I have table in following structure...
ReturnID SumbitID Status
1 1 1
1 NULL 2
2 2 3
3 3 1
3 3 1
I want this output.....
RetunrID TotalAttempt Success
1 2 1
2 1 0
3 2 2
Count Of ReturnID is TotalAttempt, when ReturnID = sumbitID and status =1 Then it count as success...
Thanks in Advance!
Something like this
SELECT
T.ReturnID
, COUNT(T.ReturnID) AS TotalAttempt
, SUM(CASE WHEN T.ReturnID = T.sumbitID AND T.Status = 1 THEN 1 ELSE 0 END) AS Status
FROM dbo.MyTable T
group by T.ReturnID
Check this:
SELECT T.ReturnID , COUNT(T.ReturnID) AS TotalAttempt , SUM(CASE WHEN T.ReturnID = T.sumbitID AND T.Statusa = 1 THEN 1 ELSE 0 END) AS Status FROM #table T GROUP BY T.ReturnID
Related
I want to select count of ids where all status is greater than equal to 1
I want something like this
SELECT count(ID)FROM table where all status >= 1
ID
status
1
1
1
2
1
1
1
1
1
1
1
1
2
0
2
1
2
0
2
1
1
1
3
1
3
1
3
1
3
2
3
2
3
2
As you can see in the table where ID = 1,3 has all status greater than equal to 1.
I want to select the count of those IDs.
For example, I want count 2 from the above table.
How can I get those ids count where all status is >=1
Edited:Edited table
To get the ID where all status are >=1 you can use
SELECT ID
FROM Table1
GROUP BY ID
HAVING MIN([status]) >= 1
to count them you can use
SELECT COUNT(*)
FROM (SELECT ID
FROM Table1
GROUP BY ID
HAVING MIN([status]) >= 1) T
For a SQL Server based report,
Table:
CID Date ID Service Days
1 3/7/2016 1 Individual 3
2 4/5/2016 2 Individual 4
3 5/24/2016 1 Individual 3
4 4/4/2016 4 Group 2
5 4/4/2016 4 Group 2
6 2/18/2016 4 Group 2
7 5/5/2016 5 Group 1
8 5/5/2016 5 Group 1
I used this code:
SELECT
ID,
Service,
COUNT(WHEN Days = 4 THEN 1 END) AS '4Days',
COUNT(WHEN Days = 3 THEN 1 END) AS '3Days',
COUNT(WHEN Days = 2 THEN 1 END) AS '2Days',
COUNT(WHEN Days = 1 THEN 1 END) AS '1Day'
FROM Table T1
GROUP BY
ID,
Service
which gives me this Output:
ID Service 4Days 3Days 2Days 1Day
1 Individual 0 2 0 0
2 Individual 1 0 0 0
4 Group 0 0 3 0
5 Group 0 0 0 2
What I want to do is not count the Group services as separate services for separate individuals, but just as one service per group. A Count Distinct used with the Date or ID could help me do that but I don't know how to make that play with the Individual services where I just wanna count them individually and not using DISTINCT. So the desired output is:
ID Service 4Days 3Days 2Days 1Day
1 Individual 0 2 0 0
2 Individual 1 0 0 0
4 Group 0 0 2 0
5 Group 0 0 0 1
I'll edit the post in case I oversimplified the problem since this is dummy data.
Looks like you could use distinct this way if you wanted:
count(distinct
case when Days = 1 then case when Service = 'Group' then 1 else "Date" end end
) as [1Day]
Depending on your indexing it's possible that introducing another column in the query would change the query plan. I suspect that probably isn't the case though.
If I am not wrong for '2Days' column service type 'Group' count should be '2' if our grouping based on 'Date' column, if so then try this:
SELECT
ID,
Service,
CASE WHEN MAX(t.days) = 4 THEN MAX(t.date) ELSE 0 END AS '4Days',
CASE WHEN MAX(t.days) = 3 THEN MAX(t.date) ELSE 0 END AS '3Days',
CASE WHEN MAX(t.days) = 2 THEN MAX(t.date) ELSE 0 END AS '2Days',
CASE WHEN MAX(t.days) = 1 THEN MAX(t.date) ELSE 0 END AS '1Day'
FROM table T1
OUTER APPLY (SELECT days,
COUNT(DISTINCT(date)) date
FROM Table WHERE days = t1.days GROUP BY days) t
GROUP BY id, service
ORDER BY ID
Based on your last edit, this is the most straight forward way I could think of to handle the query:
with cte as (
select id, service, days
from table t1
where service = 'Individual'
union all
select id, service, days
from table t1
where service = 'Group'
group by id, service, days, date
)
select id,
service,
count(case when days = 4 then 'X' end) as [4Days],
count(case when days = 3 then 'X' end) as [3Days],
count(case when days = 2 then 'X' end) as [2Days],
count(case when days = 1 then 'X' end) as [1Day]
from cte
group by id, service
I have a table as below .
Id Product1 Product2 Product3
1 1 null null
2 1 2 null
3 3 1 null
4 2 3 1
Now I would like to write a query to count the entries with Product 1.
In the above case it is 2 for Product1 , 1 for Product2 and 1 for Product 3
I came up with a query something like this.
select
Count(Product1) +
Count(Product2) +
Count(Product3)
from
Table
where Product1 = 1
But it gives me an inaccurate result.
Is there a way I can get the number of occurances for each product ?
Group By
Desired output would be something like this :
ProductID Product1Count Product2Count Product3Count
1 2 1 1
2 1 1 0
3 1 1 0
select Count(CASE WHEN Product1 = 1 THEN 1 END) +
Count(CASE WHEN Product2 = 1 THEN 1 END) +
Count(CASE WHEN Product3 = 1 THEN 1 END)
from Table1
where 1 IN (Product1, Product2, Product3)
SQLFIddle Demo
The reason for adding this line: where 1 IN (Product1, Product2, Product3) is to count only on the affected rows, thus making it more faster than running through all records.
Something like this would work:
SELECT
SUM(CASE WHEN Product1 = 1 THEN 1 ELSE 0 END)
+ SUM(CASE WHEN Product2 = 1 THEN 1 ELSE 0 END)
+ SUM(CASE WHEN Product3 = 1 THEN 1 ELSE 0 END)
FROM Table
Let's say I have this table MyTbl
Record Id_try Id Type IsOk DateOk
1 1 MYDB00125 A 0 NULL
2 1 MYDB00125 B 1 2012-07-19 20:10:05.000
3 1 MYDB00125 A 0 2012-07-25 14:10:05.000
4 2 MYDB00125 A 0 2012-07-19 22:10:05.000
5 1 MYDB00254 B 0 2012-07-19 22:10:05.000
6 1 MYDB00254 A 0 NULL
7 3 MYDB00125 A 1 2012-07-19 22:15:05.000
8 3 MYDB00125 B 1 2012-07-19 22:42:53.000
9 1 MYDB00323 A 1 2012-07-22 00:15:05.00 0
10 1 MYDB00323 C 0 NULL
And I want a group by that brings me for each Id and Type my last "Id_Try Record".
SELECT Id, MAX(Id_Try), MyTbl.Type, IsOK, MAX(DateOk) from MyTbl
GROUP BY Id, MyTbl.Type, IsOK
Won't do, because It'll bring me the last Id_Try AND the last date (Date of record 3 in the example). And I don't care if its the last date or not, I need the date of the last Id_Try.
Is this only solved by a subselect? or a having clause could do?
This is the result expected:
Record Id_try Id Type IsOk DateOk
5 1 MYDB00254 B 0 2012-07-19 22:10:05.000
6 1 MYDB00254 A 0 NULL
7 3 MYDB00125 A 1 2012-07-19 22:15:05.000
8 3 MYDB00125 B 1 2012-07-19 22:42:53.000
9 1 MYDB00323 A 1 2012-07-22 00:15:05.00 0
10 1 MYDB00323 B 0 NULL
I think you will need to break this into two pieces:
with maxIDTry as
(
SELECT MAX(Id_try) as maxId, ID
FROM MyTable
GROUP BY ID
)
SELECT * FROM MyTable as mt
INNER JOIN maxIDTry as max
ON mt.id_try = max.maxId AND mt.id = max.id
I think you want this:
select * FROM
(
select *, row_number() over (partition by id,type order by Id_try desc) as position from mytbl
) foo
where position = 1
order by record
http://www.sqlfiddle.com/#!3/95742/5
Your sample result set lists
9 1 MYDB00323 A 1 2012-07-22 00:15:05.00 0
10 1 MYDB00323 A 0 NULL
But that doesn't make sense since you're saying the ID and the Id_try have the same value. I assume you meant for Id_try to be 2 maybe? Otherwise I think my results match up.
Hope this helps.
SELECT A.Record, A.Id_try, A.Id, A.Type, A.IsOk, A.DateOk
FROM MyTbl A INNER JOIN (
SELECT MAX(Id_Try) Id_Try, Id, B1.Type
from MyTbl B1
GROUP BY Id, B1.Type) AS B
ON A.Id_Try = B.Id_Try AND A.Id = B.Id AND A.Type = B.Type
ORDER BY A.RECORD
I have the first 4 columns of data, and I wan't to use the Ranking functions in the SQL 2008 R2 to derive the fifth column. What's the best way to partition the data into subgroups based on the nextiteminsubgroup and previousiteminsubgroup fields?
Group OrderInGroup NextItemInSubGroup PreviousItemInSubGroup SubGroup
1 1 1 0 1
1 2 1 1 1
1 3 1 1 1
1 4 0 1 1
1 5 0 0 2
1 6 0 0 3
1 7 1 0 4
1 8 1 1 4
1 9 0 1 4
2 1 0 0 1
2 2 0 0 2
2 3 0 0 3
2 4 1 0 4
2 5 0 1 4
3 1 0 0 1
4 1 0 0 1
4 2 0 0 2
4 3 0 0 3
A recursive CTE solution:
DECLARE #t TABLE
([Group] INT
,OrderInGroup INT
,NextItemInSubGroup INT
,PreviousItemInSubGroup INT
,SubGroup INT
)
INSERT #t
VALUES
(1,1,1,0,1),(1,2,1,1,1),(1,3,1,1,1),(1,4,0,1,1),(1,5,0,0,2),(1,6,0,0,3),
(1,7,1,0,4),(1,8,1,1,4),(1,9,0,1,4),(2,1,0,0,1),(2,2,0,0,2),(2,3,0,0,3),
(2,4,1,0,4),(2,5,0,1,4),(3,1,0,0,1),(4,1,0,0,1),(4,2,0,0,2),(4,3,0,0,3)
;WITH recCTE
AS
(
SELECT [Group], OrderInGroup,NextItemInSubGroup , PreviousItemInSubGroup, 1 AS subgroup
FROM #t
WHERE OrderInGroup = 1
UNION ALL
SELECT r.[Group], t.OrderInGroup,t.NextItemInSubGroup , t.PreviousItemInSubGroup,
CASE WHEN r.NextItemInSubGroup = 1 THEN r.subgroup ELSE r.subgroup + 1 END
FROM recCTE AS r
JOIN #t AS t
ON t.[Group] = r.[Group]
AND t.OrderInGroup = r.OrderInGroup + 1
)
SELECT * FROM recCTE
ORDER BY [Group],OrderInGroup ;
P.S. it's best practice to avoid using SQL keywords (e.g. GROUP) as table/column names
Seems like 0 and 0 restart the ranking.
Select
Rank() Over (
Partition By
[Group]
, Case When [NextItemInSubGroup] + [PreviousItemInSubGroup] = 0
Then 0
Else 1
End
Order By [OrderInGroup]
) as [SubGroup]
From Your_Table;