CODEWITHSUNDEEP
bashsed
Josepas
Josepas

Reputation: 349

changing date in a log

today I have this file: a20 is (off) Jan 15 and I turned into this: a20 is (on) Jan 16 but the second date is whatever day the a script runs

date=$(date) with this sed commands:

sed -i '/'$1'/ s/off/on/' /home/josepas/Mantenimiento

sed -i '/'$1'/ s/).*/) /' /home/josepas/Mantenimiento

sed -i "/$1/ s/$/$date/g" /home/josepas/Mantenimiento

I want to make this a single sed that searches for the line that starts with a(somenumber) I heard about the option -e for multiple sed commands, hope you can help me

Upvotes: 0

Views: 170

Answers (2)

potong
potong

Reputation: 58381

This might work for you:

date="Jan 16"
echo "a20 is (off) Jan 15" |
sed '/^a[0-9]*/s/(off).*/(on) '"$date"'/'
a20 is (on) Jan 16

if you want to use the "tomorrow date" and you're using GNU sed, then:

echo "a20 is (off) Jan 15" | 
sed '/^a[0-9]\+/s/^\(.*\)(off) \(.*\)/echo "\1(on) $(date -d '\''\2 tomorrow'\'' '\''+%b %d'\'')"/e'
a20 is (on) Jan 16

Upvotes: 2

sgibb
sgibb

Reputation: 25736

You could concatenate s commands with ; and do some grouping by surrounding your commands with {/}. e.g.:

echo -e "a20 is (off) Jan 15\na21 is (off) Jan 15" | sed "/a20/ { s/off/on/; s/).*/) /; s/$/$(date)/ }"

Upvotes: 1

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