I have date stored in a field like this: 31st Dec 2013 but I need to alter it to something like this: 2013-12-31 00:00:00
I have tried lots of variations of DATE_FORMAT and strtotime but got nowhere. How do I go about this please?
Thanks
The DATE_FORMAT() function is used to display date/time data in different formats.
Try this:
DATE_FORMAT(your_date,'%Y %m %d %T:%f')
Use a date_format inside another date_format. The inner one detects the format you are giving, the outer one outputs what you need.
Related
I am using TO_DATE in one of my PostgreSQL functions and it is throwing errors like date/time field value out of range: "2021901". This is happening for the months of January to September as I need to add zeros in front of them. So I tried to execute a simple select query there as follows as I am using the same syntax in function.
SELECT TO_DATE(2021::varchar||09::varchar||'01','YYYYMMDD')
This is also giving me the error
ERROR: date/time field value out of range: "2021901"
SQL state: 22008
Now if I change the month to October, November, or December it works fine, but for all the other months, it is showing this error. I am actually new to Postgres and not sure how to fix this. It would be very much helpful if someone can point me in the right direction. Thanks
If your input values are numbers (integer), another alternative is to use make_date()
make_date(2021,9,1)
A better and easier way would be to just provide the date correctly and use TO_DATE, so as example do this:
SELECT TO_DATE('2021-09-01', 'YYYY-MM-DD')
If you really want to go your way, you can force a leading zero by using TO_CHAR, like this:
SELECT TO_DATE(2021::varchar||TO_CHAR(09, 'fm00')||'01','YYYYMMDD')
But I recommend to take the first propose.
I am newbie to KDB. I have a KDB table which I am querying as:
select[100] from table_name
now this table has got some date columns which have dates stored in this format
yyyy.mm.dd
I wish to query that table and retrieve the date fields in specific format (like mm/dd/yyyy). If this would've been any other RDBMS table this is what i would have done:
select to_date(date_field,'mm/dd/yyyy') from table_name
I need kdb equivalent of above. I've tried my best to go through the kdb docs but unable to find any function / example / syntax to do that.
Thanks in advance!
As Anton said KDB doesn't have an inbuilt way to specify the date format. However you can extract the components of the date individually and rearrange as you wish.
For the example table t with date column:
q)t
date
----------
2008.02.04
2015.01.02
q)update o:{"0"^"/"sv'flip -2 -2 4$'string`mm`dd`year$\:x}date from t
date o
-----------------------
2008.02.04 "02/04/2008"
2015.01.02 "01/02/2015"
From right to left inside the function: we extract the month,day and year components with `mm`dd`year$:x before stringing the result. We then pad the month and day components with a null character (-2 -2 4$') before each and add the "/" formatting ("/"sv'flip). Finally the leading nulls are filled with "0" ("0"^).
Check out this GitHub library for datetime formatting. It supports the excel way of formatting date and time. Though it might not be the right fit for formatting a very large number of objects (but if distinct dates are very less then a keyed table and lj can be used for lookup).
q).dtf.format["mm/dd/yyyy"; 2016.09.23]
"09/23/2016"
q).dtf.format["dd mmmm yyyy"; 2016.09.03] // another example
"03 September 2016"
I don't think KDB has built-in date formatting features.
The most reliable way is to format date by yourself.
For example
t: ([]date: 10?.z.d);
update dateFormatted: {x: "." vs x; x[1],"/",x[2],"/",x[0]} each string date from t
gives
date dateFormatted
------------------------
2012.07.21 "07/21/2012"
2001.05.11 "05/11/2001"
2008.04.25 "04/25/2008"
....
Or, more efficient way to do the same formatting is
update dateFormatted: "/"sv/:("."vs/:string date)[;1 2 0] from t
now qdate is available for datetime parsing and conversion
I m having some issues with converting a regular timestamp in cassandra with Apache Nifi.
My use case is following:
I have a csv file with a date in it looking like this ('2015010109') and I want to put it in cassandra by converting this string ('2015010109') to an proper format: 2015-01-01 09:00 -> yyyy/MM/dd HH:mm (I dont exactly need the minutes, but I guess it is more useful for later usage)
So far I got this propertie in my UpdateAttribute processor when trying to convert this string to a timestamp:
date : ${csvfiledate:toDate("yyyyMMddHH","GMT"):format("yyyy-MM-dd-HH")}
but then there is an error occuring in my PutCassandraQL processor: Unable to coerce '2015-01-01-09' to a formatted date (long).
I tried something along
date : ${csvfiledate:toDate("yyyyMMddHH","GMT"):format("yyyy-MM-dd HH-mmZ")} aswell, but the same error is occuring.
It seems like you need to have a specific timestamp type for cassandra as you can see here:
http://docs.datastax.com/en/archived/cql/3.0/cql/cql_reference/timestamp_type_r.html
But it isnt working so far, maybe you got some tipps.
Thanks in advance.
You've got it backwards..
toDate parameters are used to describe how to parse the date. format function is used to describe how the date output should be. So the expression should be:
${csvfiledate:toDate('yyyy-MM-dd-HH','GMT'):format('yyyyMMddHH')}
I have a large data and in that one field be like Wed Sep 15 19:17:44 +0100 2010 and I need to insert that field in Hive.
I am getting troubled for choosing data type. I tried both timestamp and date but getting null values when loading from CSV file.
The data type is a String as it is text. If you want to convert it, I would suggest a TIMESTAMP. However you will need to do this conversion yourself while loading the data or (even better) afterwards.
To convert to a timestamp, you can use the following syntax:
CAST(FROM_UNIXTIME(UNIX_TIMESTAMP(<date_column>,'FORMAT')) as TIMESTAMP)
Your format seems complex though. My suggestion is to load it as a string and then just do a simple query on the first record until you get it working.
SELECT your_column as string_representation,
CAST(FROM_UNIXTIME(UNIX_TIMESTAMP(<date_column>,'FORMAT')) as TIMESTAMP) as timestamp_representation
FROM your_table
LIMIT 1
You can find more information on the format here: http://docs.oracle.com/javase/6/docs/api/java/text/SimpleDateFormat.html
My advice would be to concat some substrings first and try to convert only the day, month, year part before you look at time and timezone et cetera.
I have imported some data into SAS from some Excel spreadsheets sent to me. When I view the output from the imported table, the date appears as "01APR2014" and maintains chronological order. When I view the column properties the type is "Date" and the length is 8. Both the format and informat are DATE9.
I need to be able to convert this date to week-year and month-year, but no matter what I try I always get Jan, 1960.
Using proc sql, I used the below to get the week-year,
"(put(datepart(a.fnlz_date),weeku3.))|| "-" ||(put(datepart(a.fnlz_date),year.)) as FNLZD_WK_YR,"
but all I got was "W00-1960". I've used the formula above successfully many times before with SAS datetime values.
For month-yr, using proc sql, I tried
"datepart(a.fnlz_date) as DT_FNLZD format=monyy.,"
but the only value returned is "JAN60".
I also tried using SUBSTR, but got an error saying it requires a character argument, so SAS must see it as a number at least.
My question; does anyone know a way to get the week-yr and/or month-yr from this format? If so, how? I'm not opposed to using a data step, but I haven't been able to get that to work either.
Thanks in advance for any help or insight provided.
datepart converts datetimes to dates. Not helpful here.
If you're just displaying this, then you have a few options, particularly for month. You can just change the format of the variable (This changes what's displayed, but not the underlying value; consider this a value label).
When you use this like this (again, it looks like you got most of the way there):
proc sql;
select datevar format=monyy5. from table;
quit;
Just don't include that datepart function call as that's not appropriate unless you have a datetime. (Date=# of days since 1/1/1960, Datetime = # of seconds since 1/1/1960:00:00:00).
That will display it with MONYY5. format, which would be MAY10 for May, 2010. You have some other format options, see the documentation on formats by category for more details.
I can't think of a Week format that matches what you want (there are week formats, like WEEKW., as you clearly found, but I don't know that they do exactly what you want. So, if you want to build one yourself, you can either build a custom picture format, or you can make a string.
Building a custom picture format isn't too hard; see the documentation on Picture formats or google SAS Date Picture Format.
proc format;
picture weekyear (default=8)
low-high = 'W%0U-%Y' (datatype=date) ;
quit;
Now you can use that as a normal format.
To get at the week/etc. to build values, you can also use functions week(), month(), etc., if that's easier.
Since the data was already in a date format, I only needed to drop the DATEPART function that only works with datetime values. So, for month-yr,
"a.fnlz_date as fnlz_mnth format=monyy.,"
gives me the results I'm looking for.
Cheers!