How do I get the min date of a timestamp field?
I tried to use
select min(myDatefield) from mytable
but this code doesn't returns the minimum date, but all dates.
Any clue why?
Try this, this might work for you.
SELECT *
FROM
(SELECT MCSSP_AUDIT_ACTIVITY_DATE
FROM MCSSP_MESG_AUDIT_BK
ORDER BY MCSSP_AUDIT_ACTIVITY_DATE ASC
)
WHERE rownum = 1;
Related
I have o table name table_1 with 4 columns id, text, fromDate, toDate. The table represents the working experience.I want to create a function which will return the row with columns id, text where the employee worked more recently. This means I need column toDate to be closest to today.
Here is a demonstration of my code:
Select (abs("toDate"-now())) as date_diff
from table_1
Select id,text
from table_1
where (abs("toDate"-now()))=select min(date_diff)
Is this correct or is there something better I can do?
I wil try something like this:
Select id,text
from table_1
where "toDate" = ( select max ("toDate") from table_1 )
It will provide you the latest "toDate" value.
Try this:
select * from table_1
order by to_date desc
limit 1
I need to select the rows for which the difference between max(date) and the date just before max(date) is smaller than 366 days. I know about SELECT MAX(date) FROM table to get the last date from now, but how could I get the date before?
I would need a query of this kind:
SELECT code, MAX(date) - before_date FROM troncon WHERE MAX(date) - before_date < 366 ;
NB : before_date does not refer to anything and is to be replaced by a functionnal stuff.
Edit : Example of the table I'm testing it on:
CREATE TABLE troncon (code INTEGER, ope_date DATE) ;
INSERT INTO troncon (code, ope_date) VALUES
('C086000-T10001', '2014-11-11'),
('C086000-T10001', '2014-11-11'),
('C086000-T10002', '2014-12-03'),
('C086000-T10002', '2014-01-03'),
('C086000-T10003', '2014-08-11'),
('C086000-T10003', '2014-03-03'),
('C086000-T10003', '2012-02-27'),
('C086000-T10004', '2014-08-11'),
('C086000-T10004', '2013-12-30'),
('C086000-T10004', '2013-06-01'),
('C086000-T10004', '2012-07-31'),
('C086000-T10005', '2013-10-01'),
('C086000-T10005', '2012-11-01'),
('C086000-T10006', '2014-04-01'),
('C086000-T10006', '2014-05-15'),
('C086000-T10001', '2014-07-05'),
('C086000-T10003', '2014-03-03');
Many thanks!
The sub query contains all rows joined with the unique max date, and you select only ones which there differente with the max date is smaller than 366 days:
select * from
(
SELECT id, date, max(date) over(partition by code) max_date FROM your_table
) A
where max_date - date < interval '366 day'
PS: As #a_horse_with_no_name said, you can partition by code to get maximum_date for each code.
I have a table with date attribute and i need to do a query that gets the MIN date and the next of the MIN date
And I tried that :
select min(SC.TIMESTAMP) as minDate, result.TIMESTAMP
from Event SC
INNER JOIN
(SELECT TIMESTAMP from Event
HAVING TIMESTAMP > min(SC.TIMESTAMP)
) as result on result.BUSINESSID1 = SC.BUSINESSID1
where SC.BUSINESSSTEP = 'CONTAINER_PLACING_EVENT'
and SC.LOCATIONCODE = '1';
Could you please advice how to do that ?
Thanks in Advance
Perhaps you can rearrange your query into this form:
select
min(TS), min(TS2)
from
event,
(select TS as TS2 from event where TS > (select min(TS) from event))
Add extra criteria as desired. I would try to rewrite yours, but it isn't entirely clear what the criteria for the count are supposed to be. If you are expecting more than one row (for example, the min and min2 of each LOCATIONCODE) then you will probably want a GROUP BY in there.
Also, I wouldn't call a column TIMESTAMP as it is a reserved word.
You can use the ROW_NUMBER() OLAP Function:
SELECT *
FROM (
SELECT
TIMESTAMP
,ROW_NUMBER() OVER (
PARTITION BY BUSINESSSTEP, LOCATIONCODE
ORDER BY TIMESTAMP ASC
) AS RN
FROM EVENT
WHERE BUSINESSSTEP = 'CONTAINER_PLACING_EVENT'
AND LOCATIONCODE = '1'
) A
WHERE RN < 3
This will return as rows instead of columns, but it should get you what you want. If you think your original query would have returned multiple rows (for multiple entities), you can change the PARTITION BY clause to include the column that makes them distinct.
If I've been recording data with timestamp. How can I query for today's data? Something similar to this example.
SELECT * FROM table WHERE recorded_time = current_date;
Thanks
DATE_TRUNC('day', recorded_time) = CURRENT_DATE
should do it.
So each record on my table has a datetime timestamp column.
How do I return the age (in hours), of these records in the database?
Thank you!
select DATEDIFF(hour,timestamp,getdate()) as hours_old
from myTable
datediff(hour,[Your Column], getdate())
http://msdn.microsoft.com/en-us/library/ms189794.aspx
Use the datediff function.
select datediff(hour, getdate(), dob) as age
from ...
Since datediff(hour,'2000-01-01 00:59','2000-01-01 01:00') returns 1, due to the (counterintuitive) way datediff works, you may want something more accurate:
select DATEDIFF(minute,the_timestamp,getdate()) / 60
from TheTable