I am curious why this does not work:
SELECT *
FROM TABLE
WHERE DATE_TIME_COLUMN BETWEEN
current date - int((dayofweek(current date))-1)
AND
current date + int(7-(dayofweek(current date)))
When this gives me exactly what I want:
select current date - int((dayofweek(current date))-1) days from sysibm.sysdummy1)
select current date + int(7-(dayofweek(current date))) days from sysibm.sysdummy1)
The above two will yield the correct dates that I want my specific date time column to be in between. What am I missing here?
SELECT *
FROM TABLE
WHERE DATE_TIME_COLUMN BETWEEN
current date - ((dayofweek(current date))-1) DAYS
AND
current date + (7-(dayofweek(current date))) DAYS
You have to signify that you are reducing DAYS from current date as shown above.
"Does not work" is a little vague :). Your first query is missing the DAY part to tell DB2 what part you're doing the date math on.
However, it kind of sounds like you might want to use the WEEK scalar function:
SELECT *
FROM TABLE
WHERE WEEK(DATE_TIME_COLUMN) = WEEK(CURRENT_DATE)
AND YEAR(DATE_TIME_COLUMN) = YEAR(CURRENT_DATE)
Related
I am running the below query to get data recorded in the past 24 hours. I need the same data recorded starting midnight (DATE > 12:00 AM) and also data recorded starting beginning of the month. Not sure if using between will work or if there is better option. Any suggestions.
SELECT COUNT(NUM)
FROM TABLE
WHERE
STATUS = 'CNLD'
AND
TRUNC(TO_DATE('1970-01-01','YYYY-MM-DD') + OPEN_DATE/86400) = trunc(sysdate)
Output (Just need Count). OPEN_DATE Data Type is NUMBER. the output below displays count in last 24 hours. I need the count beginning midnight and another count starting beginning of the month.
The query you've shown will get the count of rows where OPEN_DATE is an 'epoch date' number representing time after midnight this morning*. The condition:
TRUNC(TO_DATE('1970-01-01','YYYY-MM-DD') + OPEN_DATE/86400) = trunc(sysdate)
requires every OPEN_DATE value in your table (or at least all those for CNLD rows) to be converted from a number to an actual date, which is going to be doing a lot more work than necessary, and would stop a standard index against that column being used. It could be rewritten as:
OPEN_DATE >= (trunc(sysdate) - date '1970-01-01') * 86400
which converts midnight this morning to its epoch equivalent, once, and compares all the numbers against that value; using an index if there is one and the optimiser thinks it's appropriate.
To get everything since the start of the month you could just change the default behaviour of trunc(), which is to truncate to the 'DD' element, to truncate to the start of the month instead:
OPEN_DATE >= (trunc(sysdate, 'MM') - date '1970-01-01') * 86400
And the the last 24 hours, subtract a day from the current time instead of truncating it:
OPEN_DATE >= ((sysdate - 1) - date '1970-01-01') * 86400
db<>fiddle with some made-up data to get 72 back for today, more for the last 24 hours, and more still for the whole month.
Based on your current query I'm assuming there won't be any future-dated values, so you don't need to worry about an upper bound for any of these.
*Ignoring leap seconds...
It sounds like you have a column that is of data type TIMESTAMP and you only want to select rows where that TIMESTAMP indicates that it is today's date? And as a related problem, you want to find those that are the current month, based on some system values like CURRENT TIMESTAMP and CURRENT DATE? If so, let's call your column TRANSACTION_TIMESTAMP instead of (reserved word) DATE. Your first query could be:
SELECT COUNT(NUM)
FROM TABLE
WHERE
STATUS = 'CLND'
AND
DATE(TRANSACTION_TIMESTAMP)=CURRENT DATE
The second example of finding all for the current month up to today's date could be:
SELECT COUNT(NUM)
FROM TABLE
WHERE
STATUS = 'CLND'
AND
YEAR(DATE(TRANSACTION_TIMESTAMP)=YEAR(CURRENT DATE) AND
MONTH(DATE(TRANSACTION_TIMESTAMP)=MONTH(CURRENT DATE) AND
DAY(DATE(TRANSACTION_TIMESTAMP)<=DAY(CURRENT DATE)
I am using Presto. I have an integer column (let's call the column 'mnth_nbr') showing year and month as: yyyymm. For instance, 201901. I want to have records showing all dates AFTER 201901 as well as 2 months before the given date. In this example, it would return 201811, 201812, 201901, 201902, 201903, etc. Keep in mind that my data type here is integer.
This is what I have so far (I do a self join):
select ...
from table 1 as first_table
left join table 1 as second_table
on first_table.mnth_nbr = second_table.mnth_nbr
where first_table.mnth_nbr <= second_table.mnth_nbr
I know this gives me all dates AFTER 201901, including 201901. But, I don't know how to add the 2 previous months (201811 and 201812)as explained above.
As far as the documentation, Presto DB date_parse function expects a MySQL-like date format specifier.
So the proper condition for your use case should be :
SELECT ...
FROM mytable t
WHERE
date_parse(cast(t.mnth_nbr as varchar), '%Y%m') >= date '2019-01-01' - interval '2' month
Edit
As commented by Piotr, a more optimized expression (index-friendly) would be :
WHERE
mnth_nbr >= date_format(date '2019-01-01' - interval '2', '%Y%m')
Something like this would help. first parse your int to date
date_parse(cast(first_table.mnth_nbr as varchar), 'yyyymm') > date '2019-01-01' - interval '2' month
please keep in mind that you may encounter with indexing issues with this approach.
I am using Hive (which is similar to SQL, but the syntax can be little different for the SQL users). I have looked at the other stackoverflow, but they seems to be in the SQL with different syntax.
I am trying to the get the first day of the month through this query. This one gives me today's day. For example, if today is 2015-04-30, then result would be 2015-04-01. Thanks!
select
cust_id,
FROM_UNIXTIME(UNIX_TIMESTAMP(),'yyyy-MM-dd') as first_day_of_month_transaction
--DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0) as first_day_of_month_transaction --SQL format. Not compatible in Hive.
from
customers;
Try this
date_format(current_date,'yyyy-MM-01')
To get the first day of the month, you can use:
date_add(<date>,
1 - day(<date>) )
Applied to your expression:
date_add(FROM_UNIXTIME(UNIX_TIMESTAMP(), 'yyyy-MM-dd'),
1 - day(FROM_UNIXTIME(UNIX_TIMESTAMP(), 'yyyy-MM-dd'))
)
But this will work for any column in the right format.
SELECT TRUNC(rpt.statement_date,'MM') will give you the first day of month.
You can use this query to calculate the First day of the current month
Select date_add(last_day(add_months(current_date, -1)),1);
Instead of current_date, you can use the date field name.
last_day => Gives the last day of current month
add_months with -1 => Gives previous month
date_add with 1 => Add one day
Another way - select concat(substring(current_date,1,7),'-01');
I am trying to get the last of month, and in order to that i have written the following, to calculate the no. of days between today and the last date.
select datediff(DAY,GETDATE(),dateadd(m,1,getdate()))-GETDATE()
the bold part gives me the no. of days between today and a month from today, say 30 or 31. and then I am subtracting today's date from 30 or 31, which is " -getdate() "
The output for the above query is
1786-06-06 11:44:30.540
Could you please explain what is happening in the query? I am not looking for a solution, I would like to know how is SQL-Server interpreting the query.
Thanks. :)
The bold part of the expressions does not return a date, it returns a number of days:
31
Convert that to a datetime:
SELECT CONVERT(DATETIME, 31);
This is 31 days after day 0 (1900-01-01):
1900-02-01
Now, subtract GETDATE() as an integer (41512 days after day 0):
SELECT 31 - 41512 = -41481
Now add -41481 days to day 0:
SELECT DATEADD(DAY, -41481, 0);
-- or
SELECT DATEADD(DAY, -41481, '19000101');
Or:
SELECT CONVERT(DATETIME, 31 - CONVERT(INT, GETDATE()));
Now, I strongly recommend a couple of things:
Don't use implicit date math. #date_var_or_col - 1 for example fails with new data types like DATE and DATETIME2.
Don't use shorthand like m. If you mean MONTH, just take the massive productivity hit and type out MONTH. To see why, tell me if this provides the results you expect:
SELECT DATEPART(y, GETDATE()), DATEPART(w, GETDATE());
I am subtracting today's date from 30 or 31, which is " -getdate() "
Sounds like you understand exactly what is happening, but maybe don't understand the results.
You are implicitly converting GETDATE() to a number, which represents the number of days (and fractional days) since 1/1/1900 12:00:00 AM
When you "subtract" GETDATE() (41,511 as of 8/27/2013) from 30 or 31 you get an answer of -41,480, or 41,480 days before 1/1/1900, which would be about 6/6/1786 (plus or minus a few hours for the fractional part).
how to extract day of month from date, and select only odd days in Informix SQL?
Day of month is simply the DAY(date_or_datetime_column) function, related to the MONTH() and YEAR() functions. Getting only odd numbers is done with a simple modulo 2 expression.
So I think you only need:
SELECT date_col, DAY(date_col)
FROM table
WHERE MOD(DAY(date_col), 2) = 1
Hope that's useful.