167

Using awk or sed how can I select lines which are occurring between two different marker patterns? There may be multiple sections marked with these patterns.

For example: Suppose the file contains:

abc
def1
ghi1
jkl1
mno
abc
def2
ghi2
jkl2
mno
pqr
stu

And the starting pattern is abc and ending pattern is mno So, I need the output as:

def1
ghi1
jkl1
def2
ghi2
jkl2

I am using sed to match the pattern once:

sed -e '1,/abc/d' -e '/mno/,$d' <FILE>

Is there any way in sed or awk to do it repeatedly until the end of file?

0

10 Answers 10

254

Use awk with a flag to trigger the print when necessary:

$ awk '/abc/{flag=1;next}/mno/{flag=0}flag' file
def1
ghi1
jkl1
def2
ghi2
jkl2

How does this work?

  • /abc/ matches lines having this text, as well as /mno/ does.
  • /abc/{flag=1;next} sets the flag when the text abc is found. Then, it skips the line.
  • /mno/{flag=0} unsets the flag when the text mno is found.
  • The final flag is a pattern with the default action, which is to print $0: if flag is equal 1 the line is printed.

For a more detailed description and examples, together with cases when the patterns are either shown or not, see How to select lines between two patterns?.

13
  • 47
    If you want to print everything between and including the pattern then you can use awk '/abc/{a=1}/mno/{print;a=0}a' file.
    – scai
    Commented Nov 7, 2013 at 8:08
  • 8
    Yes, @scai ! or even awk '/abc/{a=1} a; /mno/{a=0}' file - with this, putting a condition before the /mno/ we make it evaluate the line as true (and print it) before setting a=0. This way we can avoid writing print.
    – fedorqui
    Commented Nov 7, 2013 at 9:43
  • 21
    @scai @fedorqui For including pattern output, you can do awk '/abc/,/mno/' file
    – Jotne
    Commented Dec 4, 2013 at 6:44
  • 2
    @EirNym that is a weird scenario that can be handled on very different ways: which lines would you like to print? Probably awk 'flag; /PAT1/{flag=1; next} /PAT1/{flag=0}' file would make.
    – fedorqui
    Commented Apr 24, 2017 at 8:28
  • 3
    For newbies like me, there is a doc. 1. A awk "rule" contains a "pattern" and an "action", either of which (but not both) may be omitted. So [pattern] { action } or pattern [{ action }]. 2. An action consists of one or more awk statements, enclosed in braces (‘{…}’). —— So the ending flag is abbr of flag {print $0}
    – Weekend
    Commented Jan 7, 2021 at 8:40
56

Using sed:

sed -n -e '/^abc$/,/^mno$/{ /^abc$/d; /^mno$/d; p; }'

The -n option means do not print by default.

The pattern looks for lines containing just abc to just mno, and then executes the actions in the { ... }. The first action deletes the abc line; the second the mno line; and the p prints the remaining lines. You can relax the regexes as required. Any lines outside the range of abc..mno are simply not printed.

6
  • 1
    @JonathanLeffler can I know what is the purpose of using -e Commented Dec 6, 2016 at 4:33
  • 1
    @KasunSiyambalapitiya: Mostly it means I like to use it. Formally, it specifies that the next argument is (part of) the script that sed should execute. If you want or need to use several arguments to include the entire script, then you must use -e before each such argument; otherwise, it's optional (but explicit). Commented Dec 6, 2016 at 4:41
  • Nice! (I prefer sed over awk.) When using complex regular expressions, it would be nice not to have to repeat them. Isn't it possible to delete the first / last line of the "selected" range? Or to first apply the d to all lines up to the first match, and then another d to all lines starting with the second match?
    – hans_meine
    Commented Dec 8, 2016 at 10:12
  • (Replying to my own comment.) If there's only one section to be cut, I could tentatively solve this e.g. for LaTeX using sed -n '1,/\\begin{document}/d;/\\end{document}/d;p'. (This is cheating a little bit, since the second part does not delete up to the document end, and I would not know how to cut multiple parts as the OP asked for.)
    – hans_meine
    Commented Dec 8, 2016 at 10:50
  • @JonathanLeffler what is the reason for inserting the $ mark, as in /^abc$ and others Commented Jan 25, 2017 at 4:58
20

This might work for you (GNU sed):

sed '/^abc$/,/^mno$/{//!b};d' file

Delete all lines except for those between lines starting abc and mno

3
  • !d;//d golfs 2 characters better :-) stackoverflow.com/a/31380266/895245 Commented Jul 13, 2015 at 9:54
  • 1
    This is awesome. The {//!b} prevents the abc and mno from being included in the output, but I can't figure out how. Could you explain?
    – Brendan
    Commented Feb 16, 2017 at 17:44
  • 2
    @Brendan the instruction //!b reads if the current line is neither one of the lines that match the range, break and therefore print those lines otherwise all other lines are deleted.
    – potong
    Commented Feb 17, 2017 at 1:14
16

From the previous response's links, the one that did it for me, running ksh on Solaris, was this:

sed '1,/firstmatch/d;/secondmatch/,$d'
  • 1,/firstmatch/d: from line 1 until the first time you find firstmatch, delete.
  • /secondmatch/,$d: from the first occurrance of secondmatch until the end of file, delete.
  • Semicolon separates the two commands, which are executed in sequence.
2
  • Just curious, why does the range limiter (1,) come before /firstmatch/? I'm guessing this could also be phrased '/firstmatch/1,d;/secondmatch,$d'?
    – Luke Davis
    Commented Jun 25, 2018 at 0:40
  • 3
    With "1,/firstmatch/d" you are saying "from line 1 until the first time you find 'firstmatch', delete". Whereas, with "/secondmatch/,$d" you say "from the first occurrance of 'secondmatch' until the end of file, delete". the semicolon separates the two commands, which are executed in sequence.
    – FanDeLaU
    Commented Dec 20, 2018 at 17:18
15
sed '/^abc$/,/^mno$/!d;//d' file

golfs two characters better than ppotong's {//!b};d

The empty forward slashes // mean: "reuse the last regular expression used". and the command does the same as the more understandable:

sed '/^abc$/,/^mno$/!d;/^abc$/d;/^mno$/d' file

This seems to be POSIX:

If an RE is empty (that is, no pattern is specified) sed shall behave as if the last RE used in the last command applied (either as an address or as part of a substitute command) was specified.

2
  • 1
    I think the second solution will end up with nothing as the second command is also a range. However kudos for the first.
    – potong
    Commented Jul 13, 2015 at 14:20
  • @potong true! I have to study more why the first one works. Thanks! Commented Jul 13, 2015 at 14:22
3

something like this works for me:

file.awk:

BEGIN {
    record=0
}

/^abc$/ {
    record=1
}

/^mno$/ {
    record=0;
    print "s="s;
    s=""
}

!/^abc|mno$/ {
    if (record==1) {
        s = s"\n"$0
    }   
}

using: awk -f file.awk data...

edit: O_o fedorqui solution is way better/prettier than mine.

1
3

Don_crissti's answer from Show only text between 2 matching pattern?

firstmatch="abc"
secondmatch="cdf"
sed "/$firstmatch/,/$secondmatch/!d;//d" infile

which is much more efficient than AWK's application, see here.

2
  • I don't think linking the time comparisons makes much sense here, since the requirements of the questions are quite different, hence the solutions.
    – fedorqui
    Commented Sep 11, 2015 at 15:11
  • 2
    I disagree because we should have some criterias to compare answers. Only a few has SED applications. Commented Sep 11, 2015 at 16:10
2
perl -lne 'print if((/abc/../mno/) && !(/abc/||/mno/))' your_file
1
  • Good to know perl equivalent as it is a pretty good alternative to both awk and sed.
    – akhan
    Commented Mar 8, 2017 at 23:46
0

I tried to use awk to print lines between two patterns while pattern2 also match pattern1. And the pattern1 line should also be printed.

e.g. source

package AAA
aaa
bbb
ccc
package BBB
ddd
eee
package CCC
fff
ggg
hhh
iii
package DDD
jjj

should has an ouput of

package BBB
ddd
eee

Where pattern1 is package BBB, pattern2 is package \w*. Note that CCC isn't a known value so can't be literally matched.

In this case, neither @scai 's awk '/abc/{a=1}/mno/{print;a=0}a' file nor @fedorqui 's awk '/abc/{a=1} a; /mno/{a=0}' file works for me.

Finally, I managed to solve it by awk '/package BBB/{flag=1;print;next}/package \w*/{flag=0}flag' file, haha

A little more effort result in awk '/package BBB/{flag=1;print;next}flag;/package \w*/{flag=0}' file, to print pattern2 line also, that is,

package BBB
ddd
eee
package CCC
0

This can also be done with logical operations and increment/decrement operations on a flag:

awk '/mno/&&--f||f||/abc/&&f++' file
2
  • I'm absolutely certain that i've used awk in the past for this problem, and it was nothing like this complex.
    – Owl
    Commented Mar 28, 2022 at 10:45
  • 1
    Obviously the accepted answer in awk that predates my answer by more than 7 years is much more readable, and I saw that answer before I posted mine. I'm just throwing this one here because it is one byte shorter than the accepted answer even after renaming its variable flag to f, in the spirit of some good ol' code golf fun. :-)
    – blhsing
    Commented Mar 30, 2022 at 7:14

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