How to replace the nth line in a text file using cmd?

Question

cmd/bat

Hello. Subject. It is necessary to replace the 13th line in the file with a previously known one. I write
for /f "skip=12" in (settings.ini) do (echo dparm=BIND=0,lck='ru',APP='GPS',HST='%COMPUTERNAME%')
As planned, I should skip 12 lines from the beginning of the file and replace with whatever is in parentheses. In fact, zero response. It may be necessary to interrupt the cycle, as well. the replacement is made once, but I'm not sure, this is an assumption, I had no business with smd before.

Answer 1

In general, I climbed more on the forums and did it. Initially, the task was to replace a piece in the file with the name of the computer on which the file is running. But I was smart, I decided to replace the whole line, so I fiddled around for a long time. Everything turned out to be more prosaic. In the example below, pc-name is the chunk to be replaced by the computer name retrieved via %computername%.
@echo off
setlocal enabledelayedexpansion
Set infile=settings.ini
Set find=pc-name
Set replace=%COMPUTERNAME%
@echo off
setlocal enabledelayedexpansion
set COUNT=0
for /F "tokens=* delims=, eol=#" %%n in (!infile!) do (
set LINE=%%n
set TMPR=!LINE:%find%=%replace%!
Echo !TMPR!>>TMP.TXT
)
move TMP.TXT %infile%

Answer 2

Is the content in the 12th line known?
as an option
You can PowerShell

$FilePath = C:\path_to_settings.ini
(Get-Content $FilePath) -replace 'parametr =.*','parametr = "YOUR TEXT"' | Out-File $FilePath

Answer 3

On PowerShell, just
Run PoSH from a script on the command line: (contents of a bat file)
And no one forces you to write monstrous designs, right? :)