Objective: with a shortcut, e.g. alt+d, insert
yyyy-mm-dd Ddd
Example of Ddd: Mon
I have the following code for an Autokey script
from datetime import datetime
keyboard.send_keys(datetime.now().strftime('%Y-%m-%d'))
That only gives me the date.
How should I expand the script to get the 3 character day too?
See docs.python.org/2/library/… then use %a in the format string: e.g. datetime.now().strftime('%Y-%m-%d %a')
Related
I'm having trouble converting a couple of sting fields to date with Talend Open Studio. All date fields are coming to me via csv in the format "MM/dd/yyyy hh:mm:ss aa" For example "03/20/2018 2:40:03 PM"
Many date fields are fine, and conversion with the t-map component and the talend.date (parse date) is working as it should. The problem occurs in two fields where many almost all of the dates are coming in as "12/30/1899 00:00:00 AM"
I'm using the tmap to push the file to a toutputdelimited (CSV). The format of the fields with "12/30/1899 00:00:00 AM" is coming back as "#######################." Any ideas on how to handle the situation would be great. The output date format is "yyyy-MM-dd HH:mm:ss"
Again, all other fields are converting as they should. Its only the special case "12/30/1899 00:00:00 AM" that is causing problems.
Thans!
Since your dates can be of different formats, you need to read your column as a String type in your tFileInputDelimited, then inside a tMap, check the date type (a simple way to check it is by testing its length) and parse it using the correct date format:
tFileInputDelimited -- tMap -- output
In a tMap expression, you can have something like this :
row.created_ts.length() == 22 ? TalendDate.parseDate("yyyy-MM-dd hh:mm:ssX", row.yourdate) : TalendDate.parseDate("yyyy-MM-dd hh:mm:ss.SSSSSSX", row.yourdate)
What is the command for displaying the time and date in qbasic? Could the syntax for the commands be given as well? And an explanation if possible?
You can use DATE$ and TIME$
These can also set the date and time as well.
The command for printing the time(current system time) is time$
The time$ is actually a function, in this case, no parameter is needed.
And the code is...
PRINT TIME$
The time is printed in hh: mm: ss format(hour: minutes: seconds).
And therefore the output would be something like this:
14:55:28
For printing the current system date, we use date$ function which is also a string function
The code is:
PRINT DATE$
The date is printed in mm-dd-yyyy format or month-day-year(American date format).
Hence the output will be:
02-17-2018
Hope it helps...
The QB date/time functions are:
DATE$ returns the date in a string in the form MM-DD-YYYY
TIME$ returns the time in a string in the form HH:MM:SS
When used as a command the date$ and time$ can be assigned to set the system date and time, for example DATE$ = "12-10-1990" or TIME$ = "12:10:10"
If the year is a leap year then the 29th day of February could be set. Otherwise if it is not a leap year then a syntax error will occur trying to set the date in February to the 29th.
Hi I have a date conversion problem in SAS,
I imported an excel file which has the following dates.,
2012-01-09
2011-01-31
2010-06-28
2005-06-10
2012-09-19
2012-09-19
2007-06-12
2012-09-20
2004-11-01
2007-03-27
2008-06-23
2006-04-20
2012-09-20
2010-07-14
after I imported the dates have changed like this
40917
40574
40357
38513
41171
41171
39245
41172
38292
39168
39622
38827
41172
40373
I have used the input function to convert the dates but it gives a strange result.,
the code I used.,
want_date=input(have_date, anydtdte12.);
informat want_date date9.; format have_date date9.;run;
I get very stange and out of the World dates., any idea how can I convert these?
You can encourage SAS to convert the data as date during the import, although this isn't necessarily a panacea.
proc import file=whatever out=whatever dbms=excel replace;
dbdsopts=(dbSasType=( datevar=date ) );
run;
where datevar is your date column name. This tells SAS to expect this to be a date and to try to convert it.
See So Your Data Are in Excel for more information, or the documentation.
From : http://www2.sas.com/proceedings/sugi29/068-29.pdf
Times are counted internally in SAS as seconds since midnight and
date/time combinations are calculated as the number of seconds since
midnight 1 January 1960.
Excel also uses simple numerical values for dates and times
internally. For the date values the difference with the SAS date is
only the anchor point. Excel uses 1 January 1900 as day one.
So add a constant.
EXAMPLES:
SAS_date = Excel_date - 21916;
SAS_time = Excel_time * 86400;
SAS_date_time = (Excel_date_time - 21916) * 86400;
As Justin wrote you need to correct for the different zero date (SAS vs. Excel).
Then you just need to apply a format (if you want to get a date variable to do calculations):
want_date = have_date-21916;
format want_date date9.;
Or convert it to a string:
want_date = put(have_date-21916, date9.);
In either case you can choose the date format you prefer.
in DB2 I've got date values stored as varchar in the form 'DD-Mon-YY' (e.g. 25-Jun-13). I'd like to convert these into DB2 compatible date formats using the TO_DATE function but every date conversion format I've tried gives me an error.
e.g. I've tried TO_DATE('25-Jun-13', 'YYYYMMDD') and TO_DATE('25-Jun-13', DDMMYYYY) and I always get something like "25-Jun-13" cannot be interpreted using format string "DDMMYYYY" for the TIMESTAMP_FORMAT function.
Does anyone know if there is a format string that I can use?
I think you're looking for:
DATE(TIMESTAMP_FORMAT(#your_date, 'DD-Mon-YY'))
Aside: TO_DATE is just a synonym for TIMESTAMP_FORMAT.
to_date('string', 'template')
the date format of the string must match the template
MM is month as number
MON is MONth as abbreviation
Example
select to_date('20130625', 'YYYYMMDD') as YYYYMMDD
,to_date('25-Jun-2013', 'DD-MON-YYYY') as DDMONYYYY
from sysibm.sysdummy1
YYYYMMDD DDMONYYYY
2013-06-25 00:00:00.0 2013-06-25 00:00:00.0
Try that
TO_DATE(#your_string, 'yyyy/mm/dd')
I'm trying to use joda-time to parse a date string of the form YYYY-MM-DD. I have test code like this:
DateTimeFormatter dateDecoder = DateTimeFormat.forPattern("YYYY-MM-DD");
DateTime dateTime = dateDecoder.parseDateTime("2005-07-30");
System.out.println(dateTime);
Which outputs:
2005-01-30T00:00:00.000Z
As you can see, the DateTime object produced is 30 Jan 2005, instead of 30 July 2005.
Appreciate any help. I just assumed this would work because it's one of the date formats listed here.
The confusion is with what the ISO format actually is. YYYY-MM-DD is not the ISO format, the actual resulting date is.
So 2005-07-30 is in ISO-8601 format, and the spec uses YYYY-MM-DD to describe the format. There is no connection between the use of YYYY-MM-DD as a pattern in the spec and any piece of code. The only constraint the spec places is that the result consists of a 4 digit year folowed by a dash followed by a 2 digit month followed by a dash followed by a two digit day-of-month.
As such, the spec could have used $year4-$month2-$day2, which would equally well define the output format.
You will need to search and replace any input pattern to convert "Y" to "y" and "D" to "d".
I've also added some enhanced documentation of formatting.
You're answer is in the docs: http://www.joda.org/joda-time/apidocs/org/joda/time/format/DateTimeFormat.html
The string format should be something like: "yyyy-MM-dd".
The date format described in the w3 document and JodaTime's DateTimeFormat are different.
More specifically, in DateTimeFormat, the pattern DD is for Day in year, so the value for DD of 30 is the 30th day in the year, ie. January 30th. As the formatter is reading your date String, it sets the month to 07. When it reads the day of year, it will overwrite that with 01 for January.
You need to use the pattern strings expected by DateTimeFormat, not the ones expected by the w3 dat and time formats. In this case, that would be
DateTimeFormatter dateDecoder = DateTimeFormat.forPattern("yyyy-MM-dd");