CODEWITHSUNDEEP
regexsed
More Than Five
More Than Five

Reputation: 10419

Subtracting a value in a sed expression

I have a bunch of dates in a logfile that use the day of year format instead of day of month format:

So I see:

2013-10-302

instead of 

2013-10-29

I want them in day of month format. So I do:

sed "s/2013-10-302/2013-10-29/

But a problem with this approach is that I have several dates I need to convert. I could do something like if the month is 10, then subtract 273 from the day in year. If the month is 9, subtract something else.

So even if there is no better approach, any tips on how I would just do it this way? i.e. so if is month 10 subtract 273?

Thansk

Upvotes: 1

Views: 455

Answers (2)

jkshah
jkshah

Reputation: 11703

Try using awk

awk 'BEGIN {
    FS=OFS="-"
} {
    cmd="date -d "$1"-"$2"-01 +%j"; 
    cmd | getline offset; 
    $3=sprintf("%02d",$3-offset+1); 
    print
}' file

Test

Input file

2013-10-302
2013-11-310
2013-09-268

Output

2013-10-29
2013-11-06
2013-09-25

Upvotes: 1

fedorqui
fedorqui

Reputation: 289605

You are getting the date in %Y-%m-%j format, where %j stands for day of the year.

What if you do this?

$ date -d "Jan 1 2013 + 302 days" "+%F"
2013-10-30

Put it in context, if you have a file like

$ cat a
2013-10-302

You can "clean" it up with the following steps:

$ cat a
2013-10-302

Get the old date and the 3rd field for days since 1 Jan:

$ old_date=$(cat a)
$ echo $old_date
2013-10-302
$ day=$(cut -d- -f3 <<< "$old_date")
$ echo $day
302

Calculate the real date and update the file:

$ new_date=$(date -d "Jan 1 2013 + $day days" "+%F")
$ echo $old_date, $new_date
2013-10-302, 2013-10-30
$ sed -i "s/$old_date/$new_date/g" a
$ cat a
2013-10-30

Upvotes: 1

Related Questions