Windows batch command(s) to read first line from text file

Here's a general-purpose batch file to print the top n lines from a file like the GNU head utility, instead of just a single line.

@echo off

if [%1] == [] goto usage
if [%2] == [] goto usage

call :print_head %1 %2
goto :eof

REM
REM print_head
REM Prints the first non-blank %1 lines in the file %2.
REM
:print_head
setlocal EnableDelayedExpansion
set /a counter=0

for /f ^"usebackq^ eol^=^

^ delims^=^" %%a in (%2) do (
        if "!counter!"=="%1" goto :eof
        echo %%a
        set /a counter+=1
)

goto :eof

:usage
echo Usage: head.bat COUNT FILENAME

For example:

Z:\>head 1 "test file.c"
; this is line 1

Z:\>head 3 "test file.c"
; this is line 1
    this is line 2
line 3 right here

It does not currently count blank lines. It is also subject to the batch-file line-length restriction of 8 KB.

You might give this a try:

@echo off

for /f %%a in (sample.txt) do (
  echo %%a
  exit /b
)

edit Or, say you have four columns of data and want from the 5th row down to the bottom, try this:

@echo off

for /f "skip=4 tokens=1-4" %%a in (junkl.txt) do (
  echo %%a %%b %%c %%d
)

Thanks to thetalkingwalnut with answer Windows batch command(s) to read first line from text file I came up with the following solution:

@echo off
for /f "delims=" %%a in ('type sample.txt') do (
echo %%a
exit /b
)

Slightly building upon the answers of other people. Now allowing you to specify the file you want to read from and the variable you want the result put into:

@echo off
for /f "delims=" %%x in (%2) do (
set %1=%%x
exit /b
)

This means you can use the above like this (assuming you called it getline.bat)

c:\> dir > test-file
c:\> getline variable test-file
c:\> set variable  
variable= Volume in drive C has no label.

powershell Get-Content file.txt -Head 1

This one is much quicker than the other powershell examples above, where the full file is read.

One liner, useful for stdout redirect with ">":

@for /f %%i in ('type yourfile.txt') do @echo %%i & exit

To cicle a file (file1.txt, file1[1].txt, file1[2].txt, etc.):

START/WAIT C:\LAERCIO\DELPHI\CICLADOR\dprCiclador.exe C:\LAERCIUM\Ciclavel.txt

rem set/p ciclo=< C:\LAERCIUM\Ciclavel.txt:
set/p ciclo=< C:\LAERCIUM\Ciclavel.txt

rem echo %ciclo%:
echo %ciclo%

And it's running.

Here is a workaround using powershell:

powershell (Get-Content file.txt)[0]

(You can easily read also a range of lines with powershell (Get-Content file.txt)[0..3])

If you need to set a variable inside a batch script as the first line of file.txt you may use:

for /f "usebackq delims=" %%a in (`powershell ^(Get-Content file.txt^)[0]`) do (set "head=%%a")

In Windows PowerShell below cmd can be used to get the first line and replace it with a static value

powershell -Command "(gc txt1.txt) -replace (gc txt1.txt)[0], 'This is the first line' | Out-File -encoding ASCII txt1.txt"

Reference

How can you find and replace text in a file using the Windows command-line environment?

Print 1st line only (no need to read entire file):

set /p a=< file.txt & echo !a!

To print one line at a time; user to press a key for next line:
(After printing required lines, press Ctrl+C to stop.)

for /f "delims=" %a in (downing.txt) do echo %a & pause>nul

To print 1st n lines (without user intervention):

type nul > tmp & fc tmp "%file%" /lb %n% /t | find /v "?" | more +2

Tested on Win 10 CMD.