Forgive my SQL knowledge, but I have a Person table with following data -
Id Name
---- ------
1 a
2 b
3 b
4 c
and I want the following result -
Name Total
------ ------
b 2
If I use the GROUP BY query -
SELECT Name, Total=COUNT(*) FROM Person GROUP BY Name
It gives me -
Name Total
------ ------
a 1
b 2
c 1
But I want only the one with maximum count. How do I get that?
If you want ties
SELECT top (1) with ties Name, COUNT(*) AS [count]
FROM Person
GROUP BY Name
ORDER BY count(*) DESC
The easiest way to do this in SQL Server would be to use the top syntax:
SELECT TOP 1 Name, COUNT(*) AS Total
FROM Person
GROUP BY Name
ORDER BY 2 DESC
The answer is:
WITH MaxGroup AS (
SELECT Name, COUNT(*) AS Total
FROM Person
GROUP BY Name)
SELECT Name, Total
FROM MaxGroup
WHERE Total = (SELECT MAX(Total) FROM MaxGroup)
try this...
SELECT Name, COUNT(*)
FROM Person
GROUP BY Name
having COUNT(*)=( SELECT max(COUNT(*)) FROM Person GROUP BY Name) ;
Related
I have table as:
ID PAYOR_NAME
---------- ------------
4 AETNAU
4 AETNA
2 UMR
3 CIGNA
1 METLIFE
Id needs to be one to one mapping with payor_name.But Id 4 is associated with multiple payor_name ,so it is considered as duplicate. So,I tried to find the duplicate by using:
select id, count(*) duplic_data
from (
select distinct id, payor_name
from offc.payor_collec
order by id) t1
group by id;
It is giving me the duplicate Id,But i am wondering is there any also way where we can find duplicates in one to one mappings?
There are multiple ways of finding it:
using exists
using group by and having
using analytical function, if you want all column data for that duplicate values (same as exists)
-
select id, payor_name, cnt as count_
from (
select id, payor_name,
count(1) over (partition by id) as cnt
from offc.payor_collec) t1
Where cnt > 1;
It will give you following result:
ID PAYOR_NAME COUNT_
---------- ------------ -------
4 AETNAU 2
4 AETNA 2
Cheers!!
How about simply using exists:
select pc.id, pc.payor_name, count(*)
from offc.payor_collec pc
where exists (select 1
from offc.payor_collec pc2
where pc2.id = pc.id and pc2.payor_name <> pc.payor_name
)
group by pc.id, pc.payor_name
order by pc.id, count(*) desc;
This also orders by the most frequent value, which might be helpful in figuring out the best name.
This could be a way:
select ID, count(*)
from offc.payor_collec
group by ID
having count(distinct PAYOR_NAME) > 1
Another option would be using a subquery containing having clause with having count(ID)>1
select *
from payor_collec
where ID in
( select ID
from payor_collec t
group by ID
having count(ID)>1 )
I have a following tables:
TABLE A:
ID ID NAME PRICE CODE
00001 B 1000 1
00002 A 2000 1
00003 C 3000 1
Here is the SQL I use:
Select Min (ID),
Min (ID NAME),
Sum(PRICE)
From A
GROUP BY CODE
Here is what I get:
ID ID NAME PRICE
00001 A 6000
As you can see, ID NAME don't match up with the min row value. I need them to match up.
I would like the query to return the following
ID ID NAME PRICE
00001 B 6000
What SQL can I use to get that result?
If you want one row, use limit or fetch first 1 row only:
select a.*
from a
order by a.price asc
fetch first 1 row only;
If, for some reason, you want the sum() of all prices, then you can use window functions:
select a.*, sum(a.price) over () as sum_prices
from a
order by a.price asc
fetch first 1 row only;
You can use row_number() function :
select min(id), max(case when seq = 1 then id_name end) as id_name, sum(price) as price, code
from (select t.*, row_number() over (partition by code order by id) seq
from table t
) t
group by code;
you can also use sub-query
select t1.*,t2.* from
(select ID,Name from t where ID= (select min(ID) from t)
) as t1
cross join (select sum(Price) as total from t) as t2
https://dbfiddle.uk/?rdbms=postgres_10&fiddle=a496232b552390a641c0e5c0fae791d1
id name total
1 B 6000
I need to select the top score of all combined attempts by a player and I need to use a WITH clause.
create table scorecard(
id integer primary key,
player_name varchar(20));
create table scores(
id integer references scorecard,
attempt integer,
score numeric
primary key(id, attempt));
Sample Data for scorecard:
id player_name
1 Bob
2 Steve
3 Joe
4 Rob
Sample data for scores:
id attempt score
1 1 50
1 2 45
2 1 10
2 2 20
3 1 40
3 2 35
4 1 0
4 2 95
The results would simply look like this:
player_name
Bob
Rob
But would only be Bob if Rob had scored less than 95 total. I've gotten so far as to have the name and the total scores that they got in two columns using this:
select scorecard.player_name, sum(scores.score)
from scorecard
left join scores
on scorecard.id= scores.id
group by scorecard.name
order by sum(scores.score) desc;
But how do I just get the names of the highest score (or scores if tied).
And remember, it should be using a WITH clause.
Who ever told you to "use a WITH clause" was missing a more efficient solution. To just get the (possibly multiple) winners:
SELECT c.player_name
FROM scorecard c
JOIN (
SELECT id, rank() OVER (ORDER BY sum(score) DESC) AS rnk
FROM scores
GROUP BY 1
) s USING (id)
WHERE s.rnk = 1;
A plain subquery is typically faster than a CTE. If you must use a WITH clause:
WITH top_score AS (
SELECT id, rank() OVER (ORDER BY sum(score) DESC) AS rnk
FROM scores
GROUP BY 1
)
SELECT c.player_name
FROM scorecard c
JOIN top_score s USING (id)
WHERE s.rnk = 1;
SQL Fiddle.
You could add a final ORDER BY c.player_name to get a stable sort order, but that's not requested.
The key feature of the query is that you can run a window function like rank() over the result of an aggregate function. Related:
Postgres window function and group by exception
Get the distinct sum of a joined table column
Can try something like follows.
With (SELECT id, sum(score) as sum_scores
FROM scores
group by id) as sumScoresTable,
With (SELECT max(score) as max_scores
FROM scores
group by id) as maxScoresTable
select player_name
FROM scorecard
WHERE scorecard.id in (SELECT sumScoresTable.id
from sumScoresTable
where sumScoresTable.score = (select maxScoresTable.score from maxScoresTable)
Try this code:
WITH CTE AS (
SELECT ID, RANK() OVER(ORDER BY SumScore DESC) As R
FROM (
SELECT ID, SUM(score) AS SumScore
FROM scores
GROUP BY ID )
)
SELECT player_name
FROM scorecard
WHERE ID IN (SELECT ID FROM CTE WHERE R = 1)
I have a table with 2 fields:
ID Name
-- -------
1 Alpha
2 Beta
3 Beta
4 Beta
5 Charlie
6 Charlie
I want to group them by name, with 'count', and a row 'SUM'
Name Count
------- -----
Alpha 1
Beta 3
Charlie 2
SUM 6
How would I write a query to add SUM row below the table?
SELECT name, COUNT(name) AS count
FROM table
GROUP BY name
UNION ALL
SELECT 'SUM' name, COUNT(name)
FROM table
OUTPUT:
name count
-------------------------------------------------- -----------
alpha 1
beta 3
Charlie 2
SUM 6
SELECT name, COUNT(name) AS count, SUM(COUNT(name)) OVER() AS total_count
FROM Table GROUP BY name
Without specifying which rdbms you are using
Have a look at this demo
SQL Fiddle DEMO
SELECT Name, COUNT(1) as Cnt
FROM Table1
GROUP BY Name
UNION ALL
SELECT 'SUM' Name, COUNT(1)
FROM Table1
That said, I would recomend that the total be added by your presentation layer, and not by the database.
This is a bit more of a SQL SERVER Version using Summarizing Data Using ROLLUP
SQL Fiddle DEMO
SELECT CASE WHEN (GROUPING(NAME) = 1) THEN 'SUM'
ELSE ISNULL(NAME, 'UNKNOWN')
END Name,
COUNT(1) as Cnt
FROM Table1
GROUP BY NAME
WITH ROLLUP
Try this:
SELECT ISNULL(Name,'SUM'), count(*) as Count
FROM table_name
Group By Name
WITH ROLLUP
all of the solution here are great but not necessarily can be implemented for old mysql servers (at least at my case). so you can use sub-queries (i think it is less complicated).
select sum(t1.cnt) from
(SELECT column, COUNT(column) as cnt
FROM
table
GROUP BY
column
HAVING
COUNT(column) > 1) as t1 ;
Please run as below :
Select sum(count)
from (select Name,
count(Name) as Count
from YourTable
group by Name); -- 6
The way I interpreted this question is needing the subtotal value of each group of answers. Subtotaling turns out to be very easy, using PARTITION:
SUM(COUNT(0)) OVER (PARTITION BY [Grouping]) AS [MY_TOTAL]
This is what my full SQL call looks like:
SELECT MAX(GroupName) [name], MAX(AUX2)[type],
COUNT(0) [count], SUM(COUNT(0)) OVER(PARTITION BY GroupId) AS [total]
FROM [MyView]
WHERE Active=1 AND Type='APP' AND Completed=1
AND [Date] BETWEEN '01/01/2014' AND GETDATE()
AND Id = '5b9xxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx' AND GroupId IS NOT NULL
GROUP BY AUX2, GroupId
The data returned from this looks like:
name type count total
Training Group 2 Cancelation 1 52
Training Group 2 Completed 41 52
Training Group 2 No Show 6 52
Training Group 2 Rescheduled 4 52
Training Group 3 NULL 4 10535
Training Group 3 Cancelation 857 10535
Training Group 3 Completed 7923 10535
Training Group 3 No Show 292 10535
Training Group 3 Rescheduled 1459 10535
Training Group 4 Cancelation 2 27
Training Group 4 Completed 24 27
Training Group 4 Rescheduled 1 27
You can use union to joining rows.
select Name, count(*) as Count from yourTable group by Name
union all
select "SUM" as Name, count(*) as Count from yourTable
For Sql server you can try this one.
SELECT ISNULL([NAME],'SUM'),Count([NAME]) AS COUNT
FROM TABLENAME
GROUP BY [NAME] WITH CUBE
with cttmp
as
(
select Col_Name, count(*) as ctn from tab_name group by Col_Name having count(Col_Name)>1
)
select sum(ctn) from c
You can use ROLLUP
select nvl(name, 'SUM'), count(*)
from table
group by rollup(name)
Use it as
select Name, count(Name) as Count from YourTable
group by Name
union
Select 'SUM' , COUNT(Name) from YourTable
I am using SQL server and the following should work for you:
select cast(name as varchar(16)) as 'Name', count(name) as 'Count'
from Table1
group by Name
union all
select 'Sum:', count(name)
from Table1
I required having count(*) > 1 also. So, I wrote my own query after referring some the above queries
SYNTAX:
select sum(count) from (select count(`table_name`.`id`) as `count` from `table_name` where {some condition} group by {some_column} having count(`table_name`.`id`) > 1) as `tmp`;
Example:
select sum(count) from (select count(`table_name`.`id`) as `count` from `table_name` where `table_name`.`name` IS NOT NULL and `table_name`.`name` != '' group by `table_name`.`name` having count(`table_name`.`id`) > 1) as `tmp`;
You can try group by on name and count the ids in that group.
SELECT name, count(id) as COUNT FROM table group by name
After the query, run below to get the total row count
select ##ROWCOUNT
select sum(s) from
(select count(Col_name) as s from Tab_name group by Col_name having count(*)>1)c
I tried this with solutions avaialble online, but none worked for me.
Table :
Id rank
1 100
1 100
2 75
2 45
3 50
3 50
I want Ids 1 and 3 returned, beacuse they have duplicates.
I tried something like
select * from A where rank in (
select rank from A group by rank having count(rank) > 1
This also returned ids without any duplicates. Please help.
Try this:
select id from table
group by id, rank
having count(*) > 1
select id, rank
from
(
select id, rank, count(*) cnt
from rank_tab
group by id, rank
having count(*) > 1
) t
This general idea should work:
SELECT id
FROM your_table
GROUP BY id
HAVING COUNT(*) > 1 AND COUNT(DISTINCT rank) = 1
In plain English: get every id that exists in multiple rows, but all these rows have the same value in rank.
If you want ids that have some duplicated ranks (but not necessarily all), something like this should work:
SELECT id
FROM your_table
GROUP BY id
HAVING COUNT(*) > COUNT(DISTINCT rank)