date command in a bash shell script

Question

At work all the days config files are generated fresh and appended with a session number. The company went public on Feb 16, and the 86400 is seconds in one day. The session number is generated by subtracting the company start day from seconds_since_last_day and adding a few zero's

That is the key to interacting with the days config files. I get this - However I do not understand the

date -ud "$distance days ago 00:00:00".

Is it the number of seconds since 1970?

if $session; then
        # return the session of the last day
        seconds_since_day_one=`date -ud "Feb 16 2002" +"%s"`
        seconds_since_last_day=`date -ud "$distance days ago 00:00:00" +"%s"`
        days_between=`printf "%010d" $(( (seconds_since_last_day -  seconds_since_day_one) / 86400 ))`
        # Truncate on the left to 9 bytes to leave room
        # to append the engine suffix for your environment
        echo $days_between | awk '{l=length($1); print substr( $1, (l-8), l )}'

Mattias Ahnberg · Accepted Answer · 2012-01-09 23:48:09Z

date -ud "$distance days ago 00:00:00" in itself just prints the date a certain amount of days ago in a quite readable format, but when you add the FORMAT string to control the output +"%s" does indeed mean the number in so called Unix Time (number of seconds since 1970-01-01 00:00:00 UTC).

If the variable $distance is set to a number it shows the date that number of days ago, if its set to 0 it means today, 1 it means yesterday, 2 the day before yesterday and so on. To better understand these formats and relative keywords there are rather good documentations in (amongst other places) the GNU coreutils package.

Check these URLs:
http://www.gnu.org/software/coreutils/manual/html_node/Relative-items-in-date-strings.html#Relative-items-in-date-strings
http://www.gnu.org/software/coreutils/manual/html_node/Date-input-formats.html
http://www.gnu.org/software/coreutils/manual/html_node/date-invocation.html#date-invocation

Wikipedia explanation of Unix Time:
http://en.wikipedia.org/wiki/Unix_time

Elias Dorneles · Accepted Answer · 2012-01-09 23:12:22Z

0

The option -d to date provides a generic string to obtain the date.

So, for example, date -d yesterday will print yesterday's date, and date -d 'yesterday 12:00 AM' will print yesterday's date with the time set to 12:00 AM.

So, date -d 6 days ago 00:00:00 will print the date from 6 days ago, with the time set to 00:00:00. I hope it answers your question.

The format +"%s" tells date to print the number of seconds from 1970, instead the date.

answered Jan 9, 2012 at 23:12

Elias Dorneles

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5 Comments

capser Over a year ago

if date -d 6 days ago will print the date from six days ago, and date -ud "$distance days ago 00:00:00" = Sun Jan 8 00:00:00 UTC 2012 (yesterdays date) and `date -ud "$distance days ago 00:00:00" +"%s" will give me - 1325980800 what is the value of $distance? Because I dont see $distance having any value. and this shell script gives me the correct session.

capser Over a year ago

because if distance was null i thought that it would break the date command, but it does not seem to be breaking anything.

capser Over a year ago

I forgot the 'fi' at the end of the script.

Mattias Ahnberg Over a year ago

If $distance is an empty string it defaults to yesterday (1 days ago), if its set to 0 it means "today".

Elias Dorneles Over a year ago

that's right, I just tested with date -ud' days ago', if it's empty it defaults to yesterday.

user982733 · Accepted Answer · 2012-01-10 04:31:25Z

0

mktime and strftime in awk can be used to get the date of the time. http://www.gnu.org/software/gawk/manual/html_node/Time-Functions.html

For instance, strftime("%A",mktime("YYYY MM DD 00 00 00")) should give you the day.

answered Jan 10, 2012 at 4:31

user982733

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