This is my table
date count of subscription per date
---- ----------------------------
21-03-2016 10
22-03-2016 30
23-03-2016 40
Please need your help, I need to get the result like below table, summation second row with first row, same thing for another rows:
date count of subscription per date
---- ----------------------------
21-03-2016 10
22-03-2016 40
23-03-2016 80
You can do a cumulative sum using the ANSI standard analytic SUM() function:
select date, sum(numsubs) over (order by date) as cume_numsubs
from t;
Select sum(col1) over(order by date rows between unbounded preceding and current row) cnt from mytable;
SELECT t.date, (
SELECT SUM(numsubs)
FROM mytable t2
WHERE t2.date <= t.date
) AS cnt
FROM mytable t
Related
I have below audit table. Need to find Min Row where year is Year Getdate()) compared to min Row where Year is Year(getdate())-1
You can try to use ROW_NUMBER window function PARTITION BY ID & year of the date column.
SELECT *
FROM (
SELECT *,ROW_NUMBER() OVER(PARTITION BY ID,YEAR(Date) ORDER BY Date) rn
FROM T
) t1
WHERE rn = 1
I'm trying to setup a data check, where we get the row count from a table for today and prior date. Since it isn't loaded on weekends or holidays, I can't say DATE-1.
I came-up with the following, to get the previous date:
SELECT
LOAD_DATE
,COUNT(LOAD_DATE) RW_COUNT
,ROW_NUMBER() OVER (ORDER BY LOAD_DATE ) AS LOAD_ROWNUM
FROM DATABASE1.TABLE1
WHERE LOAD_DATE >= DATE-6
GROUP BY 1
This produces the dates, counts and assigns a row number.
LOAD_DATE RW_COUNT LOAD_ROWNUM
2019-10-16 8259 1
2019-10-15 8253 2
2019-10-11 8256 3
2019-10-10 8243 4
I to take the two most current dates and compare them. Most current would be "current" and the 2nd most current would be "prior" . Then I would like to have something like this as the result set:
CURRENT_COUNT PRIOR_COUNT DIFF_PERCENT
8259 8253 .9927
My issue is, how do I reference the first two rows and compare them to each other? Unless I'm over-thinking this, I need two additional SELECT statements: 1 with the WHERE clause referencing row 1 and another with a WHERE referercing row 2.
How do I do that? Do I have two CTEs?
Eventually, I'll need a third SELECT dividing the two rows and checking for 10% tolerance. Help, I'm in analysis paralysis.
You can filter the result of an OLAP-function using QUALIFY:
SELECT
LOAD_DATE
,COUNT(LOAD_DATE) AS CURRENT_COUNT
-- previous day's count
,LEAD(RW_COUNT)
OVER (ORDER BY LOAD_DATE DESC) AS PRIOR_COUNT
-- if your TD version doesn't support LAG/LEAD (i.e. < 16.10)
--,MIN(RW_COUNT)
-- OVER (ORDER BY LOAD_DATE DESC
-- ROWS BETWEEN 1 FOLLOWING AND 1 FOLLOWING) AS PRIOR_COUNT
,CAST(CURRENT_COUNT AS DECIMAL(18,4)) / PRIOR_COUNT AS DIFF_PERCENT
FROM DATABASE1.TABLE1
WHERE LOAD_DATE >= DATE-6
GROUP BY 1
-- return the latest row only
QUALIFY ROW_NUMBER() OVER (ORDER BY LOAD_DATE DESC) = 1
checking for 10% tolerance:
DIFF_PERCENT BETWEEN 0.9 and 1.1
Either ANDed to the QUALIFY or within a CASE
I don't know what you want for your result set. But you can use LAG() with aggregation to get the previous value.
SELECT LOAD_DATE, COUNT(*) as RW_COUNT,
LAG(COUNT(*)) OVER (ORDER BY LOAD_DATE) as PREV_RW_COUNT
FROM DATABASE1.TABLE1
WHERE LOAD_DATE >= DATE-6
GROUP BY 1;
You may just want a difference of the two counts.
If your TD version (16.0+?) doesn't support LEAD/LAG, give this a try:
SELECT
load_date,
RW_COUNT,
MAX(RW_COUNT) OVER(
ORDER BY load_date DESC
ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING -- Get previous row's value
) AS RW_COUNT_prev
FROM (
SELECT load_date, COUNT(LOAD_DATE) RW_COUNT,
FROM DATABASE1.TABLE1
WHERE LOAD_DATE >= DATE-6
GROUP BY 1
) src
I have a requirement to get values from a table based on an offset conditions on a date column.
Say for eg: for the below attached table, if there is any dates that comes close within 15 days based on effectivedate column I should return only the first one.
So my expected result would be as below:
Here for A1234 policy, it returns 6/18/16 entry and skipped 6/12/16 entry as the offset between these 2 dates is within 15 days and I took the latest one from the list.
If you want to group rows together that are within 15 days of each other, then you have a variant of the gaps-and-islands problem. I would recommend lag() and cumulative sum for this version:
select polno, min(effectivedate), max(expirationdate)
from (select t.*,
sum(case when prev_ed >= dateadd(day, -15, effectivedate)
then 1 else 0
end) over (partition by polno order by effectivedate) as grp
from (select t.*,
lag(expirationdate) over (partition by polno order by effectivedate) as prev_ed
from t
) t
) t
group by polno, grp;
I have table with time stamp column tmstmp, this table contains log of certain events. I need to find out the max number events which occurred within 1 min interval.
Please read carefully! I do NOT want to extract the time stamps minute fraction and sum like this:
select count(*), TO_CHAR(tmstmp,'MI')
from log_table
group by TO_CHAR(tmstmp,'MI')
order by TO_CHAR(tmstmp,'MI');
It needs to take 1st record and then look ahead until it selects all records within 1 min from the 1st and sum number of records, then take 2nd and do the same etc..
And as the result there must be a recordset of (sum, starting timestamp).
Anyone has a snippet of code somewhere and care to share please?
Analytic function with a logical window can provide this information directly:
select l.tmstmp,
count(*) over (order by tmstmp range between current row and interval '59.999999' second following) cnt
from log_table l
order by 1
;
TMSTMP CNT
--------------------------- ----------
01.01.16 00:00:00,000000000 4
01.01.16 00:00:10,000000000 4
01.01.16 00:00:15,000000000 3
01.01.16 00:00:20,000000000 2
01.01.16 00:01:00,000000000 3
01.01.16 00:01:40,000000000 2
01.01.16 00:01:50,000000000 1
Please adjust the interval length for your precision. It must be the highest possible value below 1 minute.
To get the maximal minute use the subquery (and don't forget you may receive more that one record - with the MAX count):
with tst as (
select l.tmstmp,
count(*) over (order by tmstmp range between current row and interval '59.999999' second following) cnt
from log_table l)
select * from tst where cnt = (select max(cnt) from tst);
TMSTMP CNT
--------------------------- ----------
01.01.16 00:00:00,000000000 4
01.01.16 00:00:10,000000000 4
I think you can achieve your goal using a subquery in SELECT statement, as follow:
SELECT tmstmp, (
SELECT COUNT(*)
FROM log_table t2
WHERE t2.tmstmp >= t.tmstmp AND t2.tmstmp < t.tmstmp + 1 / (24*60)
) AS events
FROM log_table t;
One method uses a join and aggregation:
select t.*
from (select l.tmstmp, count(*)
from log_table l join
log_table l2
on l2.tmstmp >= l.tmstmp and
l2.tmstmp < l.tmstmp + interval '1' minute
group by l.tmpstmp
order by count(*) desc
) t
where rownum = 1;
Note: This assumes that tmstmp is unique on each row. If this is not true, then the subquery should be aggregating by some column that is unique.
EDIT:
For large data, there is a more efficient way that makes use of cumulative sums:
select tmstamp - interval 1 minute as starttm, tmstamp as endtm, cumulative
from (select tmstamp, sum(inc) over (order by tmstamp) as cumulative
from (select tmstamp, 1 as inc from log_table union all
select tmstamp + interval '1' day, -1 as inc from log_table
) t
order by sum(inc) over (order by tmstamp) desc
) t
where rownum = 1;
I'm pretty new to SQL and have this problem:
I have a filled table with a date column and other not interesting columns.
date | name | name2
2015-03-20 | peter | pan
2015-03-20 | john | wick
2015-03-18 | harry | potter
What im doing right now is counting everything for a date
select date, count(*)
from testtable
where date >= current date - 10 days
group by date
what i want to do now is counting the resulting lines and only returning them if there are less then 10 resulting lines.
What i tried so far is surrounding the whole query with a temp table and the counting everything which gives me the number of resulting lines (yeah)
with temp_count (date, counter) as
(
select date, count(*)
from testtable
where date >= current date - 10 days
group by date
)
select count(*)
from temp_count
What is still missing the check if the number is smaller then 10.
I was searching in this Forum and came across some "having" structs to use, but that forced me to use a "group by", which i can't.
I was thinking about something like this :
with temp_count (date, counter) as
(
select date, count(*)
from testtable
where date >= current date - 10 days
group by date
)
select *
from temp_count
having count(*) < 10
maybe im too tired to think of an easy solution, but i can't solve this so far
Edit: A picture for clarification since my english is horrible
http://imgur.com/1O6zwoh
I want to see the 2 columned results ONLY IF there are less then 10 rows overall
I think you just need to move your having clause to the inner query so that it is paired with the GROUP BY:
with temp_count (date, counter) as
(
select date, count(*)
from testtable
where date >= current date - 10 days
group by date
having count(*) < 10
)
select *
from temp_count
If what you want is to know whether the total # of records (after grouping), are returned, then you could do this:
with temp_count (date, counter) as
(
select date, counter=count(*)
from testtable
where date >= current date - 10 days
group by date
)
select date, counter
from (
select date, counter, rseq=row_number() over (order by date)
from temp_count
) x
group by date, counter
having max(rseq) >= 10
This will return 0 rows if there are less than 10 total, and will deliver ALL the results if there are 10 or more (you can just get the first 10 rows if needed with this also).
In your temp_count table, you can filter results with the WHERE clause:
with temp_count (date, counter) as
(
select date, count(distinct date)
from testtable
where date >= current date - 10 days
group by date
)
select *
from temp_count
where counter < 10
Something like:
with t(dt, rn, cnt) as (
select dt, row_number() over (order by dt) as rn
, count(1) as cnt
from testtable
where dt >= current date - 10 days
group by dt
)
select dt, cnt
from t where 10 >= (select max(rn) from t);
will do what you want (I think)