Related
Query
Declare #table1 TABLE (accountno varchar(max), saved_amount decimal)
INSERT INTO #table1 VALUES
('001',25),
('002',5)
Declare #table2 TABLE (accountno varchar(max), payamount decimal,ilno int)
INSERT INTO #table2 VALUES
('001',10,1),
('001',10,2),
('001',10,3),
('001',10,4),
('002',10,1),
('002',10,2);
WITH aa
AS (
SELECT a.*
,b.ilno
,b.payamount
,SUM(payamount) OVER (
PARTITION BY a.accountno ORDER BY CAST(a.accountno AS INT)
,ilno
) AS total_amount
FROM #table1 a
LEFT JOIN #table2 b ON a.accountno = b.accountno
)
,bb
AS (
SELECT accountno
,MAX(ilno) AS ilno
FROM aa
WHERE saved_amount >= total_amount
GROUP BY accountno
)
SELECT a.* FROM aa a INNER JOIN bb b on a.accountno =b.accountno AND a.ilno = b.ilno
Result
accountno | saved_amount | ilno | payamount | total_amount
----------------------------------------------------------
001 | 25 | 2 | 10 | 20
Expected Result
accountno | saved_amount | ilno | payamount | total_amount
----------------------------------------------------------
001 | 25 | 2 | 10 | 20
002 | 5 | 1 | 10 | 10
What I want is
If saved_amount is less than the first ilno, then get the first ilno else
get the highest ilno where saved_amount>=total_amount
You have a running total that you compare with the saved amount. You want the highest running total that doesn't exceed the saved amount. But in case even the initial pay amount exceeds the saved amount already, you want to default to this record. So the main task is to find a way of ranking the records. In my query I do it like this:
Prefer records where the running total does not exceed the saved amount.
Then look at the abolute of their difference and take the smallest.
There are certainly other ways that achieve the same. Maybe even methods that you find more readable. Then just adjust the order by clause in the ranking query.
with summed as
(
select
t1.*,
from #table1 t1
join
(
select
ilno,
payamount,
sum(payamount) over (partition by accountno order by ilno) as total_amount
from #table2
) on t2.accountno = t1.accountno
)
, ranked as
(
select summed.*,
row_number() over (partition by accountno
order by case when saved_amount >= total_amount then 1 else 2 end,
abs(saved_amount - total_amount)
) as rn
)
select *
from ranked
where rn = 1;
This is not the "nearest sum", as you said in the title, but the one that obeys the specified rules. So with a saved amount of 100 and paid amounts of first 1 and then 100, you'd get the record with a total of 1 (which is 99 less than the saved amount) and not the one with a total of 101 (which is only 1 more than the saved amount).
Other way to solve using flags:
first calculated one flag to point if saved_amount >= payamount for current row
calculated three more flags:
group_flag to show is there a case where saved_amount >= payamount for the given accountno
[min_ilno] and [max_ilno] for given account
Having this flags, the final result set is calculated easily. Here is the code:
WITH DataSource AS
(
SELECT a.*
,b.ilno
,b.payamount
,SUM(payamount) OVER (PARTITION BY a.accountno ORDER BY ilno) AS total_amount
,IIF(a.saved_amount >= SUM(payamount) OVER (PARTITION BY a.accountno ORDER BY ilno), 1, 0) AS [flag]
FROM #table1 a
LEFT JOIN #table2 b
ON a.accountno = b.accountno
),
DataSourceFinal AS
(
SELECT *
,MAX(flag) OVER (PARTITION BY accountno) as [group_flag]
,MIN(IIF(flag = 0 ,ilno, NULL)) OVER (PARTITION BY accountno) as [min_ilno]
,MAX(IIF(flag = 1 ,ilno, NULL)) OVER (PARTITION BY accountno) as [max_ilno]
FROM DataSource
)
SELECT accountno, saved_amount, ilno, payamount, total_amount
FROM DataSourceFinal
WHERE ([group_flag] = 1 AND [ilno] = [max_ilno])
OR ([group_flag] = 0 AND [ilno] = [min_ilno]);
and the output:
For example, i create a table about people contribue to 2 campaigns
+-------------------------------------+
| ID Name Campaign Amount (USD) |
+-------------------------------------+
| 1 A 1 10 |
| 2 B 1 5 |
| 3 C 2 7 |
| 4 D 2 9 |
+-------------------------------------+
Task: For each campaign, find the person (Name, ID) who contribute the most to
Expected result is
+-----------------------------------------+
| Campaign Name ID |
+-----------------------------------------+
| 1 A 1 |
| 2 D 4 |
+-----------------------------------------+
I used "group by Campaign" but the result have 2 columns "Campagin" and "max value" when I need "Name" and "ID"
Thanks for your help.
Edited: I fix some values, really sorry
You can use analytic functions for this:
select name, id, amount
from (select t.*, max(amount) over (partition by campaign) as max_amount
from t
) t
where amount = max_amount;
You can also do it by giving a rank/row_number partiton by campaign and order by descending order of amount.
Query
;with cte as(
select [num] = dense_rank() over(
partition by [Campaign]
order by [Amount] desc
), *
from [your_table_name]
)
select [Campaign], [Name], [ID]
from cte
where [num] = 1;
Try the next query:-
SELECT Campaign , Name , ID
FROM (
SELECT Campaign , Name , ID , MAX (Amount)
FROM MyTable
GROUP BY Campaign , Name , ID
) temp;
Simply use Where Clause with the max of amount group by Campaign:-
As following generic code:-
select a, b , c
from tablename
where d in
(
select max(d)
from tablename
group by a
)
Demo:-
Create table #MyTable (ID int , Name char(1), Campaign int , Amount int)
go
insert into #MyTable values (1,'A',1,10)
insert into #MyTable values (2,'B',1,5)
insert into #MyTable values (3,'C',2,7)
insert into #MyTable values (4,'D',2,9)
go
select Campaign, Name , ID
from #MyTable
where Amount in
(
select max(Amount)
from #MyTable
group by Campaign
)
drop table #MyTable
Result:-
Please find the below code for the same
SELECT *
FROM #MyTable T
OUTER APPLY (
SELECT COUNT(1) record
FROM #MyTable T1
where t.Campaign = t1.Campaign
and t.amount < t1.amount
)E
where E.record = 0
I have results like this
TimeDiffMin | OrdersCount
10 | 2
12 | 5
09 | 6
20 | 15
27 | 11
I would like the following
TimeDiffMin | OrdersCount
05 | 0
10 | 8
15 | 5
20 | 15
25 | 0
30 | 11
So you can see that i want the grouping of every 5 minutes and show the total order count in those 5 minutes. eg. 0-5 minutes 0 orders, 5-10 minutes 8 orders
any help would be appreciated.
current query:
SELECT TimeDifferenceInMinutes, count(OrderId) NumberOfOrders FROM (
SELECT AO.OrderID, AO.OrderDate, AON.CreatedDate AS CancelledDate, DATEDIFF(minute, AO.OrderDate, AON.CreatedDate) AS TimeDifferenceInMinutes
FROM
(SELECT OrderID, OrderDate FROM AC_Orders) AO
JOIN
(SELECT OrderID, CreatedDate FROM AC_OrderNotes WHERE Comment LIKE '%has been cancelled.') AON
ON AO.OrderID = AON.OrderID
WHERE DATEDIFF(minute, AO.OrderDate, AON.CreatedDate) <= 100 AND AO.OrderDate >= '2016-12-01'
) AS Temp1
GROUP BY TimeDifferenceInMinutes
Now, if you are open to a TVF.
I use this UDF to create dynamic Date/Time Ranges. You supply the range and increment
Declare #YourTable table (TimeDiffMin int,OrdersCount int)
Insert Into #YourTable values
(10, 2),
(12, 5),
(09, 6),
(20,15),
(27,11)
Select TimeDiffMin = cast(R2 as int)
,OrdersCount = isnull(sum(OrdersCount),0)
From (Select R1=RetVal,R2=RetVal+5 From [dbo].[udf-Range-Number](0,25,5)) A
Left Join (
-- Your Complicated Query
Select * From #YourTable
) B on TimeDiffMin >= R1 and TimeDiffMin<R2
Group By R1,R2
Order By 1
Returns
TimeDiffMin OrdersCount
5 0
10 6
15 7
20 0
25 15
30 11
The UDF if interested
CREATE FUNCTION [dbo].[udf-Range-Number] (#R1 money,#R2 money,#Incr money)
Returns Table
Return (
with cte0(M) As (Select cast((#R2-#R1)/#Incr as int)),
cte1(N) As (Select 1 From (Values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) N(N)),
cte2(N) As (Select Top (Select M from cte0) Row_Number() over (Order By (Select NULL)) From cte1 a,cte1 b,cte1 c,cte1 d,cte1 e,cte1 f,cte1 g,cte1 h )
Select RetSeq=1,RetVal=#R1 Union All Select N+1,(N*#Incr)+#R1
From cte2
)
-- Max 100 million observations
-- Select * from [dbo].[udf-Range-Number](0,4,0.25)
You can do this using a derived table to first build up your time difference windows and then joining from that to sum up all the Orders that fall within that window.
declare #t table(TimeDiffMin int
,OrdersCount int
);
insert into #t values
(10, 2)
,(12, 5)
,(09, 6)
,(20,15)
,(27,11);
declare #Increment int = 5; -- Set your desired time windows here.
with n(n)
as
( -- Select 10 rows to start with:
select n from(values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) as n(n)
),n2 as
( -- CROSS APPLY these 10 rows to get 10*10=100 rows we can use to generate incrementing ROW_NUMBERs. Use more CROSS APPLYs to get more rows:
select (row_number() over (order by (select 1))-1) * #Increment as StartMin
,(row_number() over (order by (select 1))) * #Increment as EndMin
from n -- 10 rows
cross apply n n2 -- 100 rows
--cross apply n n3 -- 1000 rows
--cross apply n n4 -- 10000 rows
)
select m.EndMin as TimeDiffMin
,isnull(sum(t.OrdersCount),0) as OrdersCount
from n2 as m
left join #t t
on(t.TimeDiffMin >= m.StartMin
and t.TimeDiffMin < m.EndMin
)
where m.EndMin <= 30 -- Filter as required
group by m.EndMin
order by m.EndMin
Query result:
TimeDiffMin OrdersCount
5 0
10 6
15 7
20 0
25 15
30 11
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 table the following data structure in SQL Server:
ID Date Allocation
1, 2012-01-01, 0
2, 2012-01-02, 2
3, 2012-01-03, 0
4, 2012-01-04, 0
5, 2012-01-05, 0
6, 2012-01-06, 5
etc.
What I need to do is get all consecutive day periods where Allocation = 0, and in the following form:
Start Date End Date DayCount
2012-01-01 2012-01-01 1
2012-01-03 2012-01-05 3
etc.
Is it possible to do this in SQL, and if so how?
In this answer, I'll assume that the "id" field numbers the rows consecutively when sorted by increasing date, like it does in the example data. (Such a column can be created if it does not exist).
This is an example of a technique described here and here.
1) Join the table to itself on adjacent "id" values. This pairs adjacent rows. Select rows where the "allocation" field has changed. Store the result in a temporary table, also keeping a running index.
SET #idx = 0;
CREATE TEMPORARY TABLE boundaries
SELECT
(#idx := #idx + 1) AS idx,
a1.date AS prev_end,
a2.date AS next_start,
a1.allocation as allocation
FROM allocations a1
JOIN allocations a2
ON (a2.id = a1.id + 1)
WHERE a1.allocation != a2.allocation;
This gives you a table having "the end of the previous period", "the start of the next period", and "the value of 'allocation' in the previous period" in each row:
+------+------------+------------+------------+
| idx | prev_end | next_start | allocation |
+------+------------+------------+------------+
| 1 | 2012-01-01 | 2012-01-02 | 0 |
| 2 | 2012-01-02 | 2012-01-03 | 2 |
| 3 | 2012-01-05 | 2012-01-06 | 0 |
+------+------------+------------+------------+
2) We need the start and end of each period in the same row, so we need to combine adjacent rows again. Do this by creating a second temporary table like boundaries but having an idx field 1 greater:
+------+------------+------------+
| idx | prev_end | next_start |
+------+------------+------------+
| 2 | 2012-01-01 | 2012-01-02 |
| 3 | 2012-01-02 | 2012-01-03 |
| 4 | 2012-01-05 | 2012-01-06 |
+------+------------+------------+
Now join on the idx field and we get the answer:
SELECT
boundaries2.next_start AS start,
boundaries.prev_end AS end,
allocation
FROM boundaries
JOIN boundaries2
USING(idx);
+------------+------------+------------+
| start | end | allocation |
+------------+------------+------------+
| 2012-01-02 | 2012-01-02 | 2 |
| 2012-01-03 | 2012-01-05 | 0 |
+------------+------------+------------+
** Note that this answer gets the "internal" periods correctly but misses the two "edge" periods where allocation = 0 at the beginning and allocation = 5 at the end. Those can be pulled in using UNION clauses but I wanted to present the core idea without that complication.
Following would be one way to do it. The gist of this solution is
Use a CTE to get a list of all consecutive start and enddates with Allocation = 0
Use the ROW_NUMBER window function to assign rownumbers depending on both start- and enddates.
Select only those records where both ROW_NUMBERS equal 1.
Use DATEDIFFto calculate the DayCount
SQL Statement
;WITH r AS (
SELECT StartDate = Date, EndDate = Date
FROM YourTable
WHERE Allocation = 0
UNION ALL
SELECT r.StartDate, q.Date
FROM r
INNER JOIN YourTable q ON DATEDIFF(dd, r.EndDate, q.Date) = 1
WHERE q.Allocation = 0
)
SELECT [Start Date] = s.StartDate
, [End Date ] = s.EndDate
, [DayCount] = DATEDIFF(dd, s.StartDate, s.EndDate) + 1
FROM (
SELECT *
, rn1 = ROW_NUMBER() OVER (PARTITION BY StartDate ORDER BY EndDate DESC)
, rn2 = ROW_NUMBER() OVER (PARTITION BY EndDate ORDER BY StartDate ASC)
FROM r
) s
WHERE s.rn1 = 1
AND s.rn2 = 1
OPTION (MAXRECURSION 0)
Test script
;WITH q (ID, Date, Allocation) AS (
SELECT * FROM (VALUES
(1, '2012-01-01', 0)
, (2, '2012-01-02', 2)
, (3, '2012-01-03', 0)
, (4, '2012-01-04', 0)
, (5, '2012-01-05', 0)
, (6, '2012-01-06', 5)
) a (a, b, c)
)
, r AS (
SELECT StartDate = Date, EndDate = Date
FROM q
WHERE Allocation = 0
UNION ALL
SELECT r.StartDate, q.Date
FROM r
INNER JOIN q ON DATEDIFF(dd, r.EndDate, q.Date) = 1
WHERE q.Allocation = 0
)
SELECT s.StartDate, s.EndDate, DATEDIFF(dd, s.StartDate, s.EndDate) + 1
FROM (
SELECT *
, rn1 = ROW_NUMBER() OVER (PARTITION BY StartDate ORDER BY EndDate DESC)
, rn2 = ROW_NUMBER() OVER (PARTITION BY EndDate ORDER BY StartDate ASC)
FROM r
) s
WHERE s.rn1 = 1
AND s.rn2 = 1
OPTION (MAXRECURSION 0)
Alternative way with CTE but without ROW_NUMBER(),
Sample data:
if object_id('tempdb..#tab') is not null
drop table #tab
create table #tab (id int, date datetime, allocation int)
insert into #tab
select 1, '2012-01-01', 0 union
select 2, '2012-01-02', 2 union
select 3, '2012-01-03', 0 union
select 4, '2012-01-04', 0 union
select 5, '2012-01-05', 0 union
select 6, '2012-01-06', 5 union
select 7, '2012-01-07', 0 union
select 8, '2012-01-08', 5 union
select 9, '2012-01-09', 0 union
select 10, '2012-01-10', 0
Query:
;with cte(s_id, e_id, b_id) as (
select s.id, e.id, b.id
from #tab s
left join #tab e on dateadd(dd, 1, s.date) = e.date and e.allocation = 0
left join #tab b on dateadd(dd, -1, s.date) = b.date and b.allocation = 0
where s.allocation = 0
)
select ts.date as [start date], te.date as [end date], count(*) as [day count] from (
select c1.s_id as s, (
select min(s_id) from cte c2
where c2.e_id is null and c2.s_id >= c1.s_id
) as e
from cte c1
where b_id is null
) t
join #tab t1 on t1.id between t.s and t.e and t1.allocation = 0
join #tab ts on ts.id = t.s
join #tab te on te.id = t.e
group by t.s, t.e, ts.date, te.date
Live example at data.SE.
Using this sample data:
CREATE TABLE MyTable (ID INT, Date DATETIME, Allocation INT);
INSERT INTO MyTable VALUES (1, {d '2012-01-01'}, 0);
INSERT INTO MyTable VALUES (2, {d '2012-01-02'}, 2);
INSERT INTO MyTable VALUES (3, {d '2012-01-03'}, 0);
INSERT INTO MyTable VALUES (4, {d '2012-01-04'}, 0);
INSERT INTO MyTable VALUES (5, {d '2012-01-05'}, 0);
INSERT INTO MyTable VALUES (6, {d '2012-01-06'}, 5);
GO
Try this:
WITH DateGroups (ID, Date, Allocation, SeedID) AS (
SELECT MyTable.ID, MyTable.Date, MyTable.Allocation, MyTable.ID
FROM MyTable
LEFT JOIN MyTable Prev ON Prev.Date = DATEADD(d, -1, MyTable.Date)
AND Prev.Allocation = 0
WHERE Prev.ID IS NULL
AND MyTable.Allocation = 0
UNION ALL
SELECT MyTable.ID, MyTable.Date, MyTable.Allocation, DateGroups.SeedID
FROM MyTable
JOIN DateGroups ON MyTable.Date = DATEADD(d, 1, DateGroups.Date)
WHERE MyTable.Allocation = 0
), StartDates (ID, StartDate, DayCount) AS (
SELECT SeedID, MIN(Date), COUNT(ID)
FROM DateGroups
GROUP BY SeedID
), EndDates (ID, EndDate) AS (
SELECT SeedID, MAX(Date)
FROM DateGroups
GROUP BY SeedID
)
SELECT StartDates.StartDate, EndDates.EndDate, StartDates.DayCount
FROM StartDates
JOIN EndDates ON StartDates.ID = EndDates.ID;
The first section of the query is a recursive SELECT, which is anchored by all rows that are allocation = 0, and whose previous day either doesn't exist or has allocation != 0. This effectively returns IDs: 1 and 3 which are the starting dates of the periods of time you want to return.
The recursive part of this same query starts from the anchor rows, and finds all subsequent dates that also have allocation = 0. The SeedID keeps track of the anchored ID through all the iterations.
The result so far is this:
ID Date Allocation SeedID
----------- ----------------------- ----------- -----------
1 2012-01-01 00:00:00.000 0 1
3 2012-01-03 00:00:00.000 0 3
4 2012-01-04 00:00:00.000 0 3
5 2012-01-05 00:00:00.000 0 3
The next sub query uses a simple GROUP BY to filter out all the start dates for each SeedID, and also counts the days.
The last sub query does the same thing with the end dates, but this time the day count isn't needed as we already have this.
The final SELECT query joins these two together to combine the start and end dates, and returns them along with the day count.
Give it a try if it works for you
Here SDATE for your DATE remains same as your table.
SELECT SDATE,
CASE WHEN (SELECT COUNT(*)-1 FROM TABLE1 WHERE ID BETWEEN TBL1.ID AND (SELECT MIN(ID) FROM TABLE1 WHERE ID > TBL1.ID AND ALLOCATION!=0)) >0 THEN(
CASE WHEN (SELECT SDATE FROM TABLE1 WHERE ID =(SELECT MAX(ID) FROM TABLE1 WHERE ID >TBL1.ID AND ID<(SELECT MIN(ID) FROM TABLE1 WHERE ID > TBL1.ID AND ALLOCATION!=0))) IS NULL THEN SDATE
ELSE (SELECT SDATE FROM TABLE1 WHERE ID =(SELECT MAX(ID) FROM TABLE1 WHERE ID >TBL1.ID AND ID<(SELECT MIN(ID) FROM TABLE1 WHERE ID > TBL1.ID AND ALLOCATION!=0))) END
)ELSE (SELECT SDATE FROM TABLE1 WHERE ID = (SELECT MAX(ID) FROM TABLE1 WHERE ID > TBL1.ID ))END AS EDATE
,CASE WHEN (SELECT COUNT(*)-1 FROM TABLE1 WHERE ID BETWEEN TBL1.ID AND (SELECT MIN(ID) FROM TABLE1 WHERE ID > TBL1.ID AND ALLOCATION!=0)) <0 THEN
(SELECT COUNT(*) FROM TABLE1 WHERE ID BETWEEN TBL1.ID AND (SELECT MAX(ID) FROM TABLE1 WHERE ID > TBL1.ID )) ELSE
(SELECT COUNT(*)-1 FROM TABLE1 WHERE ID BETWEEN TBL1.ID AND (SELECT MIN(ID) FROM TABLE1 WHERE ID > TBL1.ID AND ALLOCATION!=0)) END AS DAYCOUNT
FROM TABLE1 TBL1 WHERE ALLOCATION = 0
AND (((SELECT ALLOCATION FROM TABLE1 WHERE ID=(SELECT MAX(ID) FROM TABLE1 WHERE ID < TBL1.ID))<> 0 ) OR (SELECT MAX(ID) FROM TABLE1 WHERE ID < TBL1.ID)IS NULL);
A solution without CTE:
SELECT a.aDate AS StartDate
, MIN(c.aDate) AS EndDate
, (datediff(day, a.aDate, MIN(c.aDate)) + 1) AS DayCount
FROM (
SELECT x.aDate, x.allocation, COUNT(*) idn FROM table1 x
JOIN table1 y ON y.aDate <= x.aDate
GROUP BY x.id, x.aDate, x.allocation
) AS a
LEFT JOIN (
SELECT x.aDate, x.allocation, COUNT(*) idn FROM table1 x
JOIN table1 y ON y.aDate <= x.aDate
GROUP BY x.id, x.aDate, x.allocation
) AS b ON a.idn = b.idn + 1 AND b.allocation = a.allocation
LEFT JOIN (
SELECT x.aDate, x.allocation, COUNT(*) idn FROM table1 x
JOIN table1 y ON y.aDate <= x.aDate
GROUP BY x.id, x.aDate, x.allocation
) AS c ON a.idn <= c.idn AND c.allocation = a.allocation
LEFT JOIN (
SELECT x.aDate, x.allocation, COUNT(*) idn FROM table1 x
JOIN table1 y ON y.aDate <= x.aDate
GROUP BY x.id, x.aDate, x.allocation
) AS d ON c.idn = d.idn - 1 AND d.allocation = c.allocation
WHERE b.idn IS NULL AND c.idn IS NOT NULL AND d.idn IS NULL AND a.allocation = 0
GROUP BY a.aDate
Example