Display Line (n) in a text file In Command Prompt

Question

My text file include with 23 lines (lines include: !@$:/;" )

How can i only display line 3? in or 7? or 19?

I tried all commands was in stackoverflow

Example:

setlocal enabledelayedexpansion
@echo off
for /f "delims=" %%i in (mytext.txt) do (
if 1==1 (
set first_line=%%i
echo !first_line!
goto :eof
))

that's only show first line

Answer 1

@Compo has given a good answer. This is only to expound on it. Using aliases like GC should not be put into scripts. At the command line, sure, go ahead and reduce typing if you feel like it. Also, spelling out the parameter names provides more information and aids faster understanding.

To get only line 3.

GC .\mytext.txt -T 3|Select -L 1

Get-Content -Path '.\mytext.txt' -TotalCount 3 | Select-Object -Last 1

From the CMD console (Command prompt): (to get only line seven (7)

PowerShell "GC .\mytext.txt -T 7|Select -L 1"

PowerShell -NoProfile "Get-Content -Path '.\mytext.txt' -TotalCount 7 | Select-Object -Last 1"

To get lines 3 through 7:

$FirstLine = 3
$LastLine=7
powershell -NoProfile -Command "Get-Content -Path '.\t.txt' -TotalCount $LastLine | Select-Object -Last ($LastLine - $FirstLine + 1)"

Or, in a cmd.exe batch script.

SET "FIRSTLINE=3"
SET "LASTLINE=7"
powershell -NoProfile -Command ^
    "Get-Content -Path '.\t.txt' -TotalCount %LASTLINE% |" ^
        "Select-Object -Last (%LASTLINE% - %FIRSTLINE% + 1)"

Answer 2

@echo off
setlocal
set "FILE_TO_PROCESS=%~f1"
set /a LINE_NUMBER=%~2
set /a trim=LINE_NUMBER-1

break>"%temp%\empty"&&fc "%temp%\empty" "%FILE_TO_PROCESS%" /lb  %LINE_NUMBER% /t |more +4 | findstr /B /E /V "*****"|more +%trim%
endlocal

try with this bat (called lineNumber.bat) the first argument is the file you want to process the second is the line number:

call lineNumber.bat someFile.txt 5