T-SQL, Select first row of a set - sql-server-2005

I have a table that is setup like so:
ID int
Hash varchar
OtherID int
Some Example data:
1 3pm6Qav1Vd 23
2 3pm6Qav1Vd 2
3 3pm6Qav1Vd 19
4 3pm6Qav1Vd 17
5 RSUBe2VBtl 2
6 3pm6Qav1Vd 4
7 3pm6Qav1Vd 21
8 RSUBe2VBtl 23
9 RSUBe2VBtl 19
I would like to be able to pull out just the first row of each hash set:
1 3pm6Qav1Vd 23
5 RSUBe2VBtl 2
The rows would each be the lowest ID for each Hash. I'm using T-SQL on SQL Server 2005. I'm not really sure where to begin on this one.

SELECT t.ID, t.Hash, t.OtherID
FROM (SELECT ID, Hash, OtherID, ROW_NUMBER() OVER(PARTITION BY Hash ORDER BY ID) AS RowNum
FROM YourTable) t
WHERE t.RowNum = 1

select ID, Hash, OtherId
from (
select ID, Hash, OtherId, row_number() over (partition by Hash order by ID) as RN
from yourtable
) a
where RN = 1

Simple!
SELECT *
FROM [tableName]
WHERE ID IN (
SELECT MIN(ID)
FROM [tableName]
GROUP BY Hash
)
Hope this helps.

Do something like below
SELECT * FROM Table T1
INNER JOIN (
SELECT MIN(ID) ID FROM Table GROUP BY Hash) T2
ON T1. ID = T2.ID
Hope this helps!!

Related

Rolling Average in SQL with Partition [duplicate]

declare #t table
(
id int,
SomeNumt int
)
insert into #t
select 1,10
union
select 2,12
union
select 3,3
union
select 4,15
union
select 5,23
select * from #t
the above select returns me the following.
id SomeNumt
1 10
2 12
3 3
4 15
5 23
How do I get the following:
id srome CumSrome
1 10 10
2 12 22
3 3 25
4 15 40
5 23 63
select t1.id, t1.SomeNumt, SUM(t2.SomeNumt) as sum
from #t t1
inner join #t t2 on t1.id >= t2.id
group by t1.id, t1.SomeNumt
order by t1.id
SQL Fiddle example
Output
| ID | SOMENUMT | SUM |
-----------------------
| 1 | 10 | 10 |
| 2 | 12 | 22 |
| 3 | 3 | 25 |
| 4 | 15 | 40 |
| 5 | 23 | 63 |
Edit: this is a generalized solution that will work across most db platforms. When there is a better solution available for your specific platform (e.g., gareth's), use it!
The latest version of SQL Server (2012) permits the following.
SELECT
RowID,
Col1,
SUM(Col1) OVER(ORDER BY RowId ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Col2
FROM tablehh
ORDER BY RowId
or
SELECT
GroupID,
RowID,
Col1,
SUM(Col1) OVER(PARTITION BY GroupID ORDER BY RowId ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Col2
FROM tablehh
ORDER BY RowId
This is even faster. Partitioned version completes in 34 seconds over 5 million rows for me.
Thanks to Peso, who commented on the SQL Team thread referred to in another answer.
For SQL Server 2012 onwards it could be easy:
SELECT id, SomeNumt, sum(SomeNumt) OVER (ORDER BY id) as CumSrome FROM #t
because ORDER BY clause for SUM by default means RANGE UNBOUNDED PRECEDING AND CURRENT ROW for window frame ("General Remarks" at https://msdn.microsoft.com/en-us/library/ms189461.aspx)
Let's first create a table with dummy data:
Create Table CUMULATIVESUM (id tinyint , SomeValue tinyint)
Now let's insert some data into the table;
Insert Into CUMULATIVESUM
Select 1, 10 union
Select 2, 2 union
Select 3, 6 union
Select 4, 10
Here I am joining same table (self joining)
Select c1.ID, c1.SomeValue, c2.SomeValue
From CumulativeSum c1, CumulativeSum c2
Where c1.id >= c2.ID
Order By c1.id Asc
Result:
ID SomeValue SomeValue
-------------------------
1 10 10
2 2 10
2 2 2
3 6 10
3 6 2
3 6 6
4 10 10
4 10 2
4 10 6
4 10 10
Here we go now just sum the Somevalue of t2 and we`ll get the answer:
Select c1.ID, c1.SomeValue, Sum(c2.SomeValue) CumulativeSumValue
From CumulativeSum c1, CumulativeSum c2
Where c1.id >= c2.ID
Group By c1.ID, c1.SomeValue
Order By c1.id Asc
For SQL Server 2012 and above (much better performance):
Select
c1.ID, c1.SomeValue,
Sum (SomeValue) Over (Order By c1.ID )
From CumulativeSum c1
Order By c1.id Asc
Desired result:
ID SomeValue CumlativeSumValue
---------------------------------
1 10 10
2 2 12
3 6 18
4 10 28
Drop Table CumulativeSum
A CTE version, just for fun:
;
WITH abcd
AS ( SELECT id
,SomeNumt
,SomeNumt AS MySum
FROM #t
WHERE id = 1
UNION ALL
SELECT t.id
,t.SomeNumt
,t.SomeNumt + a.MySum AS MySum
FROM #t AS t
JOIN abcd AS a ON a.id = t.id - 1
)
SELECT * FROM abcd
OPTION ( MAXRECURSION 1000 ) -- limit recursion here, or 0 for no limit.
Returns:
id SomeNumt MySum
----------- ----------- -----------
1 10 10
2 12 22
3 3 25
4 15 40
5 23 63
Late answer but showing one more possibility...
Cumulative Sum generation can be more optimized with the CROSS APPLY logic.
Works better than the INNER JOIN & OVER Clause when analyzed the actual query plan ...
/* Create table & populate data */
IF OBJECT_ID('tempdb..#TMP') IS NOT NULL
DROP TABLE #TMP
SELECT * INTO #TMP
FROM (
SELECT 1 AS id
UNION
SELECT 2 AS id
UNION
SELECT 3 AS id
UNION
SELECT 4 AS id
UNION
SELECT 5 AS id
) Tab
/* Using CROSS APPLY
Query cost relative to the batch 17%
*/
SELECT T1.id,
T2.CumSum
FROM #TMP T1
CROSS APPLY (
SELECT SUM(T2.id) AS CumSum
FROM #TMP T2
WHERE T1.id >= T2.id
) T2
/* Using INNER JOIN
Query cost relative to the batch 46%
*/
SELECT T1.id,
SUM(T2.id) CumSum
FROM #TMP T1
INNER JOIN #TMP T2
ON T1.id > = T2.id
GROUP BY T1.id
/* Using OVER clause
Query cost relative to the batch 37%
*/
SELECT T1.id,
SUM(T1.id) OVER( PARTITION BY id)
FROM #TMP T1
Output:-
id CumSum
------- -------
1 1
2 3
3 6
4 10
5 15
Select
*,
(Select Sum(SOMENUMT)
From #t S
Where S.id <= M.id)
From #t M
You can use this simple query for progressive calculation :
select
id
,SomeNumt
,sum(SomeNumt) over(order by id ROWS between UNBOUNDED PRECEDING and CURRENT ROW) as CumSrome
from #t
There is a much faster CTE implementation available in this excellent post:
http://weblogs.sqlteam.com/mladenp/archive/2009/07/28/SQL-Server-2005-Fast-Running-Totals.aspx
The problem in this thread can be expressed like this:
DECLARE #RT INT
SELECT #RT = 0
;
WITH abcd
AS ( SELECT TOP 100 percent
id
,SomeNumt
,MySum
order by id
)
update abcd
set #RT = MySum = #RT + SomeNumt
output inserted.*
For Ex: IF you have a table with two columns one is ID and second is number and wants to find out the cumulative sum.
SELECT ID,Number,SUM(Number)OVER(ORDER BY ID) FROM T
Once the table is created -
select
A.id, A.SomeNumt, SUM(B.SomeNumt) as sum
from #t A, #t B where A.id >= B.id
group by A.id, A.SomeNumt
order by A.id
The SQL solution wich combines "ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW" and "SUM" did exactly what i wanted to achieve.
Thank you so much!
If it can help anyone, here was my case. I wanted to cumulate +1 in a column whenever a maker is found as "Some Maker" (example). If not, no increment but show previous increment result.
So this piece of SQL:
SUM( CASE [rmaker] WHEN 'Some Maker' THEN 1 ELSE 0 END)
OVER
(PARTITION BY UserID ORDER BY UserID,[rrank] ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Cumul_CNT
Allowed me to get something like this:
User 1 Rank1 MakerA 0
User 1 Rank2 MakerB 0
User 1 Rank3 Some Maker 1
User 1 Rank4 Some Maker 2
User 1 Rank5 MakerC 2
User 1 Rank6 Some Maker 3
User 2 Rank1 MakerA 0
User 2 Rank2 SomeMaker 1
Explanation of above: It starts the count of "some maker" with 0, Some Maker is found and we do +1. For User 1, MakerC is found so we dont do +1 but instead vertical count of Some Maker is stuck to 2 until next row.
Partitioning is by User so when we change user, cumulative count is back to zero.
I am at work, I dont want any merit on this answer, just say thank you and show my example in case someone is in the same situation. I was trying to combine SUM and PARTITION but the amazing syntax "ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW" completed the task.
Thanks!
Groaker
Above (Pre-SQL12) we see examples like this:-
SELECT
T1.id, SUM(T2.id) AS CumSum
FROM
#TMP T1
JOIN #TMP T2 ON T2.id < = T1.id
GROUP BY
T1.id
More efficient...
SELECT
T1.id, SUM(T2.id) + T1.id AS CumSum
FROM
#TMP T1
JOIN #TMP T2 ON T2.id < T1.id
GROUP BY
T1.id
Try this
select
t.id,
t.SomeNumt,
sum(t.SomeNumt) Over (Order by t.id asc Rows Between Unbounded Preceding and Current Row) as cum
from
#t t
group by
t.id,
t.SomeNumt
order by
t.id asc;
Try this:
CREATE TABLE #t(
[name] varchar NULL,
[val] [int] NULL,
[ID] [int] NULL
) ON [PRIMARY]
insert into #t (id,name,val) values
(1,'A',10), (2,'B',20), (3,'C',30)
select t1.id, t1.val, SUM(t2.val) as cumSum
from #t t1 inner join #t t2 on t1.id >= t2.id
group by t1.id, t1.val order by t1.id
Without using any type of JOIN cumulative salary for a person fetch by using follow query:
SELECT * , (
SELECT SUM( salary )
FROM `abc` AS table1
WHERE table1.ID <= `abc`.ID
AND table1.name = `abc`.Name
) AS cum
FROM `abc`
ORDER BY Name

MS-SQL max ID with inner join

Can't see the wood for the trees on this and I'm sure it's simple.
I'm trying to return the max ID for a related record in a joined table
Table1
NiD
Name
1
Peter
2
John
3
Arthur
Table2
ID
NiD
Value
1
1
5
2
2
10
3
3
10
4
1
20
5
2
15
Max Results
NiD
ID
Value
1
4
20
2
5
15
3
3
10
You can use row_number() for this:
select NiD, ID, Value
from (select t2.*,
row_number() over (partition by NiD order by ID desc) as seqnum
from table2 t2
) t2
where seqnum = 1;
As the question is stated, you do not need table1, because table2 has all the ids.
This is how I'd do it, I think ID and Value will be NULL when Table2 does not have a corresponding entry for a Table1 record:
SELECT NiD, ID, [Value]
FROM Table1
OUTER APPLY (
SELECT TOP 1 ID, [Value]
FROM Table2
WHERE Table1.NiD = Table2.NiD
ORDER BY [Value] DESC
) AS Top_Table2
CREATE TABLE Names
(
NID INT,
[Name] VARCHAR(MAX)
)
CREATE TABLE Results
(
ID INT,
NID INT,
VALUE INT
)
INSERT INTO Names VALUES (1,'Peter'),(2,'John'),(3,'Arthur')
INSERT INTO Results VALUES (1,1,5),(2,2,10),(3,3,10),(4,1,20),(5,2,15)
SELECT a.NID,
r.ID,
a.MaxVal
FROM (
SELECT NID,
MAX(VALUE) as MaxVal
FROM Results r
GROUP BY NID
) a
JOIN Results r
ON a.NID = r.NID AND a.MaxVal = r.VALUE
ORDER BY NID
Here's what I have used in similar situations, performance was fine, provided that the data set wasn't too large (under 1M rows).
SELECT
table1.nid
,table2.id
,table2.value
FROM table1
INNER JOIN table2 ON table1.nid = table2.nid
WHERE table2.value = (
SELECT MAX(value)
FROM table2
WHERE nid = table1.nid)
ORDER BY 1

SQL select top if columns are same

If I have a table like this:
Id StateId Name
1 1 a
2 2 b
3 1 c
4 1 d
5 3 e
6 2 f
I want to select like below:
Id StateId Name
4 1 d
5 3 e
6 2 f
For example, Ids 1,3,4 have stateid 1. So select row with max Id, i.e, 4.
; WITH CTE AS
(
SELECT *, ROW_NUMBER() OVER(PARTITION BY STATEID ORDER BY ID DESC) AS RN
)SELECT ID, STATEID, NAME FROM CTE WHERE RN = 1
You can use ROW_NUMBER() + TOP 1 WITH TIES:
SELECT TOP 1 WITH TIES
Id,
StateId,
[Name]
FROM YourTable
ORDER BY ROW_NUMBER() OVER (PARTITION BY StateId ORDER BY Id DESC)
Output:
Id StateId Name
4 1 d
6 2 f
5 3 e
Disclaimer: I gave this answer before the OP had specified an actual database, and hence avoided using window functions. For a possibly more appropriate answer, see the reply by #Tanjim above.
Here is an option using joins which should work across most RDBMS.
SELECT t1.*
FROM yourTable t1
INNER JOIN
(
SELECT StateId, MAX(Id) AS Id
FROM yourTable
GROUP BY StateId
) t2
ON t1.StateId = t2.StateId AND
t1.Id = t2.Id
The following using a subquery, to find the maximum Id for each of the states. The WHERE clause then only includes rows with ids from that subquery.
SELECT
[Id], [StateID], [Name]
FROM
TABLENAME S1
WHERE
Id IN (SELECT MAX(Id) FROM TABLENAME S2 WHERE S2.StateID = S1.StateID)

SQL query with a GROUP BY

I have a table like
Id WID AID DateValue
1 1 12 2015-07-10 15:14:46.770
2 1 13 2015-07-10 14:14:46.770
3 2 13 2015-07-10 13:14:46.770
4 2 13 2015-07-10 12:14:46.770
5 2 13 2015-07-10 11:14:46.770
Now, I want to get the Id value by grouping WIDAND AID, then taking the MAX value from DateValue.
The desired output is
Output:
Id
1
2
3
I tried something like this
SELECT Id, MAX(DateValue)
FROM Table1
GROUP BY WID, AID`
Though I don't want DateValue in the select but it is fine.
Can anyone help me on this
I think you want a query like this:
SELECT Id --or *
FROM (
SELECT *
, ROW_NUMBER() OVER (PARTITION BY WID, AID ORDER BY DateValue DESC) AS seqNum
FROM yourTable) dt
WHERE (SeqNum =1);
You can use a correlated subquery like so:
SELECT Id FROM Table1 t1
WHERE NOT EXISTS (
SELECT 1 FROM Table1 t2
WHERE t1.WID = t2.WID AND t1.AID = t2.AID AND t1.DateValue < t2.DateValue
)

Add a column with the max value of the group

I want to add an extra column, where the max values of each group (ID) will appear.
Here how the table looks like:
select ID, VALUE from mytable
ID VALUE
1 4
1 1
1 7
2 2
2 5
3 7
3 3
Here is the result I want to get:
ID VALUE max_values
1 4 7
1 1 7
1 7 7
2 2 5
2 5 5
3 7 7
3 3 7
Thank you for your help in advance!
Your previous questions indicate that you are using SQL Server, in which case you can use window functions:
SELECT ID,
Value,
MaxValue = MAX(Value) OVER(PARTITION BY ID)
FROM mytable;
Based on your comment on another answer about first summing value, you may need to use a subquery to actually get this:
SELECT ID,
Date,
Value,
MaxValue = MAX(Value) OVER(PARTITION BY ID)
FROM ( SELECT ID, Date, Value = SUM(Value)
FROM mytable
GROUP BY ID, Date
) AS t;
There is no need to use GROUP BY in subselect.
select ID, VALUE,
(select MAX(VALUE) from mytable where ID = t.ID) as MaxValue
from mytable t
Use this query.
SELECT ID
,value
,(
SELECT MAX(VALUE)
FROM GetMaxValue gmv
WHERE gmv.ID = gmv1.ID
GROUP BY ID
) as max_value
FROM GetMaxValue gmv1
ORDER BY ID
Try it with a sub select and group by, then grab the MAX of this group:
select
ID,
VALUE,
(select MAX(VALUE)
from mytable
group by ID
having ID = t.ID
) as max_values
from mytable t
Edit:
I built a SQL fiddle, which shows that my solution works, but also VDohnal is correct and doesn't need the group by, so I'll upvote his answer.