Bash从Z行开始从文件读取N行

Question

I have text file:

34
3467uy
56ytyyu
And so on

I want to read from this file N lines starting from line Z. I know how to read first N lines with (head -n N) or last with tail. I know now how to read from line Z to the end of the file with tail. But I can't find how to read specific amount starting from specific line. Help!

Answer 1

This can be accomplished using sed . For example, to emit lines 10 through 20 in a file:

sed -n '10,+10p' a_file

Answer 2

Pipe tail 's output to head :

tail -n +Z file | head -n N

For example:

$ seq 100 | tail -n +15 | head -n 5
15
16
17
18
19

Answer 3

使用awk ：下面的示例将从第3行打印到第5行（3 + 2行）。设置start_line和buffer变量以从start_line开始打印该行并buffer更多行。

awk -vstart_line=3 -vbuffer=2 'NR>=start_line && NR<=start_line+buffer' file

Answer 4

With GNU sed:

seq 1 10 | sed '2,+3!d'

Output:

2
3
4
5

From man sed :

addr1,+N : Will match addr1 and the N lines following addr1.