grep a file, but show several surrounding lines?

问题

I would like to grep for a string, but also show the preceding five lines and the following five lines as well as the matched line. How would I be able to do this?

回答1:

For BSD or GNU grep you can use -B num to set how many lines before the match and -A num for the number of lines after the match.

grep -B 3 -A 2 foo README.txt

If you want the same number of lines before and after you can use -C num.

grep -C 3 foo README.txt

This will show 3 lines before and 3 lines after.

回答2:

-A and -B will work, as will -C n (for n lines of context), or just -n (for n lines of context... as long as n is 1 to 9).

回答3:

ack works with similar arguments as grep, and accepts -C. But it's usually better for searching through code.

回答4:

grep astring myfile -A 5 -B 5

That will grep "myfile" for "astring", and show 5 lines before and after each match

回答5:

I normally use

grep searchstring file -C n # n for number of lines of context up and down

Many of the tools like grep also have really great man files too. I find myself referring to grep's man page a lot because there is so much you can do with it.

man grep

Many GNU tools also have an info page that may have more useful information in addition to the man page.

info grep

回答6:

Use grep

$ grep --help | grep -i context
Context control:
  -B, --before-context=NUM  print NUM lines of leading context
  -A, --after-context=NUM   print NUM lines of trailing context
  -C, --context=NUM         print NUM lines of output context
  -NUM                      same as --context=NUM

回答7:

Search for "17655" in "/some/file.txt" showing 10 lines context before and after (using Awk), output preceded with line number followed by a colon. Use this on Solaris when 'grep' does not support the "-[ACB]" options.

awk '

/17655/ {
        for (i = (b + 1) % 10; i != b; i = (i + 1) % 10) {
                print before[i]
        }
        print (NR ":" ($0))
        a = 10
}

a-- > 0 {
        print (NR ":" ($0))
}

{
        before[b] = (NR ":" ($0))
        b = (b + 1) % 10
}' /some/file.txt;

回答8:

ripgrep

If you care about the performance, use ripgrep which has similar syntax to grep, e.g.

rg -C5 "pattern" .

-C, --context NUM - Show NUM lines before and after each match.

There are also parameters such as -A/--after-context and -B/--before-context.

The tool is built on top of Rust's regex engine which makes it very efficient on the large data.

回答9:

$grep thestring thefile -5

-5 gets you 5 lines above and below the match 'thestring' is equivalent to -C 5 OR -A 5 -B 5

回答10:

Grep has an option called Context Line Control, you can use the --contect in that, simply,

| grep -C 5

| grep -5

Should do the trick

回答11:

Here is the @Ygor solution in awk

awk 'c-->0;$0~s{if(b)for(c=b+1;c>1;c--)print r[(NR-c+1)%b];print;c=a}b{r[NR%b]=$0}' b=3 a=3 s="pattern" myfile

^{Note: Replace a and b variables with number of lines before and after.}

It's especially useful for system which doesn't support grep's -A, -B and -C parameters.

来源：https://stackoverflow.com/questions/9081/grep-a-file-but-show-several-surrounding-lines

标签

logging

grep

command-line-interface