unix命令从单词的第一次出现和最后一次出现之间获取行并写入文件 [英] unix command to get lines from in between first and last occurence of a word and write to a file

问题描述

我想要一个Unix命令来查找第一个& ;;之间的行.单词的最后出现

I want a unix command to find the lines between first & last occurence of a word

例如:

假设我们有1000行.第十行包含单词"stackoverflow"，第五行包含单词"stackoverflow".

let's imagine we have 1000 lines. Tenth line contains word "stackoverflow", thirty fifth line also contains word "stackoverflow".

我想在10到35之间打印一行并将其写入新文件.

I want to print lines between 10 and 35 and write it to a new file.

推荐答案

您可以分两步完成.基本思想是:

You can make it in two steps. The basic idea is to:

1)获取第一个和最后一个匹配项的行号.

1) get the line number of the first and last match.

2)打印介于这些范围之间的行范围.

2) print the range of lines in between these range.

$ read first last <<< $(grep -n stackoverflow your_file | awk -F: 'NR==1 {printf "%d ", $1}; END{print $1}')
$ awk -v f=$first -v l=$last 'NR>=f && NR<=l' your_file

说明

• read first last读取两个值，并将它们存储在$first和$last中.
• grep -n stackoverflow your_file摸索并显示如下输出:number_of_line:output
• awk -F: 'NR==1 {printf "%d ", $1}; END{print $1}')打印文件中stackoverflow的第一个和最后一个匹配项的行数.

Explanation

• read first last reads two values and stores them in $first and $last.
• grep -n stackoverflow your_file greps and shows the output like this: number_of_line:output
• awk -F: 'NR==1 {printf "%d ", $1}; END{print $1}') prints the number of line of the first and last match of stackoverflow in the file.

和

• awk -v f=$first -v l=$last 'NR>=f && NR<=l' your_file打印从$first行号到$last行号的所有行.
• awk -v f=$first -v l=$last 'NR>=f && NR<=l' your_file prints all lines from $first line number till $last line number.

$ cat a
here we
have some text
stackoverflow

and other things
bla
bla
bla bla
stackoverflow
and whatever else
stackoverflow
to make more fun
blablabla

$ read first last <<< $(grep -n stackoverflow a | awk -F: 'NR==1 {printf "%d ", $1}; END{print $1}')
$ awk -v f=$first -v l=$last 'NR>=f && NR<=l' a
stackoverflow

and other things
bla
bla
bla bla
stackoverflow
and whatever else
stackoverflow

按步骤:

$ grep -n stackoverflow a
3:stackoverflow
9:stackoverflow
11:stackoverflow

$ grep -n stackoverflow a | awk -F: 'NR==1 {printf "%d ", $1}; END{print $1}'
3 11

$ read first last <<< $(grep -n stackoverflow a | awk -F: 'NR==1 {printf "%d ", $1}; END{print $1}')

$ echo "first=$first, last=$last"
first=3, last=11

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