sed: print only first occurence of a pattern match

Question

Several questions are similar to this one, but I have not found a solution that works when I want to search for a pattern over several lines. The following

sed -n '/First string/,/Second string/ p' my.file

will print all occurrences of the matched pattern, but I would like only the first occurrence. I am using GNU sed.

post your input text and expected output – RomanPerekhrest May 05 '17 at 12:19 — RomanPerekhrest, May 05 '17 at 12:19
First occurrence or first line? – schaiba May 05 '17 at 12:23 — schaiba, May 05 '17 at 12:23

ilkkachu · Answer 1 · 2017-05-05T12:33:29.137

14

Use q to explicitly quit when the end pattern is reached.

In GNU sed:

$ cat foo
foo
START
bar
END
blah
START another

$ sed -n '/START/,/END/p; /END/q' foo
START
bar
END

awk would maybe make it easier to not repeat the end pattern:

$ awk '/START/{p=1} p; /END/{exit}' foo
START
bar
END

edited May 05 '17 at 12:33

answered May 05 '17 at 12:24

ilkkachu

138,973

score 1 · Answer 2 · answered May 05 '17 at 12:28

1

If it's first occurrence you want, perhaps awk is better suited for this : awk 'NR==1,/original/{sub(/original/, "replacement")} 1' file . Taken from here.

answered May 05 '17 at 12:28

schaiba

7,631

sed: print only first occurence of a pattern match

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