could write a script program to loop throught the file and do this, but it seems like there should be a way to accomplish it using a simple one or two line powershell or bat file command.
What is the easiest way (shortest amount of code) to remove these comments while keeping the uncommented contents of the file?

Answer 1

Since you seem to equate "easy" with "short", here's a fairly simple solution:
gc .\samplefile.txt|%{$_-replace"(.*)(// .*)",'$1'}|?{$_}
if it's really that important to you :-)
A bit more verbose version (still using regex):
Get-Content .\samplefile.txt | Where-Object {
-not ([String]::IsNullOrEmpty($_.Trim()) -or $_-match"^\s*// ")
} |ForEach-Object { $_ -replace "(.*)(// .*)",'$1' }
That being said, I would (personally) go for a more verbose and easier-to-read/maintain solution:
To remove everything after //, the easiest way is to find the first occurrence of // with String.IndexOf() and then grab the first part with String.Substring():
PS C:\> $CommentedString = "Content // this is a comment"
PS C:\> $CommentIndex = $CommentedString.IndexOf('// ')
PS C:\> $CommentedString.Substring(0,$CommentIndex)
Content
For the indented comments you can also use String.Trim() to remove whitespace from the beginning and end of the string:
PS C:\> " // Indented comment" -match '^//'
True
You can use the ForEach-Object cmdlet to go through every line and apply the above:
function Remove-Comments {
param(
[string]$Path,
[string]$OutFile
)
# Read file, remove comments and blank lines
$CleanLines = Get-Content $Path |ForEach-Object {
$Line = $_
# Trim() removes whitespace from both ends of string
$TrimmedLine = $Line.Trim()
# Check if what's left is either nothing or a comment
if([string]::IsNullOrEmpty($TrimmedLine) -or $TrimmedLine -match "^// ") {
# if so, return nothing (inside foreach-object "return" acts like "coninue")
return
}
# See if non-empty line contains comment
$CommentIndex = $Line.IndexOf("// ")
if($CommentIndex -ge 0) {
# if so, remove the comment
$Line = $Line.Substring(0,$CommentIndex)
}
# return $Line to $CleanLines
return $Line
}
if($OutFile -and (Test-Path $OutFile)){
[System.IO.File]::WriteAllLines($OutFile, $CleanLines)
} else {
# No OutFile was specified, write lines to pipeline
Write-Output $CleanLines
}
}
Applied to your sample:
PS C:\> Remove-Comments D:\samplefile.txt
x = 1
x = 2
x = 3

Answer 2

Like a great many text processing problems, there is a simple solution using JREPL.BAT - a powerful regex text processing utility for the Windows command line. It is pure script (hybrid JScript/batch) that runs natively on any Windows machine from XP onward. Full documentation is embedded within the script.
jrepl "^(.*?)\s*// " "$1!=''?$1:false" /jmatch /f test.txt /o out.txt
You can overwrite the original file by specifying - as the output file:
jrepl "^(.*?)\s*// " "$1!=''?$1:false" /jmatch /f test.txt /o -
I've tested, and it gives the exact output you are looking for.
If you put the command within a batch script, then you must use call jrepl

Answer 3

Tha Batch file below do what you want. Sorry, but there is not an "easy short code" way to do this...
#echo off
setlocal EnableDelayedExpansion
rem Set the maximum number of trailing spaces as a power_of_2-1 value. For example, for 15 spaces:
set spcPow2=4
set "spaces= "
for /L %%i in (1,1,%spcPow2%) do set "spaces=!spaces!!spaces!"
set /A spcPow2-=1
rem Process all lines, excepting empty ones and lines that start with "/"
setlocal DisableDelayedExpansion
for /F "eol=/ delims=" %%a in (test.txt) do (
set "line=%%a"
rem Split line at "// " and get the first part
setlocal EnableDelayedExpansion
for /F "delims=¡" %%b in ("!line:// =¡!") do (
endlocal
set "line=%%b"
)
rem Eliminate trailing spaces
setlocal EnableDelayedExpansion
set spc=0
for /L %%b in (%spcPow2%,-1,0) do (
set /A "newSpc=spc+(1<<%%b)"
for %%n in (!newSpc!) do if "!line:~-%%n!" equ "!spaces:~-%%n!" set "spc=%%n"
)
if !spc! gtr 0 for %%n in (!spc!) do set "line=!line:~0,-%%n!"
rem Show resulting line
if defined line echo !line!
endlocal
)
EDIT: New solution added
#set #x=1 // & CScript //nologo //E:JScript "%~F0" < samplefile.txt & goto :EOF
WScript.Stdout.Write(WScript.Stdin.ReadAll().replace(/(.*)\/\/ .*/g,"$1"))
Copy previous code into a file with .BAT extension, that is, it is a Batch file!