My variable DateBecame looks like this:
19550101
It is of long type and the format is %10.0g. So YMD format.
I also have another variable DateLeft:
19961001
This is also of long type and format %10.0g.
How can I calculate the duration between the two dates?
The following works for me:
clear
input long(date1 date2)
19550101 19961001
end
generate diff = daily(string(date1, "%10.0g"), "YMD") - ///
daily(string(date2, "%10.0g"), "YMD")
list
+---------------------------------+
| date1 date2 diff |
|---------------------------------|
1. | 19550101 19961001 -15249 |
+---------------------------------+
Note that variables of long type are generally not ideal for this kind of calculations.
Type help datetime from Stata's command prompt for more information on working with dates.
Related
I have such events:
something;<id>abc123<timeStamp>2021-12-10T23:10:12.044Z<timeStamp>2021-12-10T23:08:55.278Z>
I want to extract the Id abc123 and the two timeStamps.
index = something
|rex field=_raw "id>(?<Id>[0-9a-z-]+)"
|rex "timeStamp>(?<timeStamp>[T0-9-\.:Z]+)"
| table _time Id timeStamp
This works with the query above. But what I struggle now is to convert the timeStamp-string to date format to get at the end the min(timeStamp) extracted in order to compute the difference between the event's _time and the min(timeStamp) by the id field. I am struggling because of the special format of the timestamp with T and Z included in it.
There's nothing special about those timestamps - they're in standard form. Use the strptime function to convert them.
index = something
|rex field=_raw "id>(?<Id>[^\<]+)"
|rex "timeStamp>(?<timeStamp>[^\<]+)"
| eval ts = strptime(timeStamp, "%Y-%m-%dT%H:%M:%S.%3N%Z")
| eval diff = ts - _time
| table _time Id timeStamp diff
Check out strftime.org, and the related strptime function used with eval
Something on the order of this (pulled the microseconds out of your rex, since Unix epoch time has no concept of subsecond intervals):
| rex field=_raw "timeStamp\>(?<timeStamp>[^\.]+)\.\d+Z"
| eval unixepoch=strptime(timeStamp,"%Y-%m-%dT%H:%M:%S")
I am new to SAS and I am having trouble with finding the difference between 2 dates.
I have 2 columns: checkin_date and checkout_date
the dates are in mmddyy10. format (mm/dd/yyyy).
I have used the following code:
stay_days= intck('day', checkin_day, checkout_day);
I am getting the right values for dates in the same month but wrong values for days that are across 2 months. For example, the difference between 02/06/2014 and 02/11/2014 is 5. But the difference between 1/31/2014 and 2/13/2014 is -18 which is incorrect.
I have also simply tried to subtract them both:
stay_day = checkout_day - checkin_day;
I am getting the same result for that too.
My entire code:
data hotel;
infile "XXXX\Hotel.dat";
input room_no num_guests checkin_month checkin_day checkin_year checkout_month checkout_day checkout_year internet_used $ days_used room_type $16. room_rate;
checkin_date = mdy(checkin_month,checkin_day,checkin_year);
informat checkin_date mmddyy.;
format checkin_date mmddyy10.;
checkout_date = mdy(checkout_month,checkout_day,checkout_year);
informat checkout_date mmddyy.;
format checkout_date mmddyy10.;
stay_day= intck('day', checkin_day, checkout_day);
Your problem is a typo - using wrong variables in intck() function. You are using variables "xxx_DAY" which is the DAY of month instead of the full DATE. Change to stay_day= intck('day', checkin_date, checkout_date);
Your data probably has the date values in the wrong variables. When using subtraction the order should be ENDDATE - STARTDATE. When using INTNX() function the order should be from STARTDATE to ENDDATE. In either case if the value in the STARTDATE variable is AFTER the value in the ENDDATE variable then the difference will be a negative number.
Perhaps you need to clean the data?
The only way to get -18 comparing 2014-01-31 and 2014-02-13 would be if you extracted the day of the month and subtracted them.
diff3 = day(end) - day(start);
which would be the same as subtracting 31 from 13.
Example using your dates:
data check;
input start end ;
informat start end mmddyy.;
format start end yymmdd10.;
diff1=intck('day',start,end);
diff2=end-start;
cards;
02/06/2014 02/11/2014
1/31/2014 2/13/2014
;
Results:
Obs start end diff1 diff2
1 2014-02-06 2014-02-11 5 5
2 2014-01-31 2014-02-13 13 13
I have a dataset containing a date field in the format of MM/dd/yyyy, my goal is to create a table with the same date format but with timestamp format (at the end, there should be an option to execute date functions on that, if it is int or string date function will not work.)
Things I tried:
my column name: as_of_date
1) cast(unix_timestamp(as_of_date, "MM/dd/yyyy") as timestamp)
i/p -> 01/03/2006, o/p ->2006-01-03 00:00:00
problem - I do not want extra zeros in the output. substr is not working on the date function
2) If i keep the value as string, date functions does't work.
day('01/03/2006')
input: '01/03/2006' , output:null (but expected 3)
Can you please help me a date format that already existing or help me to create a new date format for my logic.
Try with this once
use unix_timestamp function to match your input date format then use from_unixtime function to change the output format then cast to date type.
hive> select date(from_unixtime(unix_timestamp("03/06/2018", "MM/dd/yyyy"),"yyyy-MM-dd"));
+-------------+--+
| _c0 |
+-------------+--+
| 2018-03-06 |
+-------------+--+
In Impala:
hive> select from_unixtime(unix_timestamp("03/06/2018", "MM/dd/yyyy"),"yyyy-MM-dd");
+-------------+--+
| _c0 |
+-------------+--+
| 2018-03-06 |
+-------------+--+
I have some very strange looking 18 character alphanumeric datetimes in a SQL database, they seem to be using Hexadecimal?
I can find out what the dates are through the application which uses them, but I was looking for a way to convert them via a query. Do you know how I would convert these with TSQL?
000B3E4Bh01F2D947h - 29/05/2018 09:04:52
000B3E0Dh03A16C1Eh - 23/05/2018 10:22:26
000B3E4Eh0248C3D8h - 01/06/2018 10:38:43
000B3E4Eh0249B449h - 01/06/2018 10:39:44
I assume the date and time are separated as below, but I don't know to convert the individual parts if anyone can help with this? Thanks!!
000B3E4Eh (date) - 0249B449h (time)
(The dates are in dd/mm/yyyy format)
Your hex values are separated as you have assumed (with the h used as a delimiter) and represent integer values to add to a baseline date and time value.
Using your 000B3E54h0221CBFEh - 07/06/2018 09:56:09 value this translates as:
Date portion: 000B3E54
Integer Value: 736852
Time portion: 0221CBFE
Integer Value: 35769342
These integer values are then added as days to the date 0001/01/00 (which SQL Server can't handle, hence the +/-1 below) and milliseconds to 00:00:00 respectively, which you can see working in this script:
select convert(int, 0x000B3E54) as DateIntValue
,dateadd(day,convert(int, 0x000B3E54)-1,cast('00010101' as datetime2)) as DateValue
,convert(int, 0x0221CBFE) as TimeIntValue
,cast(dateadd(millisecond,convert(int, 0x0221CBFE),cast('19000101' as datetime2)) as time) as TimeValue
,cast(datediff(day,cast('00010101' as datetime2),'20180607')+1 as binary(4)) as DateHexValue
,cast(datediff(millisecond,cast('20180607' as date),cast('2018-06-07 09:56:09.342' as datetime2)) as binary(4)) as TimeHexValue
Which outputs:
+--------------+-----------------------------+--------------+------------------+--------------+--------------+
| DateIntValue | DateValue | TimeIntValue | TimeValue | DateHexValue | TimeHexValue |
+--------------+-----------------------------+--------------+------------------+--------------+--------------+
| 736852 | 2018-06-07 00:00:00.0000000 | 35769342 | 09:56:09.3420000 | 0x000B3E54 | 0x0221CBFE |
+--------------+-----------------------------+--------------+------------------+--------------+--------------+
Note the judicious use of datetime2 values to ensure the right amount of milliseconds are output/returned, as SQL Server datetime is only accurate to the nearest 3 milliseconds.
I have a table where the dates are in Julian dates, and I would like to convert these dates into Calendar dates.
Here is a sample of a Julian date I have: 2457395.
In calendar date it should be: 07012016 (DDMMYYYY).
I tried to convert the date into the number of days since (01/01/1900) but even if I don't know how to have a date with a number of days as an input.
Many thanks in advance!
Impala does not support dates, only timestamps; and it does not have many date/time formatting features; so you need to be creative, e.g.
select JULIAN_DATE,
adddate('1900-01-01 00:00:00Z', JULIAN_DATE -2415021) as AS_TIMESTAMP,
to_date(adddate('1900-01-01 00:00:00Z', JULIAN_DATE -2415021)) as AS_ISO_DATE_STRING,
from_unixtime(unix_timestamp(adddate('1900-01-01 00:00:00Z', JULIAN_DATE -2415021)), "ddMMyyyy") as AS_DMY_DATE_STRING
from WHATEVER
Sample output:
julian_date as_timestamp as_iso_date_string as_dmy_date_string
----------- ------------------- ------------------ ------------------
2457395 2016-01-07 00:00:00 2016-01-07 07012016
Well if it is a JDE Julian date for example 118163 for 12-06-2018 then below code can be used
date_add(to_timestamp(concat(substr(cast(cast(118162 + 1900000 as int) as string),1,4),'01','01'),'yyyyMMdd'),
cast(substr(cast(cast(118162 + 1900000 as int) as string),5,3) as int)-1)