replacing a specific line of the file using DOS

Question

I have a XML file and I need to replace a perticular line say 7th line with the my own line using DOS command. How to do this?

XML files don't have "lines", and there are no "DOS commands" for working with XML. Also, this isn't a "write me some code" site. We do expect you to put some effort into doing it yourself first before posting — Ken White, Oct 03 '12 at 19:39
I would pick a different scripting language, you could probably do it with a batch file, but it would be a proper pain. something like vbs or even a very simple c# console app would make this much simpler — SmithMart, Oct 04 '12 at 07:14
check this http://xmlstar.sourceforge.net/ - if you have a problems with the tool just ask . Or better try to use powershell. — npocmaka, Oct 04 '12 at 12:59
You might have some luck using vbscript. See this question: http://stackoverflow.com/questions/1115508/batch-find-and-edit-lines-in-txt-file — Br.Bill, Oct 04 '12 at 21:38

score 0 · Accepted Answer · answered Oct 07 '12 at 18:53

My proposal is based on the fact you identify the line by its number and not by its content

@echo OFF
REM Say LINE 7 to be removed
SETLOCAL ENABLEDELAYEDEXPANSION
SET LOWER=6
SET UPPER=8
SET INDEX=0
SET MYLINE="this is my new line"
SET OUTFILE=%~n1.new%~x1
>%OUTFILE% (
    for /F "tokens=*" %%L in (%1) do (SET /A INDEX += 1
        IF !INDEX! LEQ %LOWER% (echo %%L) ELSE break
    )
    echo %MYLINE%
    more +%UPPER% %1
)

echo off needed since we do a massive redirect
ENABLEDELAYEDEXPANSION needed for the counter in the loop
You just pass the xml file as an argument to the script.

HTH

replacing a specific line of the file using DOS

1 Answers1