I having below data in one table.
And I want to get NEXT out data from OUT column. So used LEAD function in below query.
SELECT ROW_NUMBER,TIMESTAMP,IN,OUT,LEAD(OUT) OVER (PARTITION BY NULL ORDER BY TIMESTAMP) AS NEXT_OUT
FROM MYTABLE;
It gives data as below NEXT_OUT column.
But I need to know the matching next column value in sequential way like DESIRED columns. Please let me know how can i achieve this in Oracle LEAD FUNCTION
THANKS
Assign row number to all INs and OUTs separately, sort the results by placing them in a single column and calculate LEADs:
WITH cte AS (
SELECT t.*
, CASE WHEN "IN" IS NOT NULL THEN COUNT("IN") OVER (ORDER BY "TIMESTAMP") END AS rn1
, CASE WHEN "OUT" IS NOT NULL THEN COUNT("OUT") OVER (ORDER BY "TIMESTAMP") END AS rn2
FROM t
)
SELECT cte.*
, LEAD("OUT") OVER (ORDER BY COALESCE(rn1, rn2), rn1 NULLS LAST) AS NEXT_OUT
FROM cte
ORDER BY COALESCE(rn1, rn2), rn1 NULLS LAST
Demo on db<>fiddle
Enumerate in the "in"s and the "out"s and use that information for matching.
select tin.*, tout.out as next_out
from (select t.*,
count(in) over (order by timestamp) as seqnum_in
from t
) tin left join
(select t.*,
count(out) over (order by timestamp) as seqnum_out
from t
) tout
on tin.in is not null and
tout.out is not null and
tin.seqnum_in = tout.seqnum_out;
Related
I have a peculiar problem at hand. I need to rank in the following manner:
Each ID gets a new rank.
rank #1 is assigned to the ID with the lowest date. However, the subsequent dates for that particular ID can be higher but they will get the incremental rank w.r.t other IDs.
(E.g. ADF32 series will be considered to be ranked first as it had the lowest date, although it ends with dates 09-Nov, and RT659 starts with 13-Aug it will be ranked subsequently)
For a particular ID, if the days are consecutive then ranks are same, else they add by 1.
For a particular ID, ranks are given in date ASC.
How to formulate a query?
You need two steps:
select
id_col
,dt_col
,dense_rank()
over (order by min_dt, id_col, dt_col - rnk) as part_col
from
(
select
id_col
,dt_col
,min(dt_col)
over (partition by id_col) as min_dt
,rank()
over (partition by id_col
order by dt_col) as rnk
from tab
) as dt
dt_col - rnk caluclates the same result for consecutives dates -> same rank
Try datediff on lead/lag and then perform partitioned ranking
select t.ID_COL,t.dt_col,
rank() over(partition by t.ID_COL, t.date_diff order by t.dt_col desc) as rankk
from ( SELECT ID_COL,dt_col,
DATEDIFF(day, Lag(dt_col, 1) OVER(ORDER BY dt_col),dt_col) as date_diff FROM table1 ) t
One way to think about this problem is "when to add 1 to the rank". Well, that occurs when the previous value on a row with the same id_col differs by more than one day. Or when the row is the earliest day for an id.
This turns the problem into a cumulative sum:
select t.*,
sum(case when prev_dt_col = dt_col - 1 then 0 else 1
end) over
(order by min_dt_col, id_col, dt_col) as ranking
from (select t.*,
lag(dt_col) over (partition by id_col order by dt_col) as prev_dt_col,
min(dt_col) over (partition by id_col) as min_dt_col
from t
) t;
I'm trying to put information to identify GROUP ID by replicating this Excel formula:
IF(OR(A2<>A1,AND(B2<>"000",B1="000")),D1+1,D1)
This formula is written when my cursor is in "D2", meaning I've referred to the newly added column value in the previous row to generate the current value.
I'd like to this with Db2 SQL, but I'm not sure how to because I'll need to do LAG function on the column I'm going to add and referring their value.
Kindly advise if having better way to do.
Thanks.
You need nested OLAP-functions, assuming ORDER BY SERIAL_NUMBER, EVENT_TIMESTAMP returns the order shown in Excel:
with cte as
(
select ...
case --IF(OR(A2<>A1,AND(B2<>"000",B1="000"))
when (lag(OPERATION)
over (order by SERIAL_NUMBER, EVENT_TIMESTAMP) = '000'
and OPERATION <> '000')
or lag(SERIAL_NUMBER,1,'')
over (order by SERIAL_NUMBER, EVENT_TIMESTAMP) <> SERIAL_NUMBER
then 1
else 0
end as flag -- start of new group
from tab
)
select ...
sum(flag)
over (order by SERIAL_NUMBER, EVENT_TIMESTAMP
rows unbounded preceding) as GROUP_ID
from cte
Your code is counting the number of "breaks" in your data, where a "break" is defined as 000 or the value in the first column changing.
In SQL, you can do this as a cumulative sum:
select t.*,
sum(case when prev_serial_number = serial_number or operation <> '000'
then 0 else 1
end) over (order by event_timestamp rows between unbounded preceding and current row) as column_d
from (select t.*,
lag(serial_number) over (order by event_timestamp) as prev_serial_number
from t
) t
I am facing a problem in a query and I am stuck for quite some time now. Here the situation: I have a table with certain records in it, which are terminated via ValidFrom and ValidTo. This table tracks data changes of another table - everytime the underlying data changes, the last valid record is terminated and an insert is performed. Following a SQL fiddle example:
http://sqlfiddle.com/#!18/15c7f/4/0
What I try to achieve is to group records with identical Flags within a timespan into one record. In my fiddle ecample: I would expect the first two records being combined to one record with ValidFrom 2017-01-01 and ValidTo 2017-01-10.
Anyways, I am severly stuck and I tried numerous approaches I found here and in another forum - but without success. One approach is included in the fiddle: evaluate the rownumber ordered by date and substract the rownumber partitioned by flag columns etc. ... but nothing works out.
Any help would be highly appreciated.
Try this query:
select keycol, min(validfrom), max(validto), flag1, flag2 from
(
select *,
sum(iscontinuation) over (partition by keycol order by validfrom rows between UNBOUNDED PRECEDING AND CURRENT ROW) [GroupingCol]
from (
select *,
case when
lag(validto) over (partition by keycol order by validfrom) = dateadd(day, -1, validfrom) and
lag(flag1) over (partition by keycol order by validfrom) = flag1 and
lag(flag2) over (partition by keycol order by validfrom) = flag2 then 0
else 1 end [IsContinuation]
from t
) a
) b group by keycol, flag1, flag2, groupingcol
You need to use min max with partition by your group.Got some help from Michal' Query . This can handle multiple flag data as well
With CTE2 as(
select *,
(case when flag1 = lag(flag1) over (order by (select null)) and
flag2 = lag(flag2) over (order by (select null)) then -1 else 0 end) +
row_number() over (order by (select null)) [GroupingCol]
from t)
Select KeyCol,Flag1,Flag2,ValidFrom,ValidTo From (
Select KeyCol,Flag1,Flag2,
min(ValidFrom) over (partition by KeyCol,Flag1,Flag2,GroupingCol) ValidFrom ,
Max(ValidTo) over (partition by KeyCol,Flag1,Flag2,GroupingCol) ValidTo,
Row_number() over (partition by KeyCol,Flag1,Flag2,GroupingCol order by keycol,ValidFrom) RN
From CTE2) A where RN=1
I have a table and I need to run a query that contains some aggregation Functions like Maximum , Average , Standard Deviation , ...
but instead of one Maximum I should return 5 largest number.
the simplified query is something like this:
SELECT OSI_KEY , MAX(VALUE) , AVG(VALUE) , STDDEV(VALUE), variance(VALUE)
FROM DATA_VALUES_5MIN_6_2013
GROUP BY OSI_KEY
ORDER BY OSI_KEY
and I need some Magical ;) Query like this:
SELECT OSI_KEY , MAX1(VALUE) ,MAX2(VALUE) ,MAX3(VALUE) ,MAX4(VALUE) , MAX5(VALUE) ,
AVG(VALUE) , STDDEV(VALUE), variance(VALUE)
FROM DATA_VALUES_5MIN_6_2013
GROUP BY OSI_KEY
ORDER BY OSI_KEY
I appreciate your considerations.
Oracle has an NTH_VALUE() function. Unfortunately, it is only an analytic function and not a window function. This leads to the strange construct of SELECT DISTINCT with a bunch of analytic functions:
SELECT DISTINCT OSI_KEY,
MAX(VALUE) OVER (PARTITION BY OSI_KEY),
NTH_VALUE(VALUE, 2) OVER (PARTITION BY OSI_KEY ORDER BY VALUE DESC) as MAX_2,
NTH_VALUE(VALUE, 3) OVER (PARTITION BY OSI_KEY ORDER BY VALUE DESC) as MAX_3,
NTH_VALUE(VALUE, 4) OVER (PARTITION BY OSI_KEY ORDER BY VALUE DESC) as MAX_4,
NTH_VALUE(VALUE, 5) OVER (PARTITION BY OSI_KEY ORDER BY VALUE DESC) as MAX_5,
AVG(VALUE) OVER (PARTITION BY OSI_KEY),
STDDEV(VALUE) OVER (PARTITION BY OSI_KEY),
variance(VALUE) OVER (PARTITION BY OSI_KEY)
FROM DATA_VALUES_5MIN_6_2013
ORDER BY OSI_KEY;
You can also do this using conditional aggregation, with a row_number() or dense_rank() in a subquery.
SELECT OSI_KEY, MaxValue FROM (
SELECT OSI_KEY, MAX(value) AS MaxValue FROM table GROUP BY OSI_KEY
)
ORDER BY MaxValue DESC
FETCH FIRST 5 ROWS ONLY;
I'm having trouble constructing a query that can find consecutive values meeting a condition. Example data below, note that Date is sorted DESC and is grouped by ID.
To be selected, for each ID, the most recent RESULT must be 'Fail', and what I need back is the earliest date in that run of 'Fails'. For ID==1, only the 1st two values are of interest (the last doesn't count due to prior 'Complete'. ID==2 doesn't count at all, failing the first condition, and for ID==3, only the first value matters.
A result table might be:
The trick seems to be doing some type of run-length encoding, but even with several attempts manipulating ROW_NUM and an attempt at the tabibitosan method for grouping consecutive values, I've been unable to gain traction.
Any help would be appreciated.
If your database supports window functions, you can do
select id, case when result='Fail' then earliest_fail_date end earliest_fail_date
from (
select t.*
,row_number() over(partition by id order by dt desc) rn
,min(case when result = 'Fail' then dt end) over(partition by id) earliest_fail_date
from tablename t
) x
where rn=1
Use row_number to get the latest row in the table. min() over() to get the earliest fail date for each id. If the first row has status Fail, you select the earliest_fail_date or else it would be null.
It should be noted that the expected result for id=1 is wrong. It should be 2016-09-20 as it is the earliest fail date.
Edit: Having re-read the question, i think this is what you might be looking for. Getting the minimum Fail date from the latest consecutive groups of Fail rows.
with grps as (
select t.*,row_number() over(partition by id order by dt desc) rn
,row_number() over(partition by id order by dt)-row_number() over(partition by id,result order by dt) grp
from tablename t
)
,maxfailgrp as (
select g.*,
max(case when result = 'Fail' then grp end) over(partition by id) maxgrp
from grps g
)
select id,
case when result = 'Fail' then (select min(dt) from maxfailgrp where id = m.id and grp=m.maxgrp) end earliest_fail_date
from maxfailgrp m
where rn=1
Sample Demo