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get yesterday in yyyymmdd format

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# 1  
get yesterday in yyyymmdd format

how can i get yesterday in yyyymmdd format? Smilie
# 2  
with gnu date you can do:
date --date="-1 days" +%Y%m%d

Last edited by Yogesh Sawant; 06-01-2010 at 10:03 AM.. Reason: added code tags
# 3  
Hammer & Screwdriver

if you have issue with the above method say you dont have gnu date... try connecting to oracle and get date from there is no direct mechanism to calculate date.

something like this ...

yesterday=`sqlplus user/password << EOF
Select to_char(sysdate-1,'YYYYMMDD') from dual

# 4  
I wrote this script to convert julian date into dd-mm-yyyy format. You could use this to get yesterday's date. You just have to get today's julian date (get this using `date +%j`) and subtract 1 from it. This will get you yesterday's julian date. Give this as the second argument to the script. The first argument is the present year in the YYYY format.

Note: I have changed the output so that it prints YYYYmmdd.

Also this script will blindly convert whatever the julian date that is given to it. No error checking is done - so if you give the julian date as 0, the output you get for this year is 20050100


check_done() {
        if [ $month -eq 1 -o $month -eq 3 -o $month -eq 5 -o $month -eq 7 -o $mo
nth -eq 8 -o $month -eq 10 -o $month -eq 12 ]
        elif [ $month -eq 2 ]
                if [ `expr $year % 100` -eq 0 -a `expr $year % 400` -eq 0 ]
                elif [ `expr $year % 100` -ne 0 -a `expr $year % 4` -eq 0 ]
        elif [ $month -eq 4 -o $month -eq 6 -o $month -eq 9 -o $month -eq 11 ]
        julday=`expr $julday - $daysofmth`
        if [ $julday -lt 0 ]
                julday=`expr $daysofmth + $julday`
        elif [ $julday -eq 0 ]
#########  main script starts here
if [ $# -ne 2 ]
        echo "Usage: fromjul <yyyy> <julian day>"
        exit 1



while [ $done -ne 1 ]
        month=`expr $month + 1`

printf "%.4d%.2d%.2d\n" $year $month $julday

# 5  
This does works exept for the 1st day of the month.
date | awk '{printf"%4d%2d%2d\n",$6,$2,($3-1)}' | sed 's/ /0/g'

# 6  
date | awk '{printf"%4d%2d%2d\n",$6,$2,($3-1)}' | sed 's/ /0/g'

i dont think the above one would work... the second parameter ($2) is displayed as such for month .. which is string format (ex: jul) but required output is in numeric format (07 - for jul)

here is the modification of the aboe one:

date '+%y:%m:%d' | awk -F":" '{printf"20%2d%2d%2d\n",$1,$2,($3-1)}' | sed 's/ /0/g'
# 7  
My env is Japanese, so this happened.
Thank you for modification!

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