how to calculate count in sql? - sql

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.

Related

How to group and pick only certain values based on a field using select query SQL

I have a table as follow
ID
ORDERNO
1
123
1
123
2
456
2
456
During every select query done via application using JDBC, only the grouped records based on ORDERNO should be picked.
That means, for example, during first select query only details related to ID = 1, but we cannot specify the ID number in where clause because we do not know how many number of IDs will be there in future. So the query should yield only one set of records; application will delete those records after picking, hence next select query will result in picking other set of records. How to achieve it?
You can use TOP WITH TIES for this
SELECT TOP (1) WITH TIES
t.ID,
t.ORDERNO
FROM YourTable t
ORDER BY
t.ID;
If you want to select and delete at the same time you could delete using an OUTPUT clause
WITH cte AS (
SELECT TOP (1) WITH TIES
t.ID,
t.ORDERNO
FROM YourTable t
ORDER BY
t.ID
)
DELETE cte
OUTPUT deleted.*;
As one option you could select on the MIN(ID) like:
SELECT *
FROM yourtable
WHERE ID = (SELECT MIN(ID) FROM yourtable);
You could also use window functions to do this:
SELECT ID, ORDERNO
FROM
(
SELECT ID, ORDERNO
DENSE_RANK() OVER (ORDER BY ID ASC) AS dr
FROM yourtable
)dt
WHERE dr = 1;
order your rows and select top n number of rows that you want :
select top (1) with ties ID, ORDERNO
from tablename
order by ID asc

Get all items with min values SQL Server

Here's what the table is like:
----------------------------------
EmployeeId Tasks_Count
1 1
2 1
3 2
4 1
5 3
I need a query to get all employees with min tasks count. Result should be like this:
---------------
EmployeeId
1
2
4
The problem is that i using a subquery to count tasks. Here's my code
SELECT *
FROM (SELECT EmployeeId,
COUNT(*) AS Tasks_count
FROM Tasks
INNER JOIN Status ON Tasks.StatusId=Status.Id
WHERE Status.Name != 'Closed'
GROUP BY EmployeeId
ORDER BY Tasks_count DESC) AS Employee_not_closed
WHERE Tasks_count IN (SELECT MIN(Tasks_count)
FROM Employee_not_closed)
Use FETCH FIRST WITH TIES:
select EmployeeId
from tablename
order by Tasks_Count
fetch first 1 row with ties
You can try below -
select * from tablename
where Tasks_Count in (select min(Tasks_Count) from tablename)
It can also be done using RANK() function like following.
;with cte as
(
select Employeeid, rank() over( order by Tasks_Count) rn
from #table
)
select * from cte where rn=1
You Can use the below code i have tested the code and its working fine.
select EmployeeId from StackOverFlow_3 where Tasks_Count in(select min(Tasks_Count) from StackOverFlow_3)
You can use a join on subquery
select m.EmployeeId
from my_table m
inner join
(
select min(task_count) min_task
from my_table
) t on t.min_task = m.task_count

Using DISTINCT and TOP in the same query

I want to use DISTINCT and TOP in the same query. I tried
SELECT DISTINCT TOP 10 * FROM TableA
but I still have a duplicate personID, so I tought to do:
SELECT DISTINCT (personID) TOP 10 * FROM TableA
But here the syntax is wrong. How do I do it correctly?
You're using a SELECT * which is pulling in all records. If you want to use a true DISTINCT only list out the column you want to receive distinct values of. If you have multiple columns then all those columns combined make up one distinct record.
SELECT distinct TOP 10 personID
FROM TableA
Note that without an ORDER BY this will return the first 10 records in no particular order. The results could be different each time you run the query.
You seem to want 10 random records for different persons. Try this:
select t.*
from (select t.*,
row_number() over (partition by personid order by (select NULL)) as seqnum
from t
) t
where seqnum = 1
In general, though, when using top you should also be using an order by to specify what you mean by "top".
It works simply if you use query like this:
SELECT DISTINCT TOP 2 name FROM [ATTENDANCE] ;
In the above query, name is the column_name and [ATTENDANCE] is the table_name.
You can also use WHERE with this to make filtering conditions.
select distinct personid from tablea sample 10
This works in teradata
Maybe you can do like this:
SELECT TOP 10 t.* FROM (SELECT distinct personID FROM TableA) t
Few options:
1:
SELECT TOP 10 personID
FROM (SELECT distinct personID FROM TableA)
2:
;with cte as
(SELECT distinct personID FROM TableA)
SELECT top 10 personID FROM cte
3:
;with cte as
(SELECT personID,row_count=row_number() over (partition by personID order by personID)
FROM TableA
) SELECT top 10 personID FROM cte where row_count=1
I see this is an old question, but I can suggest a more elegant solution:
SELECT personDetails.*
FROM
(
SELECT TOP (10)
personID
personID
FROM TableA
GROUP BY personID --> use this instead of distinct
) AS distinctIds
OUTER APPLY
(
SELECT TOP (1) * --> fetch only one result matching personID
FROM TableA
WHERE TableA.personId = distinctIds.personID
) AS personDetails
You can use GROUP BY instead of DISTINCT, and you can use OUTER APPLY with TOP(1) to fetch only one matching result.
The Matt Busche answer is correct based on the post title.
But if, as it seems from your code examples, your problem is to have all the record fields from the distinct top 10 personID, you can use that as a subquery:
SELECT TableA.*
FROM TableA
INNER JOIN (SELECT DISTINCT TOP 10 personID FROM TableA) AS B
ON TableA.ID = B.ID
SELECT DISTINCT ta.personid FROM (SELECT TOP 10 * FROM TableA) ta
ta is object of subquery and using ta object we can distinct the values
i fixed it i did
select distinct personid from (SELECT TOP 10 * FROM TableA)
If the goal is to select the top 1 record of every personID, then use
select * from TableA group by personid
Since you are doing a "group by", it will return every column, but will ignore (not display) any additional rows with the same personId

SELECT MAX of COUNT

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

SQL Group by & Max

I have a following data in a table:
id name alarmId alarmUnit alarmLevel
1 test voltage psu warning
2 test voltage psu ceasing
3 test voltage psu warning
4 test temp rcc warning
5 test temp rcc ceasing
I'd like to show only the most recent information about every colums group (alarmId,alarmUnit), so the result should look like this:
3 test voltage psu warning
5 test temp rcc ceasing
I've tried so far:
SELECT MAX(id) as id,name,alarmId,alarmUnit,alarmLevel GROUP BY alarmId,alarmUnit;
Selected IDs seem to be fine but selected rows aren't corresponding to them. Could you help me?
In Oracle, SQL Server 2005+ and PostgreSQL 8.4:
SELECT *
FROM (
SELECT *, ROW_NUMBER() OVER (PARTITION BY alarmId, alarmUnit ORDER BY id DESC) AS rn
FROM mytable
) q
WHERE rn = 1
In MySQL:
SELECT mi.*
FROM (
SELECT alarmId, alarmUnit, MAX(id) AS mid
FROM mytable
GROUP BY
alarmId, alarmUnit
) mo
JOIN mytable mi
ON mi.id = mo.mid
In PostgreSQL 8.3 and below:
SELECT DISTINCT ON (alarmId, alarmUnit) *
FROM mytable
ORDER BY
alarmId, alarmUnit, id DESC
If you want to get the row of the max, you'll probably need a sub-query. Something like:
SELECT *
FROM YourTable
WHERE id IN (
SELECT MAX(id) FROM YourTable GROUP BY alarmId, alarmUnit
)
Try:
SELECT * FROM table WHERE id IN
(SELECT MAX(id) FROM table GROUP BY alarmId, alarmUnit)
Maybe try something like the following:
SELECT id,name,alarmId,alarmUnit,alarmLevel
FROM table
WHERE id IN (SELECT Max(id) FROM table GROUP BY alarmId, alarmUnit)
You may have to include alarmId and alarmUnit in the sub query select.
select id, name, alarmID, alarmUnit, alarmLevel
from (select max(id) as id
from table
group by alarmID, alarmUnit) maxID
inner join table
on table.id = maxID.id