Looking for help with a query using SQL 2008 R2... I have a table with client and date fields. Most clients have a record for most dates, however some don't.
For example I have this data:
CLIENTID DT
1 5/1/14
1 5/2/14
2 5/3/14
3 5/1/14
3 5/2/14
I can find the missing dates for each CLIENTID by creating a temp table with all possible dates for the period and then joining that to each CLIENTID and DT and only selecting where there is a NULL.
This is what I can get easily for the date range 5/1/14 to 5/4/14:
CLIENTID DTMISSED
1 5/3/14
1 5/4/14
2 5/1/14
2 5/2/14
2 5/4/14
3 5/3/14
3 5/4/14
However I want to group each consecutive missed period together and get the beginning of each period and the length.
For example, if I use the date range 5/1/14 to 5/4/14 I'd like to get:
CLIENTID DTSTART MISSED
1 5/3/14 2
2 5/1/14 2
2 5/4/14 1
3 5/3/14 2
Thanks for helping!
It's fascinating how more elegantly and also mere efficiently this kind of problems can be solved in 2012.
First, the tables:
create table #t (CLIENTID int, DT date)
go
insert #t values
(1, '5/1/14'),
(1, '5/2/14'),
(2, '5/3/14'),
(3, '5/1/14'),
(3, '5/2/14')
go
create table #calendar (dt date)
go
insert #calendar values ('5/1/14'),('5/2/14'),('5/3/14'),('5/4/14')
go
Here's the 2008 solution:
;with x as (
select *, row_number() over(order by clientid, dt) as rn
from #calendar c
cross join (select distinct clientid from #t) x
where not exists (select * from #t where c.dt=#t.dt and x.clientid=#t.clientid)
),
y as (
select x1.*, x2.dt as x2_dt, x2.clientid as x2_clientid
from x x1
left join x x2 on x1.clientid=x2.clientid and x1.dt=dateadd(day,1,x2.dt)
),
z as (
select *, (select sum(case when x2_dt is null then 1 else 0 end) from y y2 where y2.rn<=y.rn) as grp
from y
)
select clientid, min(dt), count(*)
from z
group by clientid, grp
order by clientid
Compare it to 2012:
;with x as (
select *, row_number() over(order by dt) as rn
from #calendar c
cross join (select distinct clientid from #t) x
where not exists (select * from #t where c.dt=#t.dt and x.clientid=#t.clientid)
),
y as (
select x1.*, sum(case when x2.dt is null then 1 else 0 end) over(order by x1.clientid,x1.dt) as grp
from x x1
left join x x2 on x1.clientid=x2.clientid and x1.dt=dateadd(day,1,x2.dt)
)
select clientid, min(dt), count(*)
from y
group by clientid, grp
order by clientid
Related
I have these 3 tables:
CREATE TABLE DimEmployee(EmployeeID INT)
CREATE TABLE DimDepartment(DepartmentID INT)
CREATE TABLE DimDocteur(PositionID INT)
INSERT INTO DimEmployee(EmployeeID) VALUES (1),(2),(3)
INSERT INTO DimDepartment(DepartmentID) VALUES (1),(5),(6)
INSERT INTO DimPosition(PositionID) VALUES (7),(8),(9)
I want to randomly join the 3 tables and get output like below : (example)
First execute:
EmployeeID DepartmentID PositionID RandomDate
1 4 7 2020-07-24 00:00:00.000
2 5 9 2020-11-25 00:00:00.000
Second execute:
EmployeeID DepartmentID PositionID RandomDate
1 4 7 2020-05-04 00:00:00.000
2 5 9 2020-10-30 00:00:00.000
If you want a random join :
SELECT DP.EmployeeID, Q.Department INTO #T1
FROM DimEmployee AS DP
CROSS APPLY (SELECT TOP 1 Dd.DepartmentID FROM DimDepartment AS DD
ORDER BY NEWID() ) AS Q
SELECT *
INTO #T2
FROM #T1 AS T
CROSS APPLY (SELECT TOP 1 DP.PositionID FROM DimPosition AS DP
ORDER BY NEWID() ) AS Q
Or if you want all possibilities :
SELECT
a.EmployeeID, b.DepartmentID, c.PositionID
FROM
DimEmployee AS a
CROSS JOIN
DimDepartment AS b
CROSS JOIN
DimPosition AS c
You need to row-number each table and join on row-number:
CREATE TABLE DimEmployee(EmployeeID INT)
CREATE TABLE DimDepartment(DepartmentID INT)
CREATE TABLE DimDocteur(PositionID INT)
SELECT
emp.EmployeeID,
dep.DepartmentID,
doc.PositionID,
DATEADD(day, (ABS(CHECKSUM(NEWID())) % 65530), 0) RandomDate
FROM (
SELECT *, ROW_NUMBER() OVER(ORDER BY (SELECT 1)) rn
FROM DimEmployee
) emp
JOIN (
SELECT *, ROW_NUMBER() OVER(ORDER BY (SELECT 1)) rn
FROM DimDepartment
) dep ON dep.rn = emp.rn
JOIN (
SELECT *, ROW_NUMBER() OVER(ORDER BY (SELECT 1)) rn
FROM DimDocteur
) doc ON doc.rn = emp.rn
You can also change the ORDER BY to ORDER BY NEWID() to get a more random ordering.
This is in reference to below Question
Loop through each value to the seq num
But now Client want to see the data differently and started a new thread for this question.
below is the requirement.
This is the data .
ID seqNum DOS Service End Date
1 1 1/1/2017 1/15/2017
1 2 1/16/2017 1/16/2017
1 3 1/17/2017 1/21/2017
1 4 1/22/2017 2/13/2017
1 5 2/14/2017 3/21/2017
1 6 2/16/2017 3/21/2017
Expected outPut:
ID SeqNum DOSBeg DOSEnd
1 1 1/1/2017 1/30/2017
1 2 1/31/2017 3/1/2017
1 3 3/2/2017 3/31/2017
For each DOSBeg, add 29 and that is DOSEnd. then Add 1 to DOSEnd (1/31/2017) is new DOSBeg.
Now add 29 to (1/31/2017) and that is 3/1/2017 which is DOSEnd . Repeat this untill DOSend >=Max End Date i.e 3/21/2017.
Basically, we need episode of 29 days for each ID.
I tried with this code and it is giving me duplicates.
with cte as (
select ID, minDate as DOSBeg,dateadd(day,29,mindate) as DOSEnd
from #temp
union all
select ID,dateadd(day,1,DOSEnd) as DOSBeg,dateadd(day,29,dateadd(day,1,DOSEnd)) as DOSEnd
from cte
)
select ID,DOSBeg,DOSEnd
from cte
OPTION (MAXRECURSION 0)
Here mindate is Minimum DOS for this ID i.e. 1/1/2017
I came up with below logic and this is working fine for me. Is there any better way than this ?
declare #table table (id int, seqNum int identity(1,1), DOS date, ServiceEndDate date)
insert into #table
values
(1,'20170101','20170115'),
(1,'20170116','20170116'),
(1,'20170117','20170121'),
(1,'20170122','20170213'),
(1,'20170214','20170321'),
(1,'20170216','20170321'),
(2,'20170101','20170103'),
(2,'20170104','20170118')
select * into #temp from #table
--drop table #data
select distinct ID, cast(min(DOS) over (partition by ID) as date) as minDate
,row_Number() over (partition by ID order by ID, DOS) as SeqNum,
DOS,
max(ServiceEndDate) over (partition by ID)as maxDate
into #data
from #temp
--drop table #StartDateLogic
with cte as
(select ID,mindate as startdate,maxdate
from #data
union all
select ID,dateadd(day,30,startdate) as startdate,maxdate
from cte
where maxdate >= dateadd(day,30,startdate))
select distinct ID,startdate
into #StartDateLogic
from cte
OPTION (MAXRECURSION 0)
--final Result set
select ID
,ROW_NUMBER() over (Partition by ID order by ID,StartDate) as SeqNum
,StartDate
,dateadd(day,29,startdate) as EndDate
from #StartDateLogic
You were on the right track wit the recursive cte, but you forgot the anchor.
declare #table table (id int, seqNum int identity(1,1), DOS date, ServiceEndDate date)
insert into #table
values
(1,'20170101','20170115'),
(1,'20170116','20170116'),
(1,'20170117','20170121'),
(1,'20170122','20170213'),
(1,'20170214','20170321'),
(1,'20170216','20170321'),
(2,'20170101','20170103'),
(2,'20170104','20170118')
;with dates as(
select top 1 with ties id, seqnum, DOSBeg = DOS, DOSEnd = dateadd(day,29,DOS)
from #table
order by row_number() over (partition by id order by seqnum)
union all
select t.id, t.seqNum, DOSBeg = dateadd(day,1,d.DOSEnd), DOSEnd = dateadd(day,29,dateadd(day,1,d.DOSEnd))
from dates d
inner join #table t on
d.id = t.id and t.seqNum = d.seqNum + 1
)
select *
from dates d
where d.DOSEnd <= (select max(dateadd(month,1,ServiceEndDate)) from #table where id = d.id)
order by id, seqNum
basically, I need to retrieve the last two dates for customers who purchased in at least two different dates, implying there are some customer who had purchased only in one date, the data has the following form
client_id date
1 2016-07-02
1 2016-07-02
1 2016-06-01
2 2015-06-01
and I would like to get it in the following form
client_id previous_date last_date
1 2016-06-01 2016-07-02
remarques:
a client can have multiple entries for the same date
a client can have entries only for one date, such customer should be discarded
Rank your dates with DENSE_RANK. Then group by client_id and show the last dates (ranked #1 and #2).
select
client_id,
max(case when rn = 2 then date end) as previous_date,
max(case when rn = 1 then date end) as last_date
from
(
select
client_id,
date,
dense_rank() over (partition by client_id order by date desc) as rn
from mytable
)
group by client_id
having max(rn) > 1;
build up:
t=# create table s153 (c int, d date);
CREATE TABLE
t=# insert into s153 values (1,'2016-07-02'), (1,'2016-07-02'),(1,'2016-06-01'),(2,'2016-06-01');
INSERT 0 4
query:
t=# with a as (
select distinct c,d from s153
)
, b as (
select c,nth_value(d,1) over (partition by c order by d) last_date, nth_value(d,2) over (partition by c order by d) prev_date
from a
)
select * from b where prev_date is not null
;
c | last_date | prev_date
---+------------+------------
1 | 2016-06-01 | 2016-07-02
(1 row)
UNTESTED:
We use a common table expression to assign a row number based on the date in descending order and then only include those records having a row number <=2 and then ensure that those having 1 row are excluded by the having.
WITH CTE AS (
SELECT Distinct Client_ID
, Date
, row_number() over (partition by clientID order by date desc) rn
FROM Table)
SELECT Client_ID, min(date) previous_date, max(date) last_date)
FROM CTE
WHERE RN <=2
GROUP BY Client_ID
HAVING max(RN) > 1
All you need is a group by...
--test date
declare #tablename TABLE
(
client_id int,
[date] datetime
);
insert into #tablename
values( 1 , '2016-07-02'),
(1 , '2016-07-02'),
(1 , '2016-06-01'),
(2 , '2015-06-01');
--query
SELECT client_id,MIN([DATE]) AS [PREVIOUS_DATE], MAX([DATE]) AS [LAST_DATE]
FROM #tablename
GROUP BY client_id
Updated
-- create data
create table myTable
(
client_id integer,
given_date date
);
insert into myTable
values( 1 , '2016-07-02'),
(1 , '2016-07-02'),
(1 , '2016-06-01'),
(1 , '2016-06-03'),
(1 , '2016-06-09'),
(2 , '2015-06-01'),
(3 , '2016-06-03'),
(3 , '2016-06-09');
-- query
SELECT sub.client_id, sub.PREVIOUS_DATE, sub.LAST_DATE
FROM
(select
ROW_NUMBER() OVER (PARTITION BY a.client_id order by b.given_date desc,(MAX(b.given_date) - a.given_date)) AS ROW_NUMBER,
a.client_id,a.given_date AS PREVIOUS_DATE, MAX(b.given_date) - a.given_date AS diff, (b.given_date) AS LAST_DATE
FROM myTable AS a
JOIN myTable AS b
ON b.client_id = a.client_id
WHERE a.given_date <> b.given_date
group by a.client_id, a.given_date, b.given_date) AS sub
WHERE sub.ROW_NUMBER = 1
Below is my theater table:
create table theater
(
srno integer,
seatno integer,
available boolean
);
insert into theater
values
(1, 100,true),
(2, 200,true),
(3, 300,true),
(4, 400,false),
(5, 500,true),
(6, 600,true),
(7, 700,true),
(8, 800,true);
I want a sql which should take input as 'n' and returns me the first 'n' consecutive available seats, like
if n = 2 output should be 100,200
if n = 4 output should be 500,600,700,800
NOTE: I am trying to build an query for postgres 9.3
In SQL-Server you can do It in following:
DECLARE #num INT = 4
;WITH cte AS
(
SELECT *,COUNT(1) OVER(PARTITION BY cnt) pt FROM
(
SELECT tt.*
,(SELECT COUNT(srno) FROM theater t WHERE available <> 'true' and srno < tt.srno) AS cnt
FROM theater tt
WHERE available = 'true'
) t1
)
SELECT TOP (SELECT #num) srno, seatno, available
FROM cte
WHERE pt >= #num
OUTPUT
srno seatno available
5 500 true
6 600 true
7 700 true
8 800 true
This will find the available seats. written for sqlserver 2008+:
DECLARE #num INT = 4
;WITH CTE as
(
SELECT
srno-row_number() over (partition by available order by srno) grp,
srno, seatno, available
FROM theater
), CTE2 as
(
SELECT grp, count(*) over (partition by grp) cnt,
srno, seatno, available
FROM CTE
WHERE available = 'true'
)
SELECT top(#num)
srno, seatno, available
FROM CTE2
WHERE cnt >= #num
ORDER BY srno
Result:
srno seatno available
5 500 1
6 600 1
7 700 1
8 800 1
-- naive solution without window using functions
-- [the funny +-100 constants are caused by
-- "consecutive" seats being 100 apart]
-- -------------------------------------------
WITH bot AS ( -- start of an island --
SELECT seatno FROM theater t
WHERE t.available
AND NOT EXISTS (select * from theater x
where x.available AND x.seatno = t.seatno -100)
)
, top AS ( -- end of an island --
SELECT seatno FROM theater t
WHERE t.available
AND NOT EXISTS (select * from theater x
where x.available AND x.seatno = t.seatno +100)
)
, mid AS ( -- [start,end] without intervening gaps --
SELECT l.seatno AS bot, h.seatno AS top
FROM bot l
JOIN top h ON h.seatno >= l.seatno
AND NOT EXISTS (
SELECT * FROM theater x
WHERE NOT x.available
AND x.seatno >= l.seatno AND x.seatno <= h.seatno)
)
-- all the consecutive ranges
-- [ the end query should select from this
-- , using "cnt >= xxx" ]
SELECT bot, top
, 1+(top-bot)/100 AS cnt
FROM mid;
Result:
bot | top | cnt
-----+-----+-----
100 | 300 | 3
500 | 800 | 4
(2 rows)
thanks guys, but i have done achieved it like below,
select srno, seatno from (
select *, count(0) over (order by grp) grp1 from (
select t1.*,
sum(group_flag) over (order by srno) as grp
from (
select *,
case
when lag(available) over (order by srno) = available then null
else 1
end as group_flag
from theater
) t1 ) tx ) tr where tr.available=true and tr.grp1 >= 2 limit 2
I have a table Attendance in my database.
Date | Present
------------------------
20/11/2013 | Y
21/11/2013 | Y
22/11/2013 | N
23/11/2013 | Y
24/11/2013 | Y
25/11/2013 | Y
26/11/2013 | Y
27/11/2013 | N
28/11/2013 | Y
I want to count the most consecutive occurrence of a value Y or N.
For example in the above table Y occurs 2, 4 & 1 times. So I want 4 as my result.
How to achieve this in SQL Server?
Any help will be appreciated.
Try this:-
The difference between the consecutive date will remain constant
Select max(Sequence)
from
(
select present ,count(*) as Sequence,
min(date) as MinDt, max(date) as MaxDt
from (
select t.Present,t.Date,
dateadd(day,
-(row_number() over (partition by present order by date))
,date
) as grp
from Table1 t
) t
group by present, grp
)a
where Present ='Y'
SQL FIDDLE
You can do this with a recursive CTE:
;WITH cte AS (SELECT Date,Present,ROW_NUMBER() OVER(ORDER BY Date) RN
FROM Table1)
,cte2 AS (SELECT Date,Present,RN,ct = 1
FROM cte
WHERE RN = 1
UNION ALL
SELECT a.Date,a.Present,a.RN,ct = CASE WHEN a.Present = b.Present THEN ct + 1 ELSE 1 END
FROM cte a
JOIN cte2 b
ON a.RN = b.RN+1)
SELECT TOP 1 *
FROM cte2
ORDER BY CT DESC
Demo: SQL Fiddle
Note, the date's in the demo got altered due to the format you posted the dates in your question.