How to grep for lines above and below a certain pattern
我想搜索特定图案（比如小节线），而且打印的上方和下方（即1线）的图案或2行上方和下方的图案线条。
Foo line
Bar line
Baz line
....
Foo1 line
Bar line
Baz1 line
....
Use grep with the parameters -A and -B to indicate the number a of lines A fter and B efore you want to print around your pattern:
grep -A1 -B1 yourpattern file
An stands for n lines "after" the match. Bm stands for m lines "before" the match. If both numbers are the same, just use -C :
grep -C1 yourpattern file
Test
$ cat file
Foo line
Bar line
Baz line
hello
bye
hello
Foo1 line
Bar line
Baz1 line
Let's grep :
$ grep -A1 -B1 Bar file
Foo line
Bar line
Baz line
--
Foo1 line
Bar line
Baz1 line
To get rid of the group separator, you can use --no-group-separator :
$ grep --no-group-separator -A1 -B1 Bar file
Foo line
Bar line
Baz line
Foo1 line
Bar line
Baz1 line
From man grep :
-A NUM, --after-context=NUM
Print NUM lines of trailing context after matching lines.
Places a line containing a group separator (--) between
contiguous groups of matches. With the -o or --only-matching
option, this has no effect and a warning is given.
-B NUM, --before-context=NUM
Print NUM lines of leading context before matching lines.
Places a line containing a group separator (--) between
contiguous groups of matches. With the -o or --only-matching
option, this has no effect and a warning is given.
-C NUM, -NUM, --context=NUM
Print NUM lines of output context. Places a line containing a
group separator (--) between contiguous groups of matches. With
the -o or --only-matching option, this has no effect and a
warning is given.
grep is the tool for you, but it can be done with awk
awk '{a[NR]=$0} $0~s {f=NR} END {for (i=f-B;i<=f+A;i++) print a[i]}' B=1 A=2 s="Bar" file
NB this will also find one hit.
or with grep
grep -A2 -B1 "Bar" file
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