问题
Given a text file of unknown length, how can I read, for example all but the first 2 lines of the file? I know tail will give me the last N lines, but I don't know what N is ahead of time.
So for a file
AAAA
BBBB
CCCC
DDDD
EEEE
I want
CCCC
DDDD
EEEE
And for a file
AAAA
BBBB
CCCC
I'd get just
CCCC
回答1:
tail --help gives the following:
-n, --lines=K output the last K lines, instead of the last 10;
or use -n +K to output lines starting with the Kth
So to filter out the first 2 lines, -n +3 should give you the output you are looking for (start from 3rd).
回答2:
Assuming your version of tail supports it, you can specify starting the tail after X lines. In your case, you'd do 2+1.
tail -n +3
[mdemaria@oblivion ~]$ tail -n +3 stack_overflow.txt
CCCC
DDDD
EEEE
回答3:
A simple solution using awk:
awk 'NR > 2 { print }' file.name
回答4:
Try sed 1,2d. Replace 2 as needed.
回答5:
tail -n +linecount filename will start output at line linecount of filename, so tail -n +3 filename
should do what you want.
回答6:
Use this, supposing the first sample is called sample1.dat then tail --lines=3 sample1.dat which would print all lines from the 3rd line to the last line.
For the second sample, again suppose it is called sample2.dat it would be tail --lines=-1 sample2.dat which would print the last line...
回答7:
I really don't know how to do it from just tail or head but with the help of wc -l (line count) and bash expression, you can achieve that.
tail -$(( $( wc -l $FILE | grep -Eo '[0-9]+' ) - 2 )) $FILEHope this helps.
回答8:
using awk to get all but the last 2 line
awk 'FNR==NR{n=FNR}FNR<=n-3{print}' file file
awk to get all but the first 2 lines
awk 'NR>2' file
OR you can use more
more +2 file
or just bash
#!/bin/bash
i=0
while read -r line
do
[[ $i > 1 ]] && echo "$line"
((i++))
done <"file"
来源：https://stackoverflow.com/questions/3507999/whats-the-opposite-of-head-i-want-all-but-the-first-n-lines-of-a-file