Second maximum and minimum values - sql

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

Related

How do i select all columns, plus the result of the sum

I have this select:
"Select * from table" that return:
Id
Value
1
1
1
1
2
10
2
10
My goal is create a sum from each Value group by id like this:
Id
Value
Sum
1
1
2
1
1
2
2
10
20
2
10
20
I Have tried ways like:
SELECT Id,Value, (SELECT SUM(Value) FROM Table V2 WHERE V2.Id= V.Id GROUP BY IDRNC ) FROM Table v;
But the is not grouping by id.
Id
Value
Sum
1
1
1
1
1
1
2
10
10
2
10
10
Aggregation aggregates rows, reducing the number of records in the output. In this case you want to apply the result of a computation to each of your records, task carried out by the corresponding window function.
SELECT table.*, SUM(Value) OVER(PARTITION BY Id) AS sum_
FROM table
Check the demo here.
Your attempt looks correct.
Can you try the below query :
It works for me :
SELECT Id, Value,
(SELECT SUM(Value) FROM Table V2 WHERE V2.Id= V.Id GROUP BY ID) as sum
FROM Table v;
You can do it using inner join to join with selection grouped by id :
select t.*, sum
from _table t
inner join (
select id, sum(Value) as sum
from _table
group by id
) as s on s.id = t.id
You can check it here
Your select is ok if you adjust it just a little:
SELECT Id,Value, (SELECT SUM(Value) FROM Table V2 WHERE V2.Id= V.Id GROUP BY IDRNC ) FROM Table v;
GROUP BY IDRNC is a mistake and should be GROUP BY ID
you should give an alias to a sum column ...
subquery selecting the sum does not have to have self table alias to be compared with outer query that has one (this is not a mistake - works either way)
Test:
WITH
a_table (ID, VALUE) AS
(
Select 1, 1 From Dual Union All
Select 1, 1 From Dual Union All
Select 2, 10 From Dual Union All
Select 2, 10 From Dual
)
SELECT ID, VALUE, (SELECT SUM(VALUE) FROM a_table WHERE ID = v.ID GROUP BY ID) "ID_SUM" FROM a_table v;
ID VALUE ID_SUM
---------- ---------- ----------
1 1 2
1 1 2
2 10 20
2 10 20

how to find the number has more than two consecutive appearences?

The source table:
id num
-------------------
1 1
2 1
3 1
4 2
5 2
6 1
The output:(appear at least 2 times)
num times
--------------
1 3
2 2
Based on the addition logic defined in the comments it appears this is what you're after:
WITH YourTable AS(
SELECT V.id,
V.num
FROM (VALUES(1,1),
(2,1),
(3,1),
(4,2),
(5,2),
(6,1),
(7,1))V(id,num)), --Added extra row due to logic defined in comments
Grps AS(
SELECT YT.id,
YT.num,
ROW_NUMBER() OVER (ORDER BY id) -
ROW_NUMBER() OVER (PARTITION BY Num ORDER BY id) AS Grp
FROM YourTable YT),
Counts AS(
SELECT num,
COUNT(num) AS Times
FROM grps
GROUP BY grp,
num)
SELECT num,
MAX(times) AS times
FROM Counts
GROUP BY num;
This uses a CTE and ROW_NUMBER to define the groups, and then an additional CTE to get the COUNT per group. Finally you can then get the MAX COUNT per num.
I would adress this with a gaps-and-islands technique:
select num, max(cnt)
from (
select num, count(*) cnt
from (
select
id,
num,
row_number() over(order by id) rn1,
row_number() over(partition by num order by id) rn2
from mytable
) t
group by num, rn1 - rn2
) t
group by num
The most inner query computes row numbers over the whole table and within num groups; the difference between the row numbers gives you the group of adjacent records that each record belong to (you can run that subquery independently and follow how the difference evolves to understand more).
Then, the next level count the number of records in each group of adjacent records. The most outer query takes the maximum count of adjacent records in for each num.
Demo on DB Fiddle:
num | (No column name)
--: | ---------------:
1 | 3
2 | 2
this will work for you
select num,count(num) times from Tabl
group by num

Get specific row from a subquery using aggregate function

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

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.

SELECT records until new value SQL

I have a table
Val | Number
08 | 1
09 | 1
10 | 1
11 | 3
12 | 0
13 | 1
14 | 1
15 | 1
I need to return the last values where Number = 1 (however many that may be) until Number changes, but do not need the first instances where Number = 1. Essentially I need to select back until Number changes to 0 (15, 14, 13)
Is there a proper way to do this in MSSQL?
Based on following:
I need to return the last values where Number = 1
Essentially I need to select back until Number changes to 0 (15, 14,
13)
Try (Fiddle demo ):
select val, number
from T
where val > (select max(val)
from T
where number<>1)
EDIT: to address all possible combinations (Fiddle demo 2)
;with cte1 as
(
select 1 id, max(val) maxOne
from T
where number=1
),
cte2 as
(
select 1 id, isnull(max(val),0) maxOther
from T
where val < (select maxOne from cte1) and number<>1
)
select val, number
from T cross join
(select maxOne, maxOther
from cte1 join cte2 on cte1.id = cte2.id
) X
where val>maxOther and val<=maxOne
I think you can use window functions, something like this:
with cte as (
-- generate two row_number to enumerate distinct groups
select
Val, Number,
row_number() over(partition by Number order by Val) as rn1,
row_number() over(order by Val) as rn2
from Table1
), cte2 as (
-- get groups with Number = 1 and last group
select
Val, Number,
rn2 - rn1 as rn1, max(rn2 - rn1) over() as rn2
from cte
where Number = 1
)
select Val, Number
from cte2
where rn1 = rn2
sql fiddle demo
DEMO: http://sqlfiddle.com/#!3/e7d54/23
DDL
create table T(val int identity(8,1), number int)
insert into T values
(1),(1),(1),(3),(0),(1),(1),(1),(0),(2)
DML
; WITH last_1 AS (
SELECT Max(val) As val
FROM t
WHERE number = 1
)
, last_non_1 AS (
SELECT Coalesce(Max(val), -937) As val
FROM t
WHERE EXISTS (
SELECT val
FROM last_1
WHERE last_1.val > t.val
)
AND number <> 1
)
SELECT t.val
, t.number
FROM t
CROSS
JOIN last_1
CROSS
JOIN last_non_1
WHERE t.val <= last_1.val
AND t.val > last_non_1.val
I know it's a little verbose but I've deliberately kept it that way to illustrate the methodolgy.
Find the highest val where number=1.
For all values where the val is less than the number found in step 1, find the largest val where the number<>1
Finally, find the rows that fall within the values we uncovered in steps 1 & 2.
select val, count (number) from
yourtable
group by val
having count(number) > 1
The having clause is the key here, giving you all the vals that have more than one value of 1.
This is a common approach for getting rows until some value changes. For your specific case use desc in proper spots.
Create sample table
select * into #tmp from
(select 1 as id, 'Alpha' as value union all
select 2 as id, 'Alpha' as value union all
select 3 as id, 'Alpha' as value union all
select 4 as id, 'Beta' as value union all
select 5 as id, 'Alpha' as value union all
select 6 as id, 'Gamma' as value union all
select 7 as id, 'Alpha' as value) t
Pull top rows until value changes:
with cte as (select * from #tmp t)
select * from
(select cte.*, ROW_NUMBER() over (order by id) rn from cte) OriginTable
inner join
(
select cte.*, ROW_NUMBER() over (order by id) rn from cte
where cte.value = (select top 1 cte.value from cte order by cte.id)
) OnlyFirstValueRecords
on OriginTable.rn = OnlyFirstValueRecords.rn and OriginTable.id = OnlyFirstValueRecords.id
On the left side we put an original table. On the right side we put only rows whose value is equal to the value in first line.
Records in both tables will be same until target value changes. After line #3 row numbers will get different IDs associated because of the offset and will never be joined with original table:
LEFT RIGHT
ID Value RN ID Value RN
1 Alpha 1 | 1 Alpha 1
2 Alpha 2 | 2 Alpha 2
3 Alpha 3 | 3 Alpha 3
----------------------- result set ends here
4 Beta 4 | 5 Alpha 4
5 Alpha 5 | 7 Alpha 5
6 Gamma 6 |
7 Alpha 7 |
The ID must be unique. Ordering by this ID must be same in both ROW_NUMBER() functions.