remove all text after a certain character that appears on all odd lines
this works for every line in the file with sed
sed 's/;.*//'
how to edit this sed line so it just does all odd lines
3
With GNU sed:
$ sed '1~2s/;.*//' file
The address range 1~2 will match line 1, and then every second line after that.
The general case is first~step and the example in the manual is sed -n 1~2p which will print all odd lines.
This is a GNU extension.
The equivalent thing in awk:
$ awk 'NR % 2 { sub(";.*", "") } { print }' file
Or, as suggested by Arrow in comments:
$ awk -F';' 'NR % 2 { $0 = $1 } { print }' file
which has the same effect, but instead of explicitly substituting out everything after the first ; it uses ; as the field separator (with -F';') and then outputs only the first field (by replacing the whole input line with only the first field for odd rows).
answered Aug 21 '17 at 14:28
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@bob Which means that you are not using GNU
sed. What Unix are you on? On BSD systems, GNUsedis sometimes installed asgsed. – Kusalananda♦ Aug 21 '17 at 14:32 -
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This probably might work with all sed version, tested on GNU sed though
$ cat ip.txt
foo;bar-baz;xyz
a;b;c
good bad. hi there
d;e
cool;hot;normal
$ sed 's/;.*//;n' ip.txt
foo
a;b;c
good bad. hi there
d;e
cool
The n command will fetch next line but won't change anything, effectively allowing to change only odd lines
With perl
perl -pe 's/;.*// if $.%2' ip.txt
where if $.%2 adds condition to perform substitution only if line number is odd
