How do I use Head and Tail to print specific lines of a file

问题

I want to say output lines 5 - 10 of a file, as arguments passed in.

How could I use head and tail to do this?

where firstline = $2 and lastline = $3 and filename = $1.

Running it should look like this:

./lines.sh filename firstline lastline

回答1:

Aside from the answers given by fedorqui and Kent, you can also use a single sed command:

#! /bin/sh
filename=$1
firstline=$2
lastline=$3

# Basics of sed:
#   1. sed commands have a matching part and a command part.
#   2. The matching part matches lines, generally by number or regular expression.
#   3. The command part executes a command on that line, possibly changing its text.
#
# By default, sed will print everything in its buffer to standard output.  
# The -n option turns this off, so it only prints what you tell it to.
#
# The -e option gives sed a command or set of commands (separated by semicolons).
# Below, we use two commands:
#
# ${firstline},${lastline}p
#   This matches lines firstline to lastline, inclusive
#   The command 'p' tells sed to print the line to standard output
#
# ${lastline}q
#   This matches line ${lastline}.  It tells sed to quit.  This command 
#   is run after the print command, so sed quits after printing the last line.
#   
sed -ne "${firstline},${lastline}p;${lastline}q" < ${filename}

Or, to avoid any external utilites, if you're using a recent version of bash (or zsh):

#! /bin/sh

filename=$1
firstline=$2
lastline=$3

i=0
exec <${filename}  # redirect file into our stdin
while read ; do    # read each line into REPLY variable
  i=$(( $i + 1 ))  # maintain line count

  if [ "$i" -ge "${firstline}" ] ; then
    if [ "$i" -gt "${lastline}" ] ; then
      break
    else
      echo "${REPLY}"
    fi
  fi
done

回答2:

head -n XX # <-- print first XX lines
tail -n YY # <-- print last YY lines

If you want lines from 20 to 30 that means you want 11 lines starting from 20 and finishing at 30:

head -n 30 file | tail -n 11
# 
# first 30 lines
#                 last 11 lines from those previous 30

That is, you firstly get first 30 lines and then you select the last 11 (that is, 30-20+1).

So in your code it would be:

head -n $3 $1 | tail -n $(( $3-$2 + 1 ))

Based on firstline = $2, lastline = $3, filename = $1

head -n $lastline $filename | tail -n $(( $lastline -$firstline + 1 ))

回答3:

try this one-liner:

awk -vs="$begin" -ve="$end" 'NR>=s&&NR<=e' "$f"

in above line:

$begin is your $2
$end is your $3
$f is your $1

回答4:

Save this as "script.sh":

#!/bin/sh

filename="$1"
firstline=$2
lastline=$3
linestoprint=$(($lastline-$firstline+1))

tail -n +$firstline "$filename" | head -n $linestoprint

There is NO ERROR HANDLING (for simplicity) so you have to call your script as following:

./script.sh yourfile.txt firstline lastline

$ ./script.sh yourfile.txt 5 10

If you need only line "10" from yourfile.txt:

$ ./script.sh yourfile.txt 10 10

Please make sure that: (firstline > 0) AND (lastline > 0) AND (firstline <= lastline)

来源：https://stackoverflow.com/questions/15747912/how-do-i-use-head-and-tail-to-print-specific-lines-of-a-file

标签

shell

unix

arguments

tail

unix-head