I'm trying to write a SQL query for Oracle SQL in order to retrieve the last records for a certain period frequency. For example, say the frequency is Quarterly, (I'd also like monthly and annually to work), I can provide the start dates and end dates for the quarters if necessary, but I need to retrieve the last entry within each quarter. How can I do this? I've had limited luck so far without writing lots of subqueries.
Try something like:
select * from
(select t.*,
row_number() over (partition by trunc(date_field, 'MON')
order by date_field desc) rn
from my_table t)
where rn = 1
for months. (Use 'Q' or 'Y' instead of 'MON' in the trunc clause for quarters or years.)
SELECT col1, col2, col3...colN
FROM TableA
WHERE (colX = (SELECT MAX(Date) AS LastDate
FROM TableA
WHERE QuarterDate Between (BeginningDate AND EndingDate)
colX is the date column that is your date you need to be checking if in the quarter.
I think that should work.
Related
How to find if an id which was present in previous weeks but not available in current week on a rolling basis. For e.g
Week1 has id 1,2,3,4,5
Week2 has id 3,4,5,7,8
Week3 has id 1,3,5,10,11
So I found out that id 1 and 2 are missing in week 2 and id 2,4,7,8 are missing in week 3 from previous 2 weeks But how to do this on a rolling window for a large amount of data distributed over a period of 20+ years
Please find the sample dataset and expected output. I am expecting the output to be partitioned based on the week_end Date
Dataset
ID|WEEK_START|WEEK_END|APPEARING_DATE
7152|2015-12-27|2016-01-02|2015-12-27
8350|2015-12-27|2016-01-02|2015-12-27
7152|2015-12-27|2016-01-02|2015-12-29
4697|2015-12-27|2016-01-02|2015-12-30
7187|2015-12-27|2016-01-02|2015-01-01
8005|2015-12-27|2016-01-02|2015-12-27
8005|2015-12-27|2016-01-02|2015-12-29
6254|2016-01-03|2016-01-09|2016-01-03
7962|2016-01-03|2016-01-09|2016-01-04
3339|2016-01-03|2016-01-09|2016-01-06
7834|2016-01-03|2016-01-09|2016-01-03
7962|2016-01-03|2016-01-09|2016-01-05
7152|2016-01-03|2016-01-09|2016-01-07
8350|2016-01-03|2016-01-09|2016-01-09
2403|2016-01-10|2016-01-16|2016-01-10
0157|2016-01-10|2016-01-16|2016-01-11
2228|2016-01-10|2016-01-16|2016-01-14
4697|2016-01-10|2016-01-16|2016-01-14
Excepted Output
Partition1: WEEK_END=2016-01-02
ID|MAX(LAST_APPEARING_DATE)
7152|2015-12-29
8350|2015-12-27
4697|2015-12-30
7187|2015-01-01
8005|2015-12-29
Partition1: WEEK_END=2016-01-09
ID|MAX(LAST_APPEARING_DATE)
7152|2016-01-07
8350|2016-01-09
4697|2015-12-30
7187|2015-01-01
8005|2015-12-29
6254|2016-01-03
7962|2016-01-05
3339|2016-01-06
7834|2016-01-03
Partition3: WEEK_END=2016-01-10
ID|MAX(LAST_APPEARING_DATE)
7152|2016-01-07
8350|2016-01-09
4697|2016-01-14
7187|2015-01-01
8005|2015-12-29
6254|2016-01-03
7962|2016-01-05
3339|2016-01-06
7834|2016-01-03
2403|2016-01-10
0157|2016-01-11
2228|2016-01-14
Please use below query,
select ID, MAX(APPEARING_DATE) from table_name
group by ID, WEEK_END;
Or, including WEEK)END,
select ID, WEEK_END, MAX(APPEARING_DATE) from table_name
group by ID, WEEK_END;
You can use aggregation:
select t.*, max(week_end)
from t
group by id
having max(week_end) < '2016-01-02';
Adjust the date in the having clause for the week end that you want.
Actually, your question is a bit unclear. I'm not sure if a later week end would keep the row or not. If you want "as of" data, then include a where clause:
select t.id, max(week_end)
from t
where week_end < '2016-01-02'
group by id
having max(week_end) < '2016-01-02';
If you want this for a range of dates, then you can use a derived table:
select we.the_week_end, t.id, max(week_end)
from (select '2016-01-02' as the_week_end union all
select '2016-01-09' as the_week_end
) we cross join
t
where t.week_end < we.the_week_end
group by id, we.the_week_end
having max(t.week_end) < we.the_week_end;
I have a table with 3 columns: id, date and amount, but I would like to get accumulated SUM for each date (Last column).
Do you have an easy solution how to add this column?
I am trying with this:
SELECT date, sum(amount) as accumulated
FROM table group by date
WHERE max(date);
Should I user OVER() for this?
Use a window function to the total for each day:
SELECT date,
amount,
sum(amount) over (partition by date) as accumulated
FROM the_table;
However this will only work, if your dates all have the same time part (in Oracle a DATE column also contains a time). To make sure you ignore the time part, use trunc() to make sure all time parts are normalized to 00:00:00
SELECT date,
amount,
sum(amount) over (partition by trunc(date)) as accumulated
FROM the_table;
Use This:
SELECT T.ID, T.DATE, T.AMOUNT, (SELECT SUM(S.AMOUNT) FROM TABLE S WHERE S.DATE=T.DATE) ACCUMULATED
from
table T
This will give you the records from the table with a sum for all records for the date.
I haven't used MS access in over ten years. I've looked at various sites and while I can find information on min and max, I can't find the answer to this one.
I have two tables, one archival and one current. Each time it's run, I need to take the oldest month's records and append them on to the archival table, then delete them from the current table.
If I can just figure out the select statement to return the range of the oldest month, I can do the rest.
So, I need....
Select *
From MyTable
Where ????
something that qualifies it as the oldest month
order by product_Id
The following uses the fact that MS Access really does a top with ties:
select top 1 t.*
from mytable as t
order by datepart("yyyy", datecol) asc, datepart("m", datecol) asc
A more typical method would be an inner join:
select t.*
from mytable as t inner join
(select min(datecol) as mindc
from mytable
) as tt
on datepart("yyyy", t.datecol) = datepart("yyyy", tt.mindc) and
datepart("m", t.datecol) = datepart("m", tt.mindc);
My goal is to extract from my database all of the records that fall NOT on the end of the month, but on the last date that exists in each month within the data (although some of these may so happen to be the end of the month).
Currently, just to simply get the latest dates for each month, using the following code alone takes 26 minutes:
SELECT Max(DATE)
FROM Accounts
GROUP BY Year(Date), Month(Date);
The point being that for the 16M rows that we have, running this as a subquery for the FROM statement within a temp_table is just not the speed improvement we were looking for (currently it takes about 40 minutes to read in the whole table anyway).
Any suggestions?
One method is to use window functions:
select a.*
from (select a.*,
rank() over (partition by year(date), month(date)
order by day(date) desc) as seqnum
from accounts a
) a
where seqnum = 1;
Note: the use of rank() (or equivalently dense_rank()) will return all rows for the last day of each month. If you only wanted one record, you can use row_number() instead.
I would appreciate a little expert help please.
in an SQL SELECT statement I am trying to get the last day with data per month for the last year.
Example, I am easily able to get the last day of each month and join that to my data table, but the problem is, if the last day of the month does not have data, then there is no returned data. What I need is for the SELECT to return the last day with data for the month.
This is probably easy to do, but to be honest, my brain fart is starting to hurt.
I've attached the select below that works for returning the data for only the last day of the month for the last 12 months.
Thanks in advance for your help!
SELECT fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,fd.column_name
FROM super_table fd,
(SELECT TRUNC(daterange,'MM')-1 first_of_month
FROM (
select TRUNC(sysdate-365,'MM') + level as DateRange
from dual
connect by level<=365)
GROUP BY TRUNC(daterange,'MM')) fom
WHERE fd.cust_id = :CUST_ID
AND fd.coll_date > SYSDATE-400
AND TRUNC(fd.coll_date) = fom.first_of_month
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)
You probably need to group your data so that each month's data is in the group, and then within the group select the maximum date present. The sub-query might be:
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY YEAR(coll_date) * 100 + MONTH(coll_date);
This presumes that the functions YEAR() and MONTH() exist to extract the year and month from a date as an integer value. Clearly, this doesn't constrain the range of dates - you can do that, too. If you don't have the functions in Oracle, then you do some sort of manipulation to get the equivalent result.
Using information from Rhose (thanks):
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY TO_CHAR(coll_date, 'YYYYMM');
This achieves the same net result, putting all dates from the same calendar month into a group and then determining the maximum value present within that group.
Here's another approach, if ANSI row_number() is supported:
with RevDayRanked(itemDate,rn) as (
select
cast(coll_date as date),
row_number() over (
partition by datediff(month,coll_date,'2000-01-01') -- rewrite datediff as needed for your platform
order by coll_date desc
)
from super_table
)
select itemDate
from RevDayRanked
where rn = 1;
Rows numbered 1 will be nondeterministically chosen among rows on the last active date of the month, so you don't need distinct. If you want information out of the table for all rows on these dates, use rank() over days instead of row_number() over coll_date values, so a value of 1 appears for any row on the last active date of the month, and select the additional columns you need:
with RevDayRanked(cust_id, server_name, coll_date, rk) as (
select
cust_id, server_name, coll_date,
rank() over (
partition by datediff(month,coll_date,'2000-01-01')
order by cast(coll_date as date) desc
)
from super_table
)
select cust_id, server_name, coll_date
from RevDayRanked
where rk = 1;
If row_number() and rank() aren't supported, another approach is this (for the second query above). Select all rows from your table for which there's no row in the table from a later day in the same month.
select
cust_id, server_name, coll_date
from super_table as ST1
where not exists (
select *
from super_table as ST2
where datediff(month,ST1.coll_date,ST2.coll_date) = 0
and cast(ST2.coll_date as date) > cast(ST1.coll_date as date)
)
If you have to do this kind of thing a lot, see if you can create an index over computed columns that hold cast(coll_date as date) and a month indicator like datediff(month,'2001-01-01',coll_date). That'll make more of the predicates SARGs.
Putting the above pieces together, would something like this work for you?
SELECT fd.cust_id,
fd.server_name,
fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,
fd.column_name
FROM super_table fd,
WHERE fd.cust_id = :CUST_ID
AND TRUNC(fd.coll_date) IN (
SELECT MAX(TRUNC(coll_date))
FROM super_table
WHERE coll_date > SYSDATE - 400
AND cust_id = :CUST_ID
GROUP BY TO_CHAR(coll_date,'YYYYMM')
)
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)