I have a view with this information:
TableA (IDTableA, IDTableB, IDTableC, Active, date, ...)
For each register in TableA and each register in tableC, I want the register of tableB that have the max date and is active.
select IDTableA, IDtableC, IDTableB, Date, Active
from myView
where Active = 1
group by IDTableA, IDTableC
Having Max(Date)
order by IDTableA;
This query works with SQLite, but if I try this query in SQL Server I get an error that say that IDTableB in the select is not contained in the group clause.
I know that in theory the first query in the SQLite shouldn't work, but do it.
How can I do this query in SQL Server?
Thanks.
According to SQL 92, if you use GROUP BY clause, then in SELECT output expression list you can only use columns mentioned in GROUP BY list, or any other columns but they must be wrapped in aggregate functions like count(), sum(), avg(), max(), min() and so on.
Some servers like MSSQL, Postgres are strict about this rule, but for example MySQL and SQLite are very relaxed and forgiving - and this is why it surprised you.
Long story short - if you want this to work in MSSQL, adhere to SQL92 requirement.
This query in SQLServer
select IDTableA, IDtableC, IDTableB, Date, Active
from myView v1
where Active = 1
AND EXISTS (
SELECT 1
FROM myView v2
group by v2.IDTableA, v2.IDTableC
Having Max(v2.Date) = v1.Date
)
order by v1.IDTableA;
OR
Also in SQLServer2005+ you can use CTE with ROW_NUMBER
;WITH cte AS
(
select IDTableA, IDtableC, IDTableB, [Date], Active,
ROW_NUMBER() OVER(PARTITION BY IDTableA, IDTableC ORDER BY [Date] DESC) AS rn
from myView v1
where Active = 1
)
SELECT *
FROM cte
WHERE rn = 1
ORDER BY IDTableA
Try this,
select * from table1 b
where active = 1
and date = (select max(date) from table1
where idtablea = b.idtablea
and idtablec = b.idtablec
and active = 1);
SQLFIDDLE DEMO
Related
(select ID,EXTERNAL_TRANSACTION_ID,EXTERNAL_TRANSACTION_TYPE,ROW_NUMBER() OVER(PARTITION BY EXTERNAL_TRANSACTION_ID ORDER BY ID ) AS SEQNUM
from AC_POS_TRANSACTION_TRK aptt WHERE [RESULT] ='Success'
GROUP BY ID, EXTERNAL_TRANSACTION_ID,EXTERNAL_TRANSACTION_TYPE )
Hello,
On above query, I want to get rows of transaction id's which has seqnum=1 and seqnum=2
But if that transaction id has no second row (seqnum=2), I dont want to get any row for that transaction id.
Thanks!!
Something like this
Not 100% sure if this is correct without you table definition, but my understanding is that you want to EXCLUDE records if that record has an entry with seqnum=2 -- you can't use a where clause alone because that would still return seqnum = 1.
You can use an exists /not exists or in/not in clause like this
(select ID,EXTERNAL_TRANSACTION_ID,EXTERNAL_TRANSACTION_TYPE,ROW_NUMBER() OVER(PARTITION BY EXTERNAL_TRANSACTION_ID ORDER BY ID ) AS SEQNUM
from AC_POS_TRANSACTION_TRK aptt WHERE [RESULT] ='Success'
and not exists ( select 1 from AC_POS_TRANSACTION_TRK a where a.id = aptt.id
and a.seqnum = 2)
GROUP BY ID, EXTERNAL_TRANSACTION_ID,EXTERNAL_TRANSACTION_TYPE )
basically what this does is it excludes records if a record exists as specified in the NOT EXISTS query.
One option you can try is to add a count of rows per group using the same partioning critera and then filter accordingly. Not entirely sure about your query without seeing it in context and with sample data - there's no aggregation so why use group by?
However can you try something along these lines
select * from (
select ID,EXTERNAL_TRANSACTION_ID,EXTERNAL_TRANSACTION_TYPE,
Row_Number() over(partition by EXTERNAL_TRANSACTION_ID order by ID) as SEQNUM,
Count(*) over(partition by EXTERNAL_TRANSACTION_ID) Qty
from AC_POS_TRANSACTION_TRK
where [RESULT] ='Success'
)x
where SEQNUM in (1,2) and Qty>1
This should do the job.
With Qry As (
-- Your original query goes here
),
Select Qry.*
From Qry
Where Exists (
Select *
From Qry Qry1
Where Qry1.EXTERNAL_TRANSACTION_ID = Qry.EXTERNAL_TRANSACTION_ID
And Qry1.SEQNUM = 1
)
And Exists (
Select *
From Qry Qry2
Where Qry2.EXTERNAL_TRANSACTION_ID = Qry.EXTERNAL_TRANSACTION_ID
And Qry2.SEQNUM = 2
)
BTW, your original query looks problematic to me, specifically I think that instead of a GROUP BY columns those columns should be in the PARTITION BY clause of the OVER statement, but without knowing more about the table structures and what you're trying to achieve, I could not say for sure.
I have a table Activity having data like below.It contains multiple rows of CreatedBY like IVR,Raghu and IT.
But I need to get the data only when the first row of CreatedBY='IVR'.
This following query will return firstcreated row for each user (CreatedBy)-
SELECT * FROM
(
SELECT *,
ROW_NUMBER() OVER(PARTITION BY CreatedBy ORDER BY CreatedBy,[Date And Time]) RN
FROM your_table
)A
WHERE RN = 1
I suspect you want the first row per ticket_no. At least, that makes more sense as a query.
If so, in SQL Server, you can use a correlated subquery:
select a.*
from activity a
where a.createdby = 'Raghu' and
a.datetime = (select min(a2.datetime)
from activity a2
where a2.ticket_no = a.ticket_no
);
use exists
select a.*
from table a where createdby='IVR'
and datetime in
(select min(datetime) from table b where a.ticketno=b.ticketno
and createdby='IVR')
I have a database which has two tables. A Call_Info table which holds details about incoming / outgoing calls and has a unique ID named Call_ID. I have a second table which is linked and called the After_Call_Work table.
Each call will have only one After Call Work Record. The dataset is a bit messed up and for the same call there are occasionaly 3 or 4 after call work records. How can I when doing queries just retrieve the earliest After Call Work Record for that particular call ignoring the rest? I imagined using SQL function First_Value but it doesn't seem to be the right one.
Using Microsoft SQL Server 2012.
Any ideas?
You should be able to use select top, something like this:
SELECT TOP 1
FROM call_info ci JOIN after_call_work acw ON ci.call_id=acw.call_id
ORDER BY acw.work_time DESC
WHERE ci.call_id=<your_call_id>
This can be achieved by taking advantage of Window Function
WITH call_List
AS
(
SELECT Call_ID, OtherColumns, DateColumn,
ROW_NUMBER() OVER (PARTITION BY Call_ID ORDER BY DateColumn ASC) rn
FROM After_Call_Work
)
SELECT a.*, b.OtherColumns, b.DateColumn
FROM Call_Info a
INNER JOIN call_List b
ON a.Call_ID = b.Call_ID
WHERE b.rn = 1
SQLFiddle Demo
TSQL Ranking Function
WITH g AS (SELECT ROW_NUMBER() OVER (PARTITION BY callid
ORDER BY date ASC) AS row,* from after_call_work
select * from call_info cinfo inner join g on
cinfo.callid = g.callid and g.row=1
How can one filter a grouped resultset for only those groups that meet some criterion compared against the other groups? For example, only those groups that have the maximum number of constituent records?
I had thought that a subquery as follows should do the trick:
SELECT * FROM (
SELECT *, COUNT(*) AS Records
FROM T
GROUP BY X
) t HAVING Records = MAX(Records);
However the addition of the final HAVING clause results in an empty recordset... what's going on?
In MySQL (Which I assume you are using since you have posted SELECT *, COUNT(*) FROM T GROUP BY X Which would fail in all RDBMS that I know of). You can use:
SELECT T.*
FROM T
INNER JOIN
( SELECT X, COUNT(*) AS Records
FROM T
GROUP BY X
ORDER BY Records DESC
LIMIT 1
) T2
ON T2.X = T.X
This has been tested in MySQL and removes the implicit grouping/aggregation.
If you can use windowed functions and one of TOP/LIMIT with Ties or Common Table expressions it becomes even shorter:
Windowed function + CTE: (MS SQL-Server & PostgreSQL Tested)
WITH CTE AS
( SELECT *, COUNT(*) OVER(PARTITION BY X) AS Records
FROM T
)
SELECT *
FROM CTE
WHERE Records = (SELECT MAX(Records) FROM CTE)
Windowed Function with TOP (MS SQL-Server Tested)
SELECT TOP 1 WITH TIES *
FROM ( SELECT *, COUNT(*) OVER(PARTITION BY X) [Records]
FROM T
)
ORDER BY Records DESC
Lastly, I have never used oracle so apolgies for not adding a solution that works on oracle...
EDIT
My Solution for MySQL did not take into account ties, and my suggestion for a solution to this kind of steps on the toes of what you have said you want to avoid (duplicate subqueries) so I am not sure I can help after all, however just in case it is preferable here is a version that will work as required on your fiddle:
SELECT T.*
FROM T
INNER JOIN
( SELECT X
FROM T
GROUP BY X
HAVING COUNT(*) =
( SELECT COUNT(*) AS Records
FROM T
GROUP BY X
ORDER BY Records DESC
LIMIT 1
)
) T2
ON T2.X = T.X
For the exact question you give, one way to look at it is that you want the group of records where there is no other group that has more records. So if you say
SELECT taxid, COUNT(*) as howMany
GROUP by taxid
You get all counties and their counts
Then you can treat that expressions as a table by making it a subquery, and give it an alias. Below I assign two "copies" of the query the names X and Y and ask for taxids that don't have any more in one table. If there are two with the same number I'd get two or more. Different databases have proprietary syntax, notably TOP and LIMIT, that make this kind of query simpler, easier to understand.
SELECT taxid FROM
(select taxid, count(*) as HowMany from flats
GROUP by taxid) as X
WHERE NOT EXISTS
(
SELECT * from
(
SELECT taxid, count(*) as HowMany FROM
flats
GROUP by taxid
) AS Y
WHERE Y.howmany > X.howmany
)
Try this:
SELECT * FROM (
SELECT *, MAX(Records) as max_records FROM (
SELECT *, COUNT(*) AS Records
FROM T
GROUP BY X
) t
) WHERE Records = max_records
I'm sorry that I can't test the validity of this query right now.
If I have a table with columns id, name, score, date
and I wanted to run a sql query to get the record where id = 2 with the earliest date in the data set.
Can you do this within the query or do you need to loop after the fact?
I want to get all of the fields of that record..
If you just want the date:
SELECT MIN(date) as EarliestDate
FROM YourTable
WHERE id = 2
If you want all of the information:
SELECT TOP 1 id, name, score, date
FROM YourTable
WHERE id = 2
ORDER BY Date
Prevent loops when you can. Loops often lead to cursors, and cursors are almost never necessary and very often really inefficient.
SELECT TOP 1 ID, Name, Score, [Date]
FROM myTable
WHERE ID = 2
Order BY [Date]
While using TOP or a sub-query both work, I would break the problem into steps:
Find target record
SELECT MIN( date ) AS date, id
FROM myTable
WHERE id = 2
GROUP BY id
Join to get other fields
SELECT mt.id, mt.name, mt.score, mt.date
FROM myTable mt
INNER JOIN
(
SELECT MIN( date ) AS date, id
FROM myTable
WHERE id = 2
GROUP BY id
) x ON x.date = mt.date AND x.id = mt.id
While this solution, using derived tables, is longer, it is:
Easier to test
Self documenting
Extendable
It is easier to test as parts of the query can be run standalone.
It is self documenting as the query directly reflects the requirement
ie the derived table lists the row where id = 2 with the earliest date.
It is extendable as if another condition is required, this can be easily added to the derived table.
Try
select * from dataset
where id = 2
order by date limit 1
Been a while since I did sql, so this might need some tweaking.
Using "limit" and "top" will not work with all SQL servers (for example with Oracle).
You can try a more complex query in pure sql:
select mt1.id, mt1."name", mt1.score, mt1."date" from mytable mt1
where mt1.id=2
and mt1."date"= (select min(mt2."date") from mytable mt2 where mt2.id=2)