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

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

Answer 3

There are a couple of ways to do this. Your first option is to go through the for loop normally and break out of the loop once you reach the desired line.
#echo off
:: Specify which line to return
set get_line=7
:: Skip all lines before it, then print the next line and abort
set /a get_line-=1
for /F "skip=%get_line% delims=" %%A in (mytext.txt) do (
echo %%A
goto :end_loop
)
:end_loop
Your other option is to store all lines before the unwanted line in a temp variable and then display the next line.
#echo off
setlocal enabledelayedexpansion
:: Specify which line to return
set get_line=7
:: Skip all lines before it, then print the next line and abort
set /a get_line-=2
(
for /L %%A in (0,1,%get_line%) do set /p skip_line=
set /p return_line=
) <file.txt
echo !return_line!
Note that the first option is not suited for returning the first line of the script.

Answer 4

Since you said 'display', why not give PowerShell a shot:
From the PowerShell console:
GC .\mytext.txt -T 3|Select -L 1
From the CMD console (Command prompt):
PowerShell "GC .\mytext.txt -T 7|Select -L 1"
From a batch file:
#(PowerShell "GC .\mytext.txt -T 19|Select -L 1"&Pause)