I need to get a max (Amount) value of each Account from the below table
ID Account Amount
1 rx00 100
2 rx00 200
3 rx00 100
4 vxtt 50
5 vxtt 70
6 vxtt 80
I need a result table as
ID Account Amount
2 rx00 200
6 vxtt 80
Please advise to the above result
You can use ROW_NUMBER for this:
SELECT ID, Account, Amount
FROM (
SELECT ID, Account, Amount,
ROW_NUMBER() OVER (PARTITION BY Account
ORDER BY Amount DESC) AS rn
FROM mytable) AS t
WHERE t.rn = 1
If you have ties, i.e. more than one records sharing the same maximum Amount value and you want to return all these records, then use RANK instead of ROW_NUMBER.
You can use row_number to get the desired result.
DECLARE #table TABLE
(ID int, Account varchar(10),Amount int)
INSERT INTO #table
(ID,Account,Amount)
VALUES
(1,'rx00',100),
(2,'rx00',200),
(3,'rx00',100),
(4,'vxtt',50),
(5,'vxtt',70),
(6,'vxtt',80)
SELECT ID, Account, Amount
FROM (
SELECT ID, Account, Amount,
ROW_NUMBER() OVER (PARTITION BY Account
ORDER BY Amount DESC) AS rnk
FROM #table) AS t
WHERE t.rnk = 1
This will be the output:
ID Account Amount
2 rx00 200
6 vxtt 80
If you don't need the ID column in the result set , you can use the below Query.
SELECT Account,max(Amount) AS Amount FROM #table t
GROUP BY Account
This will be the output:
Account Amount
rx00 200
vxtt 80
;With cte
as
(
select id,account,amount
,row_number () over (partition by account order by amount desc) as rn
from
table
)
select * from cte where rn=1
In above Code, a unique RowNumber is assigned for every value Partitioned by account and amount (values sorted in desc order of amount).since we cant directly select Row_number or any calculated value in where clause.I used a Virtual table
Related
The data I am working with looks like below-
category_id subcategory_id date quantities
123 45 2020-02-01 500
123 45 2020-02-13 400
456 35 2020-05-09 350
456 35 2020-05-15 250
456 35 2020-06-18 200
.
.
.
n such columns
Quantities are sorted in descending order
I want to get the data (as seen above) for the first (top) 10 unique pairs of (category_id, subcategory_id). Just like we use limit 10 to get the first 10 records, I want to limit by the top 10 unique pairs of (category_id, subcategory_id) and get the all the data as seen above.
Below is for BigQuery Standard SQL
#standardSQL
SELECT * EXCEPT(rn) FROM (
SELECT *,
ROW_NUMBER() OVER(PARTITION BY category_id, subcategory_id ORDER BY quantities DESC) rn
FROM `project.dataset.table`
)
WHERE rn <= 10
Another - more BigQuery'ish alternative is below
#standardSQL
SELECT TopN.* FROM (
SELECT ARRAY_AGG(t ORDER BY quantities DESC LIMIT 10) topN
FROM `project.dataset.table` t
GROUP BY category_id, subcategory_id
) t, t.topN
If you want 10 rows, each with different category_id/subcategory_id pairs, then you can use:
select t.* except (seqnum)
from (select t.*,
row_number() over (partition by category_id, subcategory_id order by quantities desc) as seqnum
from t
) t
where seqnum = 1
order by quantities desc
limit 10;
This gets the first row (by quantities) for each id pair and then limits to the 10 largest values.
I am trying to build a query results with SQL. Here is my table:
CUST_ID ORDER_ID STORE_FREQUENCY
---------- ----------- ---------------
100 20122 500
100 20100 500
100 20100 737
200 20119 287
300 20130 434
300 20150 434
300 20130 434
300 20120 120
The expected output is:
CUST_ID UNIQUE_ORDERS TOP_STORE
--------- ----------------- ---------
100 2 737
200 1 287
300 3 434
The requirement for the output is:
TOP_STORE = Per CUST_ID, sort the STORE_FREQUENCY column by DESC and get the greatest store frequency
UNIQUE_ORDERS = Per CUST_ID, the number of unique ORDER_IDs in the column
I have started this SELECT statement, but having difficulties completing it to include the 2 columns correctly:
Select cust_id, Count(order_id) as unique_orders
From ORDERS_TABLE
Group By Order_ID
Can you help me complete the 2 columns?
Use aggregate functions such as COUNT(DISTINCT ...) and MAX()
SELECT CUST_ID, COUNT(DISTINCT ORDER_ID), MAX(STORE_FREQUENCY )
FROM TableName
GROUP BY CUST_ID
Here's a DEMO.
It seems to be that the top store should be the store with the greatest number of orders. If so, then CUST_ID 100 should have store 500 as the top store, not 737. In other words, I would expect the following output:
This requirement changes the query strategy, because we no longer can just do a single simple aggregation over the entire table. One approach is to do a separate calculation to find the top store for each customer, then join that result to a query similar to the other answers.
WITH cte AS (
SELECT CUST_ID, STORE_FREQUENCY, cnt,
ROW_NUMBER() OVER (PARTITION BY CUST_ID ORDER BY cnt DESC) rn
FROM
(
SELECT CUST_ID, STORE_FREQUENCY,
COUNT(*) OVER (PARTITION BY CUST_ID, STORE_FREQUENCY) cnt
FROM yourTable
) t
)
SELECT
t1.CUST_ID,
t1.UNIQUE_ORDERS,
t2.TOP_STORE
FROM
(
SELECT CUST_ID, COUNT(DISTINCT ORDER_ID) AS UNIQUE_ORDERS
FROM yourTable
GROUP BY CUST_ID
) t1
INNER JOIN
(
SELECT CUST_ID, STORE_FREQUENCY AS TOP_STORE
FROM cte
WHERE rn = 1
) t2
ON t1.CUST_ID = t2.CUST_ID;
Demo
I am trying to get a specific row from a subquery, but I cannot use an aggregate function in a WHERE clause and I have read that I should be using a HAVING clause but I have no idea where to start.
This is my current sql statement:
SELECT *
FROM
(
select ID, SUM(BALANCE) AS Balance FROM bankacc GROUP BY ID
)A
I will get :
ID | Balance
1 | 30
2 | 40
3 | 50
4 | 50
I need the rows with the MAX(Balance), but I have no idea where to start, please help.
With window function:
DECLARE #t TABLE ( ID INT, Amount MONEY )
INSERT INTO #t
VALUES ( 1, 10 ),
( 1, 10 ),
( 1, 10 ),
( 2, 5 ),
( 2, 20 ),
( 3, 50 )
SELECT ID ,
Amount
FROM ( SELECT ID ,
SUM(Amount) AS Amount ,
RANK() OVER ( ORDER BY SUM(Amount) DESC ) AS rn
FROM #t
GROUP BY ID
) t
WHERE rn = 1
With TOP and TIES:
SELECT TOP 1 WITH TIES
ID ,
SUM(Amount) AS Amount
FROM #t
GROUP BY ID
ORDER BY Amount desc
These versions will return rows where sum will be max, not just top 1 row.
Output:
ID Amount
3 50.00
you can wrap it in a subquery:
SELECT q.id, max(q.b)
FROM
(
select ID, SUM(BALANCE) b FROM bankacc GROUP BY ID
) q
group by q.id
or order them in dessending order and get first record:
select top 1 ID, SUM(BALANCE) b FROM bankacc GROUP BY ID order by b desc
in MySQL you need to use limit 1 instead of top 1
I think this should be simple.
-- This will return only 1 record, even if there are 2 records for MAX same amount
SELECT top 1 ID ,
Amount
FROM ( SELECT ID ,
SUM(Amount) AS Amount
FROM Table
GROUP BY ID
) t
Order by Amount desc,ID asc
Using Window function : This will return what you want.
SELECT ID ,
Amount
FROM ( SELECT ID ,
SUM(Amount) AS Amount ,
RANK() OVER ( ORDER BY SUM(Amount) DESC ) AS rnk
FROM Table
GROUP BY ID
) t
WHERE rnk = 1
I had a table like
ID UserID rupees time
1 1 200 2014-01-05
---------------------------------
2 1 500 2014-04-06
----------------------------------
3 2 10 2014-05-05
----------------------------------
4 2 20 2014-05-06
----------------------------------
I want the output lie
ID UserID Rupees time CumulativeSum
1 1 200 2014-01-05 200
-------------------------------------------------
2 1 500 2014-04-06 700
-------------------------------------------------
3 2 10 2014-05-06 10
-------------------------------------------------
4 2 20 2014-05-06 30
---------------------------------------------------
How can i get this table as purput
Please try using CTE:
;With T as(
select
*,
ROW_NUMBER() over(partition by UserId order by [time]) RN
from tbl
)
select
UserID,
rupees,
[time],
(select SUM(rupees)
from T b
where b.UserID=a.UserID and b.RN<=a.RN) CumulativeSum
from T a
For records with column value time increasing, try the below query:
select
UserID,
rupees,
[time],
(select SUM(rupees)
from tbl b
where b.UserID=a.UserID and b.[time]<=a.[time]) CumulativeSum
from tbl a
For SQL Server 2012 or later, you can use SUM() with an OVER clause that specifies a ROW clause:
declare #t table (ID int,UserID int,rupees int,[time] date)
insert into #t(ID,UserID,rupees,[time]) values
(1,1,200,'20140105'),
(2,1,500,'20140406'),
(3,2, 10,'20140505'),
(4,2, 20,'20140506')
select
*,
SUM(rupees) OVER (
PARTITION BY UserID
ORDER BY id /* or time? */
ROWS BETWEEN
UNBOUNDED PRECEDING AND
CURRENT ROW)
as total
from #t
Result:
ID UserID rupees time total
----------- ----------- ----------- ---------- -----------
1 1 200 2014-01-05 200
2 1 500 2014-04-06 700
3 2 10 2014-05-05 10
4 2 20 2014-05-06 30
DECLARE #t table (UserID INT,rupees INT,DateKey Date )
INSERT INTO #t VALUES
(1,200,'2014-01-05'),
(2,300,'2014-01-06'),
(2,800,'2014-03-06')
select UserID,
rupees,
DateKey,
(SELECT SUM(rupees)from #t t
where t.rupees <= tt.rupees) from #t tt
GROUP BY UserID,rupees,DateKey
Hope this too helps you.
DECLARE #tab TABLE (id INT,userId INT,rupees INT,[time] Date)
INSERT INTO #tab VALUES
(1,1,200 ,'2014-01-05'),
(2,1,500 ,'2014-04-06'),
(3,2,10 ,'2014-05-05'),
(4,2,20 ,'2014-05-06')
SELECT LU.id,LU.userId,LU.rupees,LU.time,SUM(b.rupees) CumulativeSum
FROM (SELECT *,ROW_NUMBER() OVER (PARTITION BY userId ORDER BY [time]) R FROM #tab) B
JOIN (SELECT *,ROW_NUMBER() OVER (PARTITION BY userId ORDER BY [time]) R FROM #tab) LU
ON B.userId = LU.userId AND B.R <= LU.R
GROUP BY LU.id,LU.userId,LU.rupees,LU.time
Result
I am assuming that you are not using SQL Server 2012, which provides the cumulative sum function. The other answers use some form of the row_number() function, but these seems totally unnecessary. I usually approach cumulative sums using correlated subqueries:
select ID, UserID, rupees, [time],
(select sum(rupees)
from table t2
where t2.UserId = t.UserId and
t2.ID <= t.ID
) as CumulativeSum
from table t;
This requires having a column that uniquely identifies each row, and that seems to be the purpose of id. For performance, I would want to have an index on table(UserId, ID, rupees).
select *, SUM(rupees) OVER (
PARTITION BY UserID
ORDER BY id) as CumSum from #tbl
Given a table with multiple rows of an int field and the same identifier, is it possible to return the 2nd maximum and 2nd minimum value from the table.
A table consists of
ID | number
------------------------
1 | 10
1 | 11
1 | 13
1 | 14
1 | 15
1 | 16
Final Result would be
ID | nMin | nMax
--------------------------------
1 | 11 | 15
You can use row_number to assign a ranking per ID. Then you can group by id and pick the rows with the ranking you're after. The following example picks the second lowest and third highest :
select id
, max(case when rnAsc = 2 then number end) as SecondLowest
, max(case when rnDesc = 3 then number end) as ThirdHighest
from (
select ID
, row_number() over (partition by ID order by number) as rnAsc
, row_number() over (partition by ID order by number desc) as rnDesc
) as SubQueryAlias
group by
id
The max is just to pick out the one non-null value; you can replace it with min or even avg and it would not affect the outcome.
This will work, but see caveats:
SELECT Id, number
INTO #T
FROM (
SELECT 1 ID, 10 number
UNION
SELECT 1 ID, 10 number
UNION
SELECT 1 ID, 11 number
UNION
SELECT 1 ID, 13 number
UNION
SELECT 1 ID, 14 number
UNION
SELECT 1 ID, 15 number
UNION
SELECT 1 ID, 16 number
) U;
WITH EX AS (
SELECT Id, MIN(number) MinNumber, MAX(number) MaxNumber
FROM #T
GROUP BY Id
)
SELECT #T.Id, MIN(number) nMin, MAX(number) nMax
FROM #T INNER JOIN
EX ON #T.Id = EX.Id
WHERE #T.number <> MinNumber AND #T.number <> MaxNumber
GROUP BY #T.Id
DROP TABLE #T;
If you have two MAX values that are the same value, this will not pick them up. So depending on how your data is presented you could be losing the proper result.
You could select the next minimum value by using the following method:
SELECT MAX(Number)
FROM
(
SELECT top 2 (Number)
FROM table1 t1
WHERE ID = {MyNumber}
order by Number
)a
It only works if you can restrict the inner query with a where clause
This would be a better way. I quickly put this together, but if you can combine the two queries, you will get exactly what you were looking for.
select *
from
(
select
myID,
myNumber,
row_number() over (order by myID) as myRowNumber
from MyTable
) x
where x.myRowNumber = 2
select *
from
(
select
myID,
myNumber,
row_number() over (order by myID desc) as myRowNumber
from MyTable
) y
where x.myRowNumber = 2
let the table name be tblName.
select max(number) from tblName where number not in (select max(number) from tblName);
same for min, just replace max with min.
As I myself learned just today the solution is to use LIMIT. You order the results so that the highest values are on top and limit the result to 2. Then you select that subselect and order it the other way round and only take the first one.
SELECT somefield FROM (
SELECT somefield from table
ORDER BY somefield DESC LIMIT 2)
ORDER BY somefield ASC LIMIT 1