I am working with a legacy system whereby it seems they store dates as integers, such as 733473. Anyone seen this format before and can think of the logic to turn this into a date? It appears to be a day number, but not sure how many days from what date!
Edit: the value I get for 01/01/2013 is 734869. It is the amount of days since 0 i believe, but does this have a name? I need to write/find a native function in MSSQL server to convert these.
You are most likely correct, it's the number of days since Jan 1, year 0000 (or possibly 1).
$ date -u -d "0000-01-01 + 733473 days"
Fri Mar 7 00:00:00 UTC 2008
However, without knowing the date it actually corresponds to, we can't know for sure which leap year system it follows.
OK not sure the name of this date type (if there is one), but the below works:
declare #dt datetime2 = '00010101'
declare #int int = 734869
select dateadd(day,#int-1,#dt)
2013-01-01 00:00:00.0000000
Related
I am getting an exif date value like
EXIFPhotoDate: 1506173228000 and
UploadDate: 1506485214000
but I know it is
EXIFPhotoDate 23/9/2017, 23:27 and
UploadDate 9/27/2017, 01:59
The former is when queried via REST and the latter is when queried via the table.
How can I get standard date/time from a value like this?
Looks like you have a number of milliseconds since January 01 1970 in UTC. If you remove the 000 from the end, you will have a Unix timestamp, that is, the number of seconds since January 01 1970 in UTC:
unixTimestamp = 1506173228000 / 1000
Once your question doesn't state any programming language, it's hard to give you further help.
I need to assign a variable with the value in the format YYYYMM e.g for today run the previous month values should be generated as
Var1 = 201607
Is there any in build method available? Could you share the steps to generate this?
The trick in these cases is to subtract one month from the day 15 of the current month:
$ date --date="$(date +%Y-%m-15) - 1 month"
Fri Jul 15 00:00:00 CEST 2016
Then it is just a matter of using the proper format:
$ date --date="$(date +%Y-%m-15) - 1 month" "+%Y%m"
201607
To store the value in a var, just use the common var=$(command) syntax.
From GNU Coreutils → 28.7 Relative items in date strings:
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="$(date +%Y-%m-15) -1 month" +'Last month was %B!'
Last month was June!
Also, take care when manipulating dates around clock changes such as
daylight saving leaps. In a few cases these have added or subtracted
as much as 24 hours from the clock, so it is often wise to adopt
universal time by setting the TZ environment variable to ‘UTC0’ before
embarking on calendrical calculations.
I would like a spreadsheet row to contain the date of today, but only on every other Thursday, changing at 9:30 am.
To give you an example:
Next thursday the 21.07.16 it shell contain "21.07.16".
Until in exactly 14 days on thursday the 4.08.16 it shell contain this date and than change to 4.08.16.
Also I would like this change to happen at 9:30 am.
I can not think of a way how to do this. Can you point me into a direction?
One has to set a starting datetime somewhere in the past, such as July 7, 2016, at 9:30am.
Then find the difference between the current and the starting datetime. Truncate this difference down to a multiple of 14, and add this value to the starting datetime.
The datetimes are represented in Sheets numerically as the number of days since December 30, 1899. In this system, 2016-07-07 9:30 is about 42558.4 So the formula would be
=42558.4 + 14*floor((now()-42558.4)/14)
The output should be formatted as a date.
A less cryptic version is
=value("2016-07-07 09:30") + 14*floor((now() - value("2016-07-07 09:30"))/14)
(value follows the local convention for parsing dates, but I hope the format I used will be understood universally.)
I am in the process of optimizing some UniVerse data access code we have which uses UniObjects. After some experimentation, it seems that using a UniSession.OConv call to parse certain things such as decimal numbers (most we have a MR4 or MR2 or MR2$) and dates (almost all are D2/) is extremely slow (I think it might make a call back to the server to parse it).
I have already built a parser for the MR*[$] codes, but I was wondering about the dates as they are stored so I can build one for D2/. Usually they seem to be stored as a 5 digit number. I thought it could be number of days since the Unix Epoch since our UniVerse server runs on HP-UX, but after finding '15766' as a last modified date and multiplying it by 86400 (seconds per day), I got March 02, 2013 which doesn't make sense as a last modified date since as far as I know that is still in the future.
Does anyone know what the time base of these date numbers are?
It is stored as a number of days. Just do a conversion on 0 and you will get the start date.
Edit:
As noted by Los, the Epoch used in UniVerse (and UniData) is 31st Dec 1967.
In Universe and any other Pick database, Dates and Times are stored as separate values.
The internal date is the number of days before of after 31/12/1967, which is day zero.
The internal time is the number of seconds after midnight. It can be stored as a decimal but is not normally.
In TCL there is a CDT command (stands for Convert Date) that converts dates from human readable to numeric and and vice versa:
CDT 9/28/2017
* Result: 18169
CDT 18169
* Result: 09/28/2017
this is my question:
I'm migrating data from a Btrieve file (.dat) through Pervasive Control Center and there is field type which is defined as integer but is a date and for example the date '31/12/2009' (seen in the legacy system) is view it as the number 733772 when I export it.
The legacy system shows the date correctly but I can't export it in the same format or at least I can't convert it. Does anybody know how to convert this number through Excel or something?
When I divided 733772 by 365.2425 (Number of days in year considering Leap year and 29 days of Feb - http://www.timeanddate.com/date/leapyear.html), it gave back 2009.
Go to format cells and changing category to date.