Replacing character before EOL in a .txt using batch - powershell

Question

I would like to search & replace the end of each line in my file.txt and delete the char before.
In other words, I have a redundant comma at end of each line, and it disturbs me when passing the data to Excel.
I can't delete all commas, since most of them are valuable. I just need to delete those at end of each line, since they make a blank cell.
I don't mind using a PowerShell gc command.

Answer 1

This is the simplest way to "delete the last char of each line" using a Batch file:
#echo off
setlocal EnableDelayedExpansion
(for /F "delims=" %%a in (file.txt) do (
set "line=%%a"
echo(!line:~0,-1!
)) > output.txt
This Batch file have a series of limitations: it removes empty lines and lines that start with semicolon, and remove exclamation marks from the lines. Each restriction can be fixed introducing a modification that eventually ends in a code similar to aschipfl's answer, but this code may be enough for your needs.

Answer 2

Here is a PowerShell script that will take file in.txt as input, remove the last character from each line, and output them to out.txt:
Get-Content "in.txt" | % { $_.Substring(0, ($_.Length - 1)) } | Out-File "out.txt"
Adding a simple test to check you are only deleting commas:
Get-Content "in.txt" | % {
if($_.Substring(($_.Length - 1), 1) -eq ",") {
$_.Substring(0, ($_.Length - 1))
} else {
$_
}
} | Out-File "out.txt"

Answer 3

Here is a pure batch-file solution:
The following code snippet removes the last character from each line in file.txt if it is , and writes the result into file_new.txt:
#echo off
setlocal EnableExtensions
> "file_new.txt" (
for /F usebackq^ delims^=^ eol^= %%L in ("file.txt") do (
setlocal DisableDelayedExpansion
set "LINE=%%L"
setlocal EnableDelayedExpansion
if "!LINE:~-1!"=="," (
echo(!LINE:~,-1!
) else (
echo(!LINE!
)
endlocal
endlocal
)
)
endlocal
The toggling of delayed environment variable expansion (see setlocal /? and set /? for help) is required to avoid trouble with some special characters.
The above approach removes empty lines from the file as for /F ignores them. To avoid this, use the following script:
#echo off
setlocal EnableExtensions
> "file_new.txt" (
for /F delims^=^ eol^= %%L in ('findstr /N /R "^" "file.txt"') do (
setlocal DisableDelayedExpansion
set "LINE=%%L"
setlocal EnableDelayedExpansion
set "LINE=!LINE:*:=!"
if "!LINE:~-1!"=="," (
echo(!LINE:~,-1!
) else (
echo(!LINE!
)
endlocal
endlocal
)
)
endlocal
This uses the fact that findstr /N /R "^" returns every line (even empty ones) of the given file and prefixes it with a line number plus :. Therefore no line appears to be empty to for /F. The line set "LINE=!LINE:*:=!" is inserted to remove that prefix (everything up to the first :) from each line prior to outputting.