removing lines between two patterns (not inclusive) with sed

Question

Ok

I know that this is trivial question but: How can i remove lines from files that are between two known patterns/words:

pattern1
garbage
pattern2

to obtain:

pattern1
pattern2

And does anyone known good(simple written!) resources for studying sed?? With many clear examples?

Possible duplicate of Using sed to delete all lines between two matching patterns, SED delete lines between two pattern matches, sed delete lines between two patterns, without the second pattern, including the first pattern, SED delete specific lines between two patterns? and friends. — jww
– jww, Commented Dec 23, 2019 at 17:44

Dennis Williamson · Accepted Answer · 2011-02-22 01:01:20Z

20

sed -n '/pattern1/{p; :a; N; /pattern2/!ba; s/.*\n//}; p' inputfile

Explanation:

/pattern1/{ # if pattern1 is found p # print it :a # loop N # and accumulate lines /pattern2/!ba # until pattern2 is found s/.*\n// # delete the part before pattern2 } p # print the line (it's either pattern2 or it's outside the block)

Edit:

Some versions of sed have to be spoon-fed:

sed -n -e '/pattern1/{' -e 'p' -e ':a' -e 'N' -e '/pattern2/!ba' -e 's/.*\n//' -e '}' -e 'p' inputfile

edited Feb 22, 2011 at 1:01

answered Feb 22, 2011 at 0:02

Dennis Williamson

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8 Comments

kasper Over a year ago

it gives me sed: -e expression #1, char 58: Unknown option to 's'. My sed version is 3.02

kasper Over a year ago

@Dennis: Oh, version from edit is doing the right thing. Thanks once more !

user332325 Over a year ago

Or with less black magic sed '/pattern1/,/pattern2/{/pattern1/b;/pattern2/b;d}' inputfile :)

Dennis Williamson Over a year ago

@pooh: Your first one doesn't work with this test data (lines with "xxx" should not be printed, everything else should):

echo -e 'aaa\npattern1\ncccxxx\ndddxxx \npattern2\nfff\nggg\npattern1\nAAAxxx\nBBBxxx \nCCCxxx\nDDDxxx\npattern2\nEEE\nFFF'

user332325 Over a year ago

@Dennis Williamson: Ah, true:) Tested on one line contents!

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t7e · Accepted Answer · 2022-09-25 03:13:06Z

17

This will work for both GNU and BSD sed:

sed '/pattern1/,/pattern2/{//!d;}' file

edited Sep 25, 2022 at 3:13

t7e

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answered Apr 22, 2012 at 20:14

potong

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2 Comments

rkd Over a year ago

If anybody else uses this with BSD sed (i.e. Mac OSX), and you get an error extra characters at the end of d command, simply add ; after the d like this: sed '/pattern1/,/pattern2/{//!d;}' file

qodeninja Over a year ago

how do you do this inclusive of the patterns?

Daniel Gallagher · Accepted Answer · 2011-02-21 22:24:27Z

This is easily done with awk:

BEGIN { doPrint = 1; } /pattern1/ { doPrint = 0; print $0; } /pattern2/ { doPrint = 1; } { if (doPrint) print $0; }

I've found the sed info is fairly easy reading, with many examples. Same thing for awk.

kurumi · Accepted Answer · 2011-02-22 01:01:05Z

4

awk '/pattern1/{g=1;next}/pattern2/{g=0;next}g' file

answered Feb 22, 2011 at 1:01

kurumi

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Comments

Inder · Accepted Answer · 2018-09-04 00:24:06Z

3

This sed code will work as well:

sed '/PATTERN1/,/PATTERN2/d' FILE

edited Sep 4, 2018 at 0:24

Inder

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answered Jul 7, 2016 at 8:10

publkaccion

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1 Comment

Sadi Over a year ago

Although not as OP wanted, this works well as the inclusive version of the answer by @potong

tylo · Accepted Answer · 2011-02-22 13:03:28Z

You may also use the Unix text editor ed:

echo ' pattern1 garbage pattern2 ' > test.txt cat <<-'EOF' | sed -e 's/^ *//' -e 's/ *$//' | ed -s test.txt &>/dev/null H /pattern1/+1,/pattern2/-1d wq EOF

For more information see: Editing files with the ed text editor from scripts

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