I have a table "well". It contains a column app_rate_unit (type: nvarchar).
My goal is to count every distinct value in the table and let the DBMS (MS Server 2005) give me the most occurring one.
This is my code:
SELECT MAX(app_rate_unit) AS MAX_APP
FROM (SELECT app_rate_unit, COUNT(*) AS co
FROM dbo.well AS w
GROUP BY app_rate_unit
) AS derivedtbl_1
The poblem with it is however, that my DBMS actually delivers the lowest count to me.
SideQuestion: How do I filter for a foreign key (in the table) and NOT NULL (in app_rate_unit) when counting?
select top 1 app_rate_unit, count(*) from dbo.well
group by app_rate_unit
order by count(*) desc
Try this
SELECT
COUNT(app_rate_unit)AS MAX_APP ,
app_rate_unit
FROM
dbo.well
WHERE
app_rate_unit IS NOT NULL
GROUP BY
app_rate_unit
ORDER BY
MAX_APP DESC
The above script will give you the count and the item. You can change the count if you are not sure only one item will have the maximum number of occurrence.
select top 1 count(*) as co from dbo.well as w group by app_rate_unit
order by count(*) desc
In PostgreSQL we can write query which using max of count as
select max(count) from (
select count(id) from Table _name group by created_by,status_id having status_id = 6 ) as Alias
eg
select max(count) from (
select count(id) from orders group by created_by,status_id having status_id = 6 ) as foo
Related
I am using SQL Server and I have a table "a"
month segment_id price
-----------------------------
1 1 100
1 2 200
2 3 50
2 4 80
3 5 10
I want to make a query which presents the original columns where the price will be the max per month
The result should be:
month segment_id price
----------------------------
1 2 200
2 4 80
3 5 10
I tried to write SQL code:
Select
month, segment_id, max(price) as MaxPrice
from
a
but I got an error:
Column segment_id is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause
I tried to fix it in many ways but didn't find how to fix it
Because you need a group by clause without segment_id
Select month, max(price) as MaxPrice
from a
Group By month
as you want results per each month, and segment_id is non-aggregated in your original select statement.
If you want to have segment_id with maximum price repeating per each month for each row, you need to use max() function as window analytic function without Group by clause
Select month, segment_id,
max(price) over ( partition by month order by segment_id ) as MaxPrice
from a
Edit (due to your lastly edited desired results) : you need one more window analytic function row_number() as #Gordon already mentioned:
Select month, segment_id, price From
(
Select a.*,
row_number() over ( partition by month order by price desc ) as Rn
from a
) q
Where rn = 1
I would recommend a correlated subquery:
select t.*
from t
where t.price = (select max(t2.price) from t t2 where t2.month = t.month);
The "canonical" solution is to use row_number():
select t.*
from (select t.*,
row_number() over (partition by month order by price desc) as seqnum
from t
) t
where seqnum = 1;
With the right indexes, the correlated subquery often performs better.
Only because it was not mentioned.
Yet another option is the WITH TIES clause.
To be clear, the approach by Gordon and Barbaros would be a nudge more performant, but this technique does not require or generate an extra column.
Select Top 1 with ties *
From YourTable
Order By row_number() over (partition by month order by price desc)
With not exists:
select t.*
from tablename t
where not exists (
select 1 from tablename
where month = t.month and price > t.price
)
or:
select t.*
from tablename inner join (
select month, max(price) as price
from tablename
group By month
) g on g.month = t.month and g.price = t.price
Here is the query to select the Config Type that has the Max number of rows.
Is there another way that is just as fast as the second query to select the data?
select CONFIG_TYPE, MAX(COUNTING) FROM
(select CONFIG_TYPE, COUNT(*) as COUNTING FROM NOTIFICATION_CONFIG GROUP BY CONFIG_TYPE)
WHERE COUNTING =
(select MAX(COUNTING) FROM
(select COUNT(*) as COUNTING FROM NOTIFICATION_CONFIG GROUP BY CONFIG_TYPE)
)GROUP BY CONFIG_TYPE
This second query will do the job in 1 search, unlike the other that does it in 2.
select CONFIG_TYPE, COUNTING FROM
(select CONFIG_TYPE, COUNT(*) as COUNTING FROM NOTIFICATION_CONFIG
GROUP BY CONFIG_TYPE ORDER BY COUNTING DESC FETCH FIRST 1 ROW ONLY)
just wondering if there was another way to do this.
Use window functions!
SELECT c.*
FROM (SELECT CONFIG_TYPE, COUNT(*) as COUNTING,
RANK() OVER (ORDER BY COUNT(*) DESC) as seqnum
FROM NOTIFICATION_CONFIG
GROUP BY CONFIG_TYPE
) c
WHERE seqnum = 1;
This uses RANK(), so it will return all maximum values (in the case of ties). If you want one arbitrary maximum returned in the case of ties, then use ROW_NUMBER() instead. Or, if you don't want ties, you can do:
SELECT CONFIG_TYPE, COUNT(*) as COUNTING
FROM NOTIFICATION_CONFIG
GROUP BY CONFIG_TYPE
ORDER BY COUNT(*) DESC
FETCH FIRST 1 ROW ONLY;
this is just messed up
select CONFIG_TYPE, MAX(COUNTING)
FROM ( select CONFIG_TYPE, COUNT(*) as COUNTING
FROM NOTIFICATION_CONFIG
GROUP BY CONFIG_TYPE
)
WHERE COUNTING = ( select MAX(COUNTING)
FROM ( select COUNT(*) as COUNTING
FROM NOTIFICATION_CONFIG
GROUP BY CONFIG_TYPE
)
)
GROUP BY CONFIG_TYPE
Is there a way to change "LIMIT 1" and get the same output? I have to get client's name, surname and a quantity of books that has the most books
SELECT stud.skaitytojas.name, stud.skaitytojas.surname,
COUNT (stud.skaitytojas.nr) AS quantity
FROM stud.egzempliorius , stud.skaitytojas
WHERE stud.egzempliorius.client = stud.skaitytojas.nr
GROUP BY stud.skaitytojas.nr
ORDER BY quantity DESC
LIMIT 1
Postgres supports the ANSI standard FETCH FIRST 1 ROW ONLY, so you can do:
SELECT s.name, s.surname, COUNT(s.nr) AS quantity
FROM stud.egzempliorius e JOIN
stud.skaitytojas s
ON e.client = s.nr
GROUP BY s.name, s.surname
ORDER BY quantity DESC
FETCH FIRST 1 ROW ONLY;
Also notice the use of table aliases and proper JOIN syntax. I also prefer to list the columns in the SELECT in the GROUP BY, although that is optional if s.nr is unique.
You can select the row with the highest quantity using row_number()
SELECT * FROM (
SELECT * , row_number() over (order by quantity desc) rn FROM (
SELECT stud.skaitytojas.name, stud.skaitytojas.surname, COUNT (stud.skaitytojas.nr) AS quantity
FROM stud.egzempliorius , stud.skaitytojas
WHERE stud.egzempliorius.client = stud.skaitytojas.nr
GROUP BY stud.skaitytojas.name, stud.skaitytojas.surname
) t
) t where rn = 1
If you want to include ties for the highest quantity, then use rank() instead.
In SQL Server 2012 I have a table called Deal_Country. This table contains a field called deal_id, which is not a primary key. The same deal_id value can be in the table multiple times.
What I would like to know is the number that the most common deal_id is in the table.
Please see example below.
deal_id
--------
ABC12
DFG34
DFG34
KNG10
ABC12
PPL11
ABC12
The answer I would like returned is 3 as the most frequent deal_id (ABC12) is shown 3 times.
I have tried the query below however I get
"cannot perform an aggregate function on an expression containing an aggregate
or a subquery."
select max(count(distinct deal_id))
from DEAL_COUNTRY
Use order by and top:
select top 1 deal_id, count(*) as numtimes
from DEAL_COUNTRY
group by deal_id
order by count(*) desc
In MySQL, you would use limit instead of top 1.
MySQL Solution:
count first, sort them from max to min and pick the top 1 record.
select deal_id, count(deal_id) deal_count
from deal_country
order by 2 desc
limit 1
This will handle a situation where there are 2 deals that are both at the top.
WITH cte AS
(
SELECT
deal_id,
count(*) as cnt
FROM DEALS
GROUP BY deal_id
)
,
cte2 AS (
SELECT
deal_id,
RANK() OVER (ORDER BY cnt desc) AS RankNumber
FROM cte
)
SELECT * FROM cte2 WHERE RankNumber = 1;
EDIT: forgot about WITH TIES. eg
SELECT TOP 1 WITH TIES
deal_id,
COUNT(*) AS cnt
FROM DEALS
GROUP BY deal_id
ORDER BY COUNT(*) DESC
I have the following table:
memberid
2
2
3
4
3
...and I want the following result:
memberid count
2 2
3 1 ---Edit by gbn: do you mean 2?
4 1
I was attempting to use:
SELECT MemberID,
COUNT(MemberID)
FROM YourTable
GROUP BY MemberID
...but now I want find which record which has maximum count. IE:
memberid count
2 2
SELECT memberid, COUNT(*) FROM TheTable GROUP BY memberid
Although, it won't work for your desired output because you have "memberid = 3" twice.
Edit: After late update to question...
SELECT TOP 1 WITH TIES --WITH TIES will pick up "joint top".
memberid, COUNT(*)
FROM
TheTable
GROUP BY
memberid
ORDER BY
COUNT(*) DESC
SELECT MemberID, COUNT(MemberID) FROM YourTable GROUP BY MemberID
What if there is a tie (or more) for the max? Do you want to display one or all?
This is how I would do this
SELECT memberid, COUNT(1)
FROM members
GROUP BY memberid
HAVING COUNT(1) = (
SELECT MAX(result.mem_count)
FROM (
SELECT memberid, COUNT(1) as mem_count
FROM members
GROUP BY memberid
) as result
)
I would love to see a more efficient approach though.
Do it like this:
SELECT memberid, COUNT(memberid) AS [count] FROM [Table] GROUP BY memberid
This should do the trick with no subselects required:
select top 1 memberid, COUNT(*) as counted
from members
group by memberid
order by counted desc
Can be done quite easy:
SELECT TOP 1 MemberId, COUNT(*) FROM YourTable GROUP BY MemberId ORDER By 2 DESC
I believe the original poster requested 2 result sets.
The only way I know of to get this (in SQL Server) is to dump the original records into a temp table and then do a SELECT and MAX on that. I do welcome an answer that requires less code!
-- Select records into a temp table
SELECT
Table1.MemberId
,CNT = COUNT(*)
INTO #Temp
FROM YourTable AS Table1
GROUP BY Table1.MemberId
ORDER BY Table1.MemberId
-- Get original records
SELECT * FROM #Temp
-- Get max. count record(s)
SELECT
Table1.MemberId
,Table1.CNT
FROM #Temp AS Table1
INNER JOIN (
SELECT CNT = MAX(CNT)
FROM #Temp
) AS Table2 ON Table2.CNT = Table1.CNT
-- Cleanup
DROP TABLE #Temp
How about this query:
SELECT TOP 1 MemberID,
COUNT(MemberID)
FROM YourTable
GROUP BY MemberID
ORDER by count(MemberID) desc
SELECT count(column_name)
FROM your_table;
You need to use a subselect:
SELECT MemberID, MAX(Count) FROM
(SELECT MemberID, COUNT(MemberID) Count FROM YourTable GROUP BY MemberID)
GROUP BY MemberID
The second group by is needed to return both, the count and the MemberID.