I need to convert Julian timestamp to Regular timestamp in UNIX using Bash.
On Tandem OS, conversion is pretty straightforward -
Example:
212186319010244541
$OLSAPP SYSTST 1> #interprettimestamp 212186319010244541
#interprettimestamp 212186319010244541 expanded to:
2455860 2011 10 25 16 10 10 244 541
I wish to do the same on UNIX environment. The conversion will be a part of a parser script. So one-liners would be greatly appreciated.
UPDATE:
INTERPRETTIMESTAMP inbuilt function on Tandem returns a space-separated list of nine numbers, consisting of the Julian day number, year, month, day, hour, minute, second, millisecond, and microsecond.
Assuming the number is as #blahdiblah says
"a value representing the number of microseconds since January 1, 4713 B.C."
Then you first need to know the Julian timestamp for 01-JAN-1970 which is the epoch for unix time. So a cludgy oracle query gives
210866803200000000
Then you could in theory just have a shell command to compute the number of seconds since 1-Jan-1970.
unixtime=$(( ( 212186319010244541 - 210866803200000000 ) / 1000000 ))
The problems with this are:
you still need to format it
your bash may not like integer arithmatic with 18 digit numbers. (think its OK in 64 bit, but not 32 bit).
Now if you have perl installed you can solve these using the bigint and POSIX modules. As a shell "one" liner it looks like
perl -mbigint -mPOSIX -e 'print( POSIX::strftime("%Y-%m-%d %T",localtime( ($ARGV[0]-210866803200000000)/1000000 ) )."\n")' 212186319010244541
Which gives
2011-10-25 15:10:10
The 1 hour difference is probably due to daylight savings differences. It could be either in the perl, or more likely the value I used for 01-Jan-1970 could be an hour out. So you may need to check both of them to be sure its right for your system.
this is MJD converter
MJD is -3506716800 than epoch
# date -d '1858-11-17 UTC' +%s
-3506716800
example:
MDJ 57153 is Mon May 11 00:00:00 UTC 2015
# date -d #`echo 57153*86400-3506716800|bc`
Mon May 11 00:00:00 UTC 2015
Related
I have a file "file_XYZ_18548".
Here the name's ending 18548 is a timestamp in days, and it is changing day by day, like "file_XYZ_18550".
I would like to get this date via variable but I couldn't find a date command to get the timestamp in days.
I can get the date result but I can't get the timestamp in reverse.
timeinday=18550
timestp=timeinday*86400
datetm=$(echo $timestp | gawk '{print(strftime("%Y-%m-%d %H:%M:%S", $0))}')
echo $datetm
2020-10-14 10:09:10
How can I get this with date command in bash scripting? Is there any way of this via awk/gawk etc..?
Given that 2020-10-14 - 18548 days corresponds to 1970-01-03, it is reasonable to guess that the 'epoch' for the day count is 1970-01-01. It is likely that day 1 was 1970-01-01, so day zero was 1969-12-31.
Converting day count to date
You can use the GNU date command like this:
daycount=18548
date -u -d "#$(( ($daycount-1) * 86400 ))" +"%Y-%m-%d %H:%M:%S"
That yields the result 2020-10-12 00:00:00. You can drop the time component of the format if you wish (you probably do; midnight isn't very exciting when it is always midnight). You can calibrate the -1 to resolve exactly what day should correspond to 18548. Just in case it isn't obvious, there are 86,400 seconds in a day (24 hours • 60 minutes per hour • 60 seconds per minute).
The $(( … )) notation is Bash's Arithmetic Expansion notation.
Converting current time to day count
If you want to convert the current time to the day offset, then you can use the %s specifier to get the seconds since 1970-01-01 00:00:00Z and divide by 86400 to get the number of days:
echo $(( $(date +'%s') / 86400 ))
which (at 2020-10-14 23:30 -06:00, aka 1602739800 seconds since the Unix Epoch) yields the result:
18550
Again, if need be, you can adjust the value of the division to account for when day 1 was in this scheme. Shell arithmetic in Bash is integer arithmetic, which is exactly what is wanted. You might need to use -u to get UTC as the time zone (as I did earlier), and so on.
I have two timestamps in formats :
Timestamp 1 (Variable - RunStartDate): Thu May 3 14:12:54 CDT 2018
Timestamp 2 (Variable - RunEndDate): Thu May 3 18:11:46 CDT 2018
I want the difference of number of hours between these two timestamps in UNIX shell. (I.e. RunEndDate - RunStartDate in hours)
Please help, I am new to UNIX and it is throwing me errors when I just try to subtract the two.
You have a few options here, such as calling out to Perl or Python and using a date/time library to do the math for you. Another option is to use the date program to convert the dates to seconds, subtract the values, and then convert back to hours. Unfortunately, you can't do floating-point math in Bash, so we'll have to call out to a helper program to do that, too.
START=$(date -d "$RunStartDate" +"%s")
END=$(date -d "$RunEndDate" +"%s")
HOURS=$(bc -l <<< "($END - $START) / 3600")
Note that this will only work on GNU systems (e.g. Linux).
When I issue bash command:
date --date="2018-03-03 12:16:13 -1hour" "+%Y:%m:%d %H:%M:%S"
I expect the result would be:
2018:03:03 11:16:13
but instead, I get:
2018:03:03 15:16:13
I wonder if this has to make with time zones, and how to avoid this behaviour.
I can reproduce this. My timezone is America/New_York
$ date --date="2018-03-03 12:16:13 - 1 hour" "+%Y:%m:%d %H:%M:%S"
2018:03:03 09:16:13
$ env TZ='Europe/Belgrade' date --date="2018-03-03 12:16:13 - 1 hour" "+%Y:%m:%d %H:%M:%S"
2018:03:03 15:16:13
The parser appears to be taking the -1 as the timezone GMT+01:00, then converting that to your local timezone.
If we rearrange the phrases to avoid the timezone parsing ambiguity, we can get your desired result:
$ date --date="- 1 hour 2018-03-03 12:16:13" "+%Y:%m:%d %H:%M:%S"
2018:03:03 11:16:13
From info coreutils 'date invocation'
When a relative item causes the resulting date to cross a boundary where the clocks were adjusted, typically for daylight saving time,
the resulting date and time are adjusted accordingly.
The fuzz in units can cause problems with relative items. For
example, '2003-07-31 -1 month' might evaluate to 2003-07-01, because
2003-06-31 is an invalid date. To determine the previous month more
reliably, you can ask for the month before the 15th of the current
month. For example:
$ date -R
Thu, 31 Jul 2003 13:02:39 -0700
$ date --date='-1 month' +'Last month was %B?'
Last month was July?
$ date --date="$(date +%Y-%m-15) -1 month" +'Last month was %B!'
Last month was June!
Also, take care when manipulating dates around clock changes such
as daylight saving leaps. In a few cases these have added or
subtracted as much as 24 hours from the clock, so it is often wise to
adopt universal time by setting the 'TZ' environment variable to
'UTC0' before embarking on calendrical calculations.
One can avoid that by putting -1 hour before the string,
$ date --date='-1 hour 2018-03-03 12:16:13' "+%Y:%m:%d %H:%M:%S"
2018:03:03 11:16:13
I have a time-stamp like 7:00:00, which means 7am.
I would like to write a short command that returns 06:45:00, or simply 06:45, preferably using date command so that I can avoid long shell script. Do you have any elegant solution?
I'm also looking for a 24h format. For example, 12:00:00 - 15 minutes = 11:45:00.
With GNU date, use 7:00:00 AM - 15 minutes as d (--date) string :
% date -d '7:00:00 AM - 15 minutes' '+%H:%M'
06:45
+%H:%M sets the output format as HH:MM.
On BSD variants Date has a -v flag which can be used to take the current timestamp and display the result of a positive or negative adjustment.
This will subtract 15mins from the current timestamp:
date -v -15M
I would like to automate a download of an image from a third party server with a CRON job and then upload the image to my website.
I have 2 issues:
First, the third party site changes the image name every day using the following logic:
http://thirdpartysite.com/ImageFinder.aspx?ReportID=FILENAME
where FILENAME is 26601 +14 for each day after 6 Oct 2014 (so 7 Oct would be 26615, 8 Oct would be 26629 etc).
How do I build this into a simple Linux bash script for use with wget?
Second, how do I upload this to my site via FTP (or similar) with the same script.
NOTE: I have permission to host the file on my site and have linked the original site / placed credit for the image.
Following the suggestions of #Abhay, first get the timestamp of Oct 6, let's store it in the variable $d0:
d0=$(date +%s -d 20141006)
Then store the timestamp of a target date, say Oct 8 and store it in $d1:
d1=$(date +%s -d 20141008)
Then you can calculate the difference and apply the required arithmetic operations in $((...)), like this:
echo $((26601 + 14 * (d1 - d0) / 60 / 60 / 24))
# outputs: 26629
The date command has one very good format: %s, which prints number of seconds since "epoch", which is the fixed date 1 January 1970, 00:00 UTC. I'll call it "timestamp". In combination with this, you can use the -d date-string, so that it prints the given date as number of seconds. Now you can take today's timestamp, subtract the timestamp of "6 Oct 2014" from it, and you get number of seconds between the two times. Now you can divide it by (60 * 60 * 24) to get it in number of days, and do further arithmetic to get the desired number, and make a file name out of it.
The date string formats that -d option takes are flexible, but as of now I am not sure whether it takes "6 Oct 2014" as is. Try a few permutations, or better, check the "info" page.