CODEWITHSUNDEEP
bashdateawk
Moseleyi
Moseleyi

Reputation: 2859

Using and manipulating date variable inside AWK

I assume that there may be a better way to do it but the only one I came up with was using AWK.

I have a file with name convention like following:

testfile_2016_03_01.txt

Using one command I am trying to shift it by one day testfile_20160229.txt

I started from using: 

file=testfile_2016_03_01.txt
IFS="_"
arr=($file)
datepart=$(echo ${arr[1]}-${arr[2]}-${arr[3]} | sed 's/.txt//')
date -d "$datepart - 1 days" +%Y%m%d

the above works fine, but I really wanted to do it in AWK. The only thing I found was how to use "date" inside AWK

new_name=$(echo ${file##.*} | awk -F'_' ' {
"date '+%Y%m%d'" | getline date;
print date
}')
echo $new_name

okay so two things happen here. For some reason $4 also contains .txt even though I removed it(?) ##.*

And the main problem is I don't know how to pass the variables to that date, the below doesn't work

`awk -F'_' '{"date '-d "2016-01-01"' '+%Y%m%d'" | getline date; print date}')

ideally I want 2016-01-01 to be variables coming from the file name $2-$3-$4 and substract 1 day but I think I'm getting way too many single and double quotes here and my brain is losing..

Upvotes: 1

Views: 260

Answers (3)

potong
potong

Reputation: 58381

This might work for you:

file='testfile_2016_03_01.txt'
IFS='_.' read -ra a <<< "$file"
date -d "${a[1]}${a[2]}${a[3]} -1 day" "+${a[0]}_%Y%m%d.${a[4]}"

Upvotes: 0

Ed Morton
Ed Morton

Reputation: 203324

WIth GNU awk for time functions:

$ file=testfile_2016_03_01.txt
$ awk -v file="$file" 'BEGIN{ split(file,d,/[_.]/); print strftime(d[1]"_%Y%m%d."d[5],mktime(d[2]" "d[3]" "d[4]" 12 00 00")-(24*60*60)) }'
testfile_20160229.txt

Upvotes: 1

anubhava
anubhava

Reputation: 785058

Equivalent awk command:

file='testfile_2016_03_01.txt'
echo "${file%.*}" |
awk -F_ '{cmd="date -d \"" $2"-"$3"-"$4 " -1 days\"" " +%Y%m%d";
          cmd | getline date; close(cmd); print date}'

20160229

Upvotes: 2

Related Questions