I am doing some calculation but my calculation is off because my date field is showing the time-stamp and i only want to use as Date only when i am doing the calculation. How can i just ignore the minutes and just use the date when doing the calculation? Here is what i have:
SELECT EF.DSCH_TS,
CASE WHEN EXTRACT (DAY FROM EF.DSCH_TS - EF.ADMT_TS)>=0 THEN 'GroupA' END AS CAL
FROM MainTable EF;
Netezza has built-in function for this by simply using:
SELECT DATE(STATUS_DATE) AS DATE,
COUNT(*) AS NUMBER_OF_
FROM X
GROUP BY DATE(STATUS_DATE)
ORDER BY DATE(STATUS_DATE) ASC
This will return just the date portion of the timetamp and much more useful than casting it to a string with TO_CHAR() because it will work in GROUP BY, HAVING, and with other netezza date functions. (Where as the TO_CHAR method will not)
Also, the DATE_TRUNC() function will pull a specific value out of Timestamp ('Day', 'Month, 'Year', etc..) but not more than one of these without multiple functions and concatenate.
DATE() is the perfect and simple answer to this and I am surprised to see so many misleading answers to this question on Stack. I see TO_DATE a lot, which is Oracle's function for this but will not work on Netezza.
With your query, assuming that you're interested in the days between midnight to midnight of the two timestamps, it would look something like this:
SELECT EF.DSCH_TS,
CASE
WHEN EXTRACT (DAY FROM (DATE(EF.DSCH_TS) - DATE(EF.ADMT_TS)))>=0 THEN 'GroupA'
END AS CAL
FROM MainTable EF;
You may want to consider rewriting your case statement to return an interval. This will allow for a little more flexibility.
SELECT EF.DSCH_TS,
CASE
WHEN age(date(EF.DSCH_TS),date(EF.ADMT_TS))>= interval '6 days'
THEN 'GroupA' END AS CAL
FROM MainTable EF;
Use date_trunc() with the first argument of 'day'. I think this is what you want:
SELECT EF.DSCH_TS,
(case when date_trunc('day', EF.DSCH_TS) >= date_trunc('day', EF.ADMT_TS) THEN 'GroupA' END) AS CAL
FROM MainTable EF;
Related
I have a data set as below,
Same is date in "YYYYMMDD" format, I wanted to convert the columns to date format and take the difference between the same.
I used to below code
SELECT to_date(statement_date_key::text, 'yyyymmdd') AS statement_date,
to_date(paid_date_key::text, 'yyyymmdd') AS paid_date,
statement_date - paid_date AS Diff_in_days
FROM Table
WHERE Diff_in_days >= 90
;
Idea is to convert both the columns to dates, take the difference between them and filter cases where difference in days is more than 90.
Later I was informed that server is supported by HiveSQL and does not support of using ":", date time, and temp tables can not be created.
I'm currently stuck on how to go about given the constraints.
Help would be much appreciated.
Sample date for reference is provided in the link
dbfiddle
Hive is a little convoluted in its use of dates. You can use unix_timestamp() and work from there:
SELECT datediff(to_date(unix_timestamp(cast(statement_date_key as varchar(10)), 'yyyyMMdd')),
to_date(unix_timestamp(cast(paid_date_key as varchar(10)), 'yyyyMMdd'))
) as diff_in_days
FROM Table;
Note that you need to use a subquery if you want to use diff_in_days in a where clause.
Also, if you have date keys, then presumably you also have a calendar table, which should make this much simpler.
Hello You Can Use Below Query It Work Well
select * from (
select convert(date, statement_date_key) AS statement_date,
convert(date, paid_date) AS paid_date,
datediff(D, convert(date, statement_date_key), convert(date, paid_date)) as Diff_in_days
from Table
) qry
where Diff_in_days >= 90
Simple way: Function unix_timestamp(string, pattern) converts string in given format to seconds passed from unix epoch, calculate difference in seconds then divide by (60*60*24) to get difference in days.
select * from
(
select t.*,
(unix_timestamp(string(paid_date_key), 'yyyyMMdd') -
unix_timestamp(string(statement_date_key), 'yyyyMMdd'))/86400 as Diff_in_days
from Table t
) t
where Diff_in_days>=90
You may want to add abs() if the difference can be negative.
One more method using regexp_replace:
select * from
(
select t.*,
datediff(date(regexp_replace(string(paid_date_key), '(\\d{4})(\\d{2})(\\d{2})','$1-$2-$3')),
date(regexp_replace(string(statement_date_key), '(\\d{4})(\\d{2})(\\d{2})','$1-$2-$3'))) as Diff_in_days
from Table t
) t
where Diff_in_days>=90
Running below query and not getting the output. Can someone please tell whats wrong in it?
Select distinct (table.datex)
from table
where table.datex =
(
CASE when extract( day from sysdate) >=19
then last_day(add_months(sysdate, -1))
else last_day(add_months(sysdate, -2))
END
)
Sample data
Datex
ID
30-JUN-21
A
31-MAY-21
B
29-JUN-21
C
Expected result
Datex
30-JUN-21
When I am passing the value hard-coded(calculated by the case) to where clause it's working fine, but when I apply the case it's not working. No error. No output is coming.
Date or datetime?
Oracle's LAST_DAY doesn't do what the name suggests, and the docs (https://docs.oracle.com/en/database/oracle/oracle-database/21/sqlrf/LAST_DAY.html#GUID-296C7C02-7FB9-4AAC-8927-6A79320CE0C6) fail to explain that, too.
Unlike several other DBMS Oracle doesn't have a date type. It only has a datetime type and they even call that inappropriately DATE. This means that a "date" in Oracle always has a time part. A date with its time part set to 00:00:00 can be considered a day's midnight (i.e. the very beginning of the day) or the whole day.
The function SYSDATE gives us a date in the sense of the DATE datatype, not in the sense of a real day, i.e. it gives us the datetime of "now", e.g. 2021-07-20 14:38:00. ADD_MONTHS changes the month in that datetime (and sometimes the year and sometimes even the day), i.e. leaves the time part untouched. LAST_DAY, too, changes the date part to get to the last day of the month, but leaves the time part untouched.
Your CASE expression hence results in something like TIMESTAMP '2021-07-20 14:38:00' and not in DATE '2021-07-20' as one might expect.
You say that you tried your query with the date you computed wth your case expression, and it worked. Did you compute the resulting day in your head or with a query? If the latter: The tool you are using may be set to only display a datetime's date part and omit the time part. This would explain why you only saw 30-JUN-21 when checking the CASE expression.
Solution
Truncate the datetime down to a whole day
Select distinct datex
from mytable
where (extract(day from sysdate) >=19 and datex = trunc(last_day(add_months(sysdate, -1))))
or (extract(day from sysdate) < 19 and datex = trunc(last_day(add_months(sysdate, -2))))
It doesn't matter whether you apply TRUNC late as in my example or right away on SYSDATE (with TRUNC(SYSDATE)) by the way. The only aim is to get rid of the time part at some point in the expression.
Don't use case in where clauses. Boolean logic can handle that.
And take a look if it is really the condition you want
Select distinct datex
from your_table
where
(
extract(day from sysdate) >=19
and datex = last_day(add_months(sysdate,-1))
)
or
(
extract(day from sysdate) < 19
and datex = last_day(add_months(sysdate,-2))
)
I am trying to use the Date_Trunc for MONTH function in a SQL statement but somehow it is not working for me. I am trying to pull entries which happen after April 1st, 2019. The raw date format from the Redshift database is this format which I am trying to group into month/year buckets: 2019-04-08T00:13:20.000Z
Input
SELECT
client_id as user_id,
session_utc as job_date --(format:2019-04-08T00:13:20.000Z)
FROM table1 as hits
WHERE job_date >= DATE_TRUNC('month', 2019-04-01)
group by 1,2;
Output
"ERROR: function date_trunc("unknown", integer) does not exist Hint: No function matches the given name and argument types. You may need to add explicit type casts."
What am I doing wrong?
The DATE_TRUNC Function - Amazon Redshift takes timestamp as input and provides a timestamp as output:
DATE_TRUNC('datepart', timestamp)
For example:
SELECT DATE_TRUNC('month', '2019-05-07'::timestamp)
2019-05-01 00:00:00
Therefore, your line should read:
WHERE job_date >= DATE_TRUNC('month', '2019-04-01'::timestamp)
If you wish to have the output as a date, append ::date:
SELECT DATE_TRUNC('month', '2019-05-07'::timestamp)::date
2019-05-01
Also, note that the date converts into a timestamp as at midnight. This can cause a difference for some comparisons. For example:
'2019-05-07 03:03:31.389324+00'::timestamp > '2019-05-07'::timestamp
will evaluate as True because it is comparing to midnight at the start of the day. This is different to comparing two dates (without timestamps).
The syntax for the function is DATE_TRUNC('datepart', timestamp), seems you need to use as DATE_TRUNC('month', session_utc)(this already truncates to the first date of April 2019 i.e. 2019-04-01 )
Assuming you are using Postgres, you need quotes around your date constant and can convert to the right types:
WHERE job_date >= DATE_TRUNC('month'::text, '2019-04-01'::date)
Here is a db<>fiddle.
I would like to filter out the data using a sub query in the interval function
Following is the query i use
SEL * FROM my_table WHERE MY_DATE < CURRENT_DATE- INTERVAL '30' MONTH;
The above query works, However i want to parameterize the period '30' using a sub query. Please suggest how to achieve this.
Thanks in Advance
Don't use interval calculations with year/month as it will fail, e.g. DATE '2016-12-31' + INTERVAL '30' MONTH results in 2019-06-31 (according to Standard SQL) which obviously doesn't exist.
SELECT *
FROM my_table
WHERE MY_DATE < ADD_MONTHS(CURRENT_DATE, (SELECT -col FROM tab));
If col is actually an INTERVAL you need to cast it to an INT.
I would like to subtract the current date from a given date in SQL or in JDBC. I would like to have the result in hours. Not sure how to treat the date in that case. Can some one give me a basic example Please.
You don't mention which database server you use - here's a sample in MySQL.
select hour(TIMEDIFF('2011-03-15 19:59:59.000001', now()))
Note: the "hour" function doesn't deal with rounding, so if you need that, you need to do some further arithmetic.
Date functions are pretty vendor-specific, so your mileage may vary....
In standard SQL
select (date '2011-03-16' - CURRENT_DATE) as days_different,
(date '2011-03-16' - CURRENT_DATE) * 24 as hours_different
days_different hours_different
--
1 24
As I write this, CURRENT_DATE = '2011-03-15'.