使用awk或sed在文件的第一列中添加增量日期

Question

I want to add date in the first column of the file which starts from "1950/01/01" (format : %m/%d/%Y) and date should increment with each row using awk or sed. When adding date, leap year also needs to be considered ..

File contents are:

1,2,3,4
3,4,5,6
4,7,8,9

Expected output:

1950/01/01,1,2,3,4
1950/02/01,3,4,5,6
1950/03/01,4,7,8,9

So far I have manage to add formatted date in the first column but not able to increment it with every row.

awk -v date="$(date -d '1950/01/01' '+%m/%d/%Y')" -F"," 'BEGIN { OFS = "," }  ; {print date, $0}' inout.csv 

Answer 1

Using Gawk

Input

$ cat file
1,2,3,4
3,4,5,6
4,7,8,9

Output

$ awk -v date="$( date -d '1950/01/01' '+%s' )" '{print strftime("%Y/%d/%m",date)","$0; date+=86400}' file
1950/01/01,1,2,3,4
1950/02/01,3,4,5,6
1950/03/01,4,7,8,9

Explanation

 awk -v date="$( date -d '1950/01/01' '+%s' )" '{ # call awk set var date with timestamp

        print strftime("%Y/%d/%m",date) "," $0;   # print date, sep comma and current line 
        date+=86400                               # one day = 86400 seconds, just increment
 }
 ' file

Answer 2

Alternatively with bash and GNU date:

a=1950/01/01
while IFS= read -r line; do echo "$a,$line"; a=$(date +%Y/%m/%d -d "$a 1 day"); done < file

Output:

1950/01/01,1,2,3,4
1950/01/02,3,4,5,6
1950/01/03,4,7,8,9