Get yesterday's date in bash on Linux, DST-safe

Question

I have a shell script that runs on Linux and uses this call to get yesterday's date in YYYY-MM-DD format:

date -d "1 day ago" '+%Y-%m-%d'

It works most of the time, but when the script ran yesterday morning at 2013-03-11 0:35 CDT it returned "2013-03-09" instead of "2013-03-10".

Presumably daylight saving time (which started yesterday) is to blame. I'm guessing the way "1 day ago" is implemented it subtracted 24 hours, and 24 hours before 2013-03-11 0:35 CDT was 2013-03-09 23:35 CST, which led to the result of "2013-03-09".

So what's a good DST-safe way to get yesterday's date in bash on Linux?

Answer 1

I think this should work, irrespective of how often and when you run it ...

date -d "yesterday 13:00" '+%Y-%m-%d'

Answer 2

Under Mac OSX date works slightly different:

For yesterday

date -v-1d +%F

For Last week

date -v-1w +%F

Answer 3

This should also work, but perhaps it is too much:

date -d @$(( $(date +"%s") - 86400)) +"%Y-%m-%d"

Answer 4

If you are certain that the script runs in the first hours of the day, you can simply do

  date -d "12 hours ago" '+%Y-%m-%d'

BTW, if the script runs daily at 00:35 (via crontab?) you should ask yourself what will happen if a DST change falls in that hour; the script could not run, or run twice in some cases. Modern implementations of cron are quite clever in this regard, though.

Answer 5

date -d "yesterday" '+%Y-%m-%d'

To use this later:

date=$(date -d "yesterday" '+%Y-%m-%d')

Answer 6

Here a solution that will work with Solaris and AIX as well.

Manipulating the Timezone is possible for changing the clock some hours. Due to the daylight saving time, 24 hours ago can be today or the day before yesterday.

You are sure that yesterday is 20 or 30 hours ago. Which one? Well, the most recent one that is not today.

echo -e "$(TZ=GMT+30 date +%Y-%m-%d)\n$(TZ=GMT+20 date +%Y-%m-%d)" | grep -v $(date +%Y-%m-%d) | tail -1

The -e parameter used in the echo command is needed with bash, but will not work with ksh. In ksh you can use the same command without the -e flag.

When your script will be used in different environments, you can start the script with #!/bin/ksh or #!/bin/bash. You could also replace the \n by a newline:

echo "$(TZ=GMT+30 date +%Y-%m-%d)
$(TZ=GMT+20 date +%Y-%m-%d)" | grep -v $(date +%Y-%m-%d) | tail -1

Answer 7

you can use

date -d "30 days ago" +"%d/%m/%Y"

to get the date from 30 days ago, similarly you can replace 30 with x amount of days

Answer 8

Just use date and trusty seconds:

As you rightly point out, a lot of the details about the underlying computation are hidden if you rely on English time arithmetic. E.g. -d yesterday, and -d 1 day ago will have different behaviour.

Instead, you can reliably depend on the (precisely documented) seconds since the unix epoch UTC, and bash arithmetic to obtain the moment you want:

date -d @$(( $(date +"%s") - 24*3600)) +"%Y-%m-%d"

This was pointed out in another answer. This form is more portable across platforms with different date command line flags, is language-independent (e.g. "yesterday" vs "hier" in French locale), and frankly (in the long-term) will be easier to remember, because well, you know it already. You might otherwise keep asking yourself: "Was it -d 2 hours ago or -d 2 hour ago again?" or "Is it -d yesterday or -d 1 day ago that I want?"). The only tricky bit here is the @.

Armed with bash and nothing else:

Bash solely on bash, you can also get yesterday's time, via the printf builtin:

 %(datefmt)T
     causes printf to output the date-time string resulting from using
     datefmt as a format string for strftime(3).  The  corresponding  argu‐
     ment  is an integer representing the number of seconds since the
     epoch.  Two special argument values may be used: -1 represents the
     current time, and -2 represents the time the shell was invoked.
     If no argument is specified, conversion behaves as if -1 had
     been  given.
     This is an exception to the usual printf behavior.

So,

# inner printf gets you the current unix time in seconds
# outer printf spits it out according to the format
printf "%(%Y-%m-%d)T\n" $(( $(printf "%(%s)T" -1) - 24*3600 ))

or, equivalently with a temp variable (outer subshell optional, but keeps environment vars clean).

(
  now=$(printf "%(%s)T" -1);
  printf "%(%Y-%m-%d)T\n" $((now - 24*3600));
)

Note: despite the manpage stating that no argument to the %()T formatter will assume a default -1, i seem to get a 0 instead (thank you, bash manual version 4.3.48)

Answer 9

You can use:

date -d "yesterday 13:55" '+%Y-%m-%d'

Or whatever time you want to retrieve will retrieved by bash.

For month:

date -d "30 days ago" '+%Y-%m-%d'

Answer 10

As this question is tagged bash "DST safe":

And using fork to date command implie delay, there is a simple and more efficient way using pure bash built-in:

printf -v tznow '%(%z %s)T' -1
TZ=${tznow% *} printf -v yesterday '%(%Y-%m-%d)T' $(( ${tznow#* } - 86400 ))
echo $yesterday

This is a lot quicker on more system friendly than having to fork to date.

From bash version 5.0, there is a new variable $EPOCHSECONDS

printf -v tz '%(%z)T' -1
TZ=$tz printf -v yesterday '%(%Y-%m-%d)T' $(( EPOCHSECONDS - 86400 ))
echo $yesterday