I need to write a variable with "date" function and I need that the format is yyyymmdd (i.e. 20151031). Also I would remove a day (i.e. 20151031 becomes 20151030) I would use bash.
Thanks!
3
Use date +%Y%m%d to print the current date.
Use date --date '1 day ago' +%Y%m%d to print yesterday's date (GNU's date).
Use date -v -1d +Y%m%d (FreeBSD).
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Ok, but if I put it into a variable I get error #!/usr/bin/env bash DATE=date --date '1 day ago' +%Y%m%d gunzip -c /var/log/cisco/cisco.log-echo $DATE.gz > /var/log/cisco/file.log root@raspberrypi:/etc/my_scripts# date --date date: option '--date' requires an argument – Federi Nov 3 '15 at 13:56
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You have to enclose your command into
$(). Example :DATE=$(date --date '1 day ago' +%Y%m%d)– Vinz Nov 3 '15 at 13:59
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bash-4.2 and above have built-in support for that (inspired from something similar in ksh93):
printf -v now '%(%s)T' -1
printf '%(%Y%m%d)T\n' "$((now - 86400))"
Would print yesterday's date (well, the date 86400 seconds ago) in YYYYMMDD format. It may not work if run at certain hours of the night at the time of winter/summer time switch (when days are 23 or 25 hours long).
It would probably be more reliable for those corner cases if written as:
printf -v now '%(%s:%H)T' -1
printf '%(%Y%m%d)T\n' "$((${now%:*} - 3600 * (12 + ${now#*:})))"
man date. thedatecommand allows you to format the output – Centimane Nov 3 '15 at 13:24man date, and yesterday is: epoch - 86400 (seconds) – Olivier Dulac Nov 3 '15 at 13:34