I'm trying to SELECT a result set which contains two columns:
The DISTINCT values of a column.
The number of times that that value appears in the column in the table.
This statement (sort of) illustrates what I'm trying to do:
SELECT DISTINCT ID AS X,
(SELECT COUNT(*) FROM #t WHERE ID = X)
FROM #t
ORDER BY X;
The desired result set would look like:
| ID | COUNT
|------|------
| 0 | 12
| 1 | 16
| 2 | 4
SELECT ID AS X,
COUNT(*) AS "COUNT"
FROM #t
GROUP BY ID
ORDER BY X
SELECT ID AS X,
COUNT(1) AS "COUNT"
FROM #t
GROUP BY ID
ORDER BY ID
Use Group By
Your query (Correlated subqueries) should work too. As per GordonLinoff, Replace X with ID in your query and it should work.
SELECT DISTINCT ID AS X,
(SELECT COUNT(*) FROM #t WHERE ID = X)
FROM #t
ORDER BY ID ;
This is just a simple agregation using GROUP BY:
SELECT ID, COUNT(*)
FROM #t
GROUP BY ID
ORDER BY ID;
Related
From this table:
Number Value
1 a
2 b
3 a
2 c
2 b
3 a
2 b
I need to get count of all duplicate rows by Number and Value, i.e. 5.
Thanks.
I think this query is what you want:
SELECT SUM(t.cnt)
FROM
(
SELECT COUNT(*) cnt
FROM table_name
GROUP BY number, value
HAVING COUNT(*) > 1
)t;
Maybe something like this?
select value,number,max(cnt) as Count_distinct from (
select *,row_number () over (partition by value,number order by number) as cnt
from #sample
)t
group by value,number
Output
+---------------------------------+
| Value | Number | Count_Distinct |
| a | 1 | 1 |
| b | 2 | 3 |
| c | 2 | 1 |
| a | 3 | 2 |
+---------------------------------+
Select
count(distinct Number) as Distinct_Numbers,
count(distinct Value) as Distinct_Values
from
Table
This shows how many distinct values are in each column. Does this help?
Give a row number partition by both the columns and order by both the columns. Then count the number of rows where row number greater than 1.
Query
;with cte as(
select [rn] = row_number() over(
partition by [Number], [Value]
order by [Number], [Value]
), *
from [your_table_name]
)
select count(*) from cte
where [rn] > 1;
I think you mean number of unique number - value pairs, you can use:
SELECT count(*)
FROM
(SELECT ROW_NUMBER() OVER (PARTITION BY number, value ORDER BY (select 1)) from mytable rnk) i
where i.rnk = 1
May be this query may help you
select * from [dbo].[Sample_table1]
;WITH
DupContactRecords(number,value,DupsCount)
AS
(
SELECT number,value, COUNT() AS TotalCount FROM [Sample_table1] GROUP BY number,value HAVING COUNT() > 1
)
--to get the duplicats
/*select * from DupContactRecords*/
SELECT sum(DupsCount) FROM DupContactRecords
For example, i create a table about people contribue to 2 campaigns
+-------------------------------------+
| ID Name Campaign Amount (USD) |
+-------------------------------------+
| 1 A 1 10 |
| 2 B 1 5 |
| 3 C 2 7 |
| 4 D 2 9 |
+-------------------------------------+
Task: For each campaign, find the person (Name, ID) who contribute the most to
Expected result is
+-----------------------------------------+
| Campaign Name ID |
+-----------------------------------------+
| 1 A 1 |
| 2 D 4 |
+-----------------------------------------+
I used "group by Campaign" but the result have 2 columns "Campagin" and "max value" when I need "Name" and "ID"
Thanks for your help.
Edited: I fix some values, really sorry
You can use analytic functions for this:
select name, id, amount
from (select t.*, max(amount) over (partition by campaign) as max_amount
from t
) t
where amount = max_amount;
You can also do it by giving a rank/row_number partiton by campaign and order by descending order of amount.
Query
;with cte as(
select [num] = dense_rank() over(
partition by [Campaign]
order by [Amount] desc
), *
from [your_table_name]
)
select [Campaign], [Name], [ID]
from cte
where [num] = 1;
Try the next query:-
SELECT Campaign , Name , ID
FROM (
SELECT Campaign , Name , ID , MAX (Amount)
FROM MyTable
GROUP BY Campaign , Name , ID
) temp;
Simply use Where Clause with the max of amount group by Campaign:-
As following generic code:-
select a, b , c
from tablename
where d in
(
select max(d)
from tablename
group by a
)
Demo:-
Create table #MyTable (ID int , Name char(1), Campaign int , Amount int)
go
insert into #MyTable values (1,'A',1,10)
insert into #MyTable values (2,'B',1,5)
insert into #MyTable values (3,'C',2,7)
insert into #MyTable values (4,'D',2,9)
go
select Campaign, Name , ID
from #MyTable
where Amount in
(
select max(Amount)
from #MyTable
group by Campaign
)
drop table #MyTable
Result:-
Please find the below code for the same
SELECT *
FROM #MyTable T
OUTER APPLY (
SELECT COUNT(1) record
FROM #MyTable T1
where t.Campaign = t1.Campaign
and t.amount < t1.amount
)E
where E.record = 0
how to get around this error :
Unable to use an aggregate or a subquery in an expression used in the
GROUP BY list of a GROUP BY clause.
here is my query :
select Id, name,dayA,monthA,yearA,
sum(x) as x,
(select SUM(x) group by month) as total,
from table_A
group by Id,name,monthA,dAyA,yearA, SUM(x)
in other words :
sample data :
id name dayA monthA yearA x
===========================
1 name1 2 3 2016 4
2 name2 2 3 2016 3
3 name1 2 3 2016 2
Expected result :
id name dayA monthA yearA x total
===================================
1 name1 2 3 2016 4 6
2 name2 2 3 2016 3 3
3 name1 2 3 2016 2 6
Thanks in advance
you're query has more problem.
(select SUM(x) group by month) as total, is it from the same table, not likely since column month is not mention inyour group by. When using sub query in a query, you must guaranteed that i will only return one record.
Based on your sample data and expected results...
create table table_A(
id int,
name varchar(25),
dayA int,
monthA int,
yearA int,
x int
)
insert into table_A
values (1,'name1',2,3,2016,4),
(2,'name2',2,3,2016,3),
(2,'name1',2,3,2016,2)
select ta.id, ta.name, ta.dayA, ta.monthA, ta.yearA, ta.x, total.Total from table_A as ta
left join
(select name, sum(x) as Total from table_A group by name) total on ta.name = total.name
group by
ta.id, ta.name, ta.dayA, ta.monthA, ta.yearA, ta.x, total.name, total.Total
May be this is what you want:
select table_A.*, TotalSums.total
from table_A
left join (select name, monthA, dayA, yearA, sum(x) as total from table_A group by name, monthA, dayA, yearA) as TotalSums
on table_A.name = TotalSums.name
and table_A.monthA = TotalSums.monthA
and table_A.dayA = TotalSums.dayA
and table_A.yearA = TotalSums.yearA
order by id
i think this is what you're looking for
select Id, main.name,dayA,main.monthA,main.yearA,
sum(x) as x,
,max(total.total) as total
from table_A as main
join (select SUM(x) total ,name ,monthA,yearA from table_A group by name,monthA,yearA) as total
on main.name = total.name
and main.monthA = total.monthA
and main.yearA = total.yearA
group by Id,main.name,monthA,dAyA,yearA
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
I have a simple query:
select id, count(*) n
from mytable
group by id
Is it possible to include also the sum(n) in the same query? So the result would look something like this:
id n
---- -----------
1 12
2 1
3 14
4 1
5 2
6 6
Sum=36
You can use a common table expression to do this:
--
; WITH cte as (SELECT id
,count(*) n
FROM mytable
GROUP BY id)
SELECT id, n FROM cte
UNION ALL
SELECT 'Sum', SUM(n) from cte
You can also use ROLLUP: (this may not be exactly correct syntax)
SELECT id
,count(*) n
FROM mytable
GROUP BY id
WITH ROLLUP