Table 1
Table 2
I need to find the Count of total number of unique stores that have "Achieved Date" not Null that achieved all of the "Achievement Ids" "enabled" on Table 2.
So far I can find the count of stores that achieved a hard coded number, but I'm not breaking through the part where I use the Count of Enabled Ids on table 2 to define what the number is.
SELECT
COUNT(*) AS count
FROM
(SELECT
StoreNumber, COUNT(*) as Achievements
FROM
StoreAchievementProgress
WHERE
AchievedDate IS NOT NULL
GROUP BY
StoreNumber) count
maybe this query
SELECT S.StoreNumber
FROM StoreAchievementProgress S
RIGHT JOIN (SELECT Id FROM Table2 WHERE Enabled=1 )T
ON T.Id=S.AchievementId
AND AchievedDate IS NOT NULL
GROUP BY S.StoreNumber
HAVING COUNT(1) = (SELECT COUNT(Id) FROM Table2 WHERE Enabled=1 )
Joining the stores with a count of their enabled achievements to how many they can get
SELECT COUNT(*) AS StoresFullAchievements
FROM
(
SELECT p.StoreNumber, COUNT(*) AS TotalEnabledAchievements
FROM StoreAchievementProgress p
JOIN Achievements a ON a.id = p.AchievementId
WHERE p.AchievedDate IS NOT NULL
AND a.Enabled = 1
GROUP BY p.StoreNumber
) AS s
JOIN
(
SELECT COUNT(*) AS TotalEnabled
FROM Achievements
WHERE Enabled = 1
) a
ON a.TotalEnabled = s.TotalEnabledAchievements
Related
I have a table in postgres with 2 fields: they are columns of ids of users who have looked at some data, under two conditions:
viewee viewer
------ ------
93024 66994
93156 93151
93163 113671
137340 93161
92992 93161
93161 93135
93156 93024
And I want to group them by both viewee and viewer field, and count the number of occurrences, and return that count
from high to low:
id count
------ -----
93161 3
93156 2
93024 2
137340 1
66994 1
92992 1
93135 1
93151 1
93163 1
I have been running two queries, one for each column, and then combining the results in my JavaScript application code. My query for one field is...
SELECT "viewer",
COUNT("viewer")
FROM "public"."friend_currentfriend"
GROUP BY "viewer"
ORDER BY count DESC;
How would I rewrite this query to handle both fields at once?
You can combine to columns from the table into a single one by using union all then use group by as below:
select id ,count(*) Count from (
select viewee id from vv
union all
select viewer id from vv) t
group by id
order by count(*) desc
Results:
This is a good place to use a lateral join:
select v.viewx, count(*)
from t cross join lateral
(values (t.viewee), (t.viewer)) v(viewx)
group by v.viewx
order by count(*) desc;
You can try this :
SELECT a.ID,
SUM(a.Total) as Total
FROM (SELECT t.Viewee AS ID,
COUNT(t.Viewee) AS Total
FROM #Temp t
GROUP BY t.Viewee
UNION
SELECT t.Viewer AS ID,
COUNT(t.Viewer) AS Total
FROM #Temp t
GROUP BY t.Viewer
) a
GROUP BY a.ID
ORDER BY SUM(a.Total) DESC
Im new to DB2 , and tried based on some similar posts, I have a table where I need to find the count of IDs based on where status=P and
the count of(primary=1) more than once.
so my result should be 2 here - (9876,3456)
Tried:
SELECT id, COUNT(isprimary) Counts
FROM table
GROUP BY id
HAVING COUNT(isprimary)=1;
Try the query below:
select ID as IDs,Count(isPrimary) as isPrimary
From Table
where Status = 'p'
Group by ID
Having Count(isPrimary) >1
You are close, I think all you need to do is to add a where clause like:
SELECT id, COUNT(*) as Counted
FROM table
WHERE PrimaryFlag = 1
AND[status] = 'P'
GROUP BY id
EDIT: if you need to count only the distinct IDs, then try:
SELECT COUNT(t.ID) FROM
(
SELECT id, COUNT(*) as Counted
FROM table
WHERE PrimaryFlag = 1
AND[status] = 'P'
GROUP BY id
) as t
I have two tables: groups(group_id), member(group_id, name) and I would like to select all the members in groups where everyone in the group has a non-null name. For example, if this is the members table:
group_id|name
1|a
1|b
2|c
2|null
3|null
3|null
then the result of the query should return:
group_id|name
1|a
1|b
I tried running
SELECT * FROM members M1
WHERE ALL(SELECT M2.name IS NOT NULL FROM members M2)
ORDER BY M1.group_id
but it didn't work.
Use bool_and():
select group_id, name
from members
where group_id in (
select group_id
from members
group by 1
having bool_and(name is not null)
);
SELECT
*
FROM groups g
INNER JOIN members m
ON g.group_id = m.group_id
WHERE NOT EXISTS (SELECT * FROM members mbr WHERE mbr.name IS NULL AND mbr.group_id = m.group_id)
Essentially, we select all records, except for those where we can find a null name record with the same group ID.
Note that I don't believe this is SARG-able, so if you have a massive database that relies on indexes, this may be a bit on the slow side.
If you just need to group and display non null value, how if just a simple
SELECT group_id, name FROM members
group by group_id, name
having name is not null and id = 1;
Another solution:
SELECT * FROM Table
WHERE group_id NOT IN (
SELECT group_id FROM table
WHERE name IS NULL
)
The following query returns the results that I need but I have to add the ID of the row to then update it. If I add the ID directly in the select statement it will return me more results then I need because each ID is unique so the DISTINCT statement see the line as unique.
SELECT DISTINCT ucpse.MemberID, ucpse.ProductID, ucpse.UserID
FROM UserCustomerProductSalaryExceptions as ucpse
WHERE EXISTS (SELECT NULL
FROM UserCustomerProductSalaryExceptions as upcse2
WHERE ucpse.userid = upcse2.userid AND ucpse.MemberID = upcse2.MemberID AND ucpse.ProductID = upcse2.ProductID
GROUP BY upcse2.UserID, upcse2.memberid, upcse2.productid
HAVING COUNT(UserID) >= 2
)
So basically I need to add ucpse.ID in the Select statement while keeping DISTINCT values for MemberID,ProductID and UserID.
Any Ideas ?
Thank you
According to you comment:
If the data has been duplicated 67 times for a given employee with a given product and a given client, I need to keep only one of thoses records. It's not important which one, so this is why I use DISTINC to obtain unique combinaison of given employee with a given product and a given client.
You can use MIN() or MAX() and GROUP BY instead of DISTINCT
SELECT MAX(ucpse.ID) AS ID, ucpse.MemberID, ucpse.ProductID, ucpse.UserID
FROM UserCustomerProductSalaryExceptions as ucpse
WHERE EXISTS (SELECT NULL
FROM UserCustomerProductSalaryExceptions as upcse2
WHERE ucpse.userid = upcse2.userid AND ucpse.MemberID = upcse2.MemberID AND ucpse.ProductID = upcse2.ProductID
GROUP BY upcse2.UserID, upcse2.memberid, upcse2.productid
HAVING COUNT(UserID) >= 2
)
GROUP BY ucpse.MemberID, ucpse.ProductID, ucpse.UserID
UPDATE:
From you comments I think the below query is what you need
DELETE FROM UserCustomerProductSalaryExceptions
WHERE ID NOT IN ( SELECT MAX(ucpse.ID) AS ID
FROM #UserCustomerProductSalaryExceptions
GROUP BY ucpse.MemberID, ucpse.ProductID, ucpse.UserID
HAVING COUNT(ucpse.ID) >= 2
)
If all you want is to delete the duplicates, this will do it:
WITH X AS
(SELECT ID,
ROW_NUMBER() OVER (PARTITION BY MemberID, ProductID, UserID ORDER BY ID) AS DupRowNum<br
FROM UserCustomerProductSalaryExceptions
)
DELETE X WHERE DupRowNum > 1
ID's not necessary - try:
UPDATE uu SET
<your settings here>
FROM UserCustomerProductSalaryExceptions uu
JOIN ( <paste your entire query above here>
) uc ON uc.MemberID=uu.MemberId AND uc.ProductID=uu.ProductId AND uc.UserID=uu.UserId
From the sound of your data structure (which I would STRONGLY advise normalizing as soon as possible), it sounds like you should be updating all the records. It sounds as if each duplicate is important because it contains some information about an employee's relation to a customer or product.
I would probably update all the records. Try this:
UPDATE UCPSE
SET
--Do your updates here
FROM UserCustomerProductSalaryExceptions as ucpse
JOIN
(
SELECT UserID, MemberID, ProductID
FROM UserCustomerProductSalaryExceptions
GROUP BY UserID, MemberID, ProductID
HAVING COUNT(UserID) >= 2
) T
ON ucpse.UserID = T.UserID AND ucpse.MemberID = T.MemberID AND ucpse.ProductID = T.ProductID
I'm looking for an efficient way to exclude rows from my SELECT statement WHERE more than one row is returned with an identical value for a certain column.
Specifically, I am selecting a bunch of accounts, but need to exclude accounts where more than one is found with the same SSN associated.
this will return all SSNs with exactly 1 row
select ssn,count(*)
from SomeTable
group by ssn
having count(*) = 1
this will return all SSNs with more than 1 row
select ssn,count(*)
from SomeTable
group by ssn
having count(*) > 1
Your full query would be like this (will work on SQL Server 7 and up)
select a.* from account a
join(
select ssn
from SomeTable
group by ssn
having count(*) = 1) s on a.ssn = s.ssn
For SQL 2005 or above you can try this:
WITH qry AS
(
SELECT a.*,
COUNT(*) OVER(PARTITION BY ssn) dup_count
FROM accounts a
)
SELECT *
FROM qry
WHERE dup_count = 1
For SQL 2000 and 7:
SELECT a.*
FROM accounts a INNER JOIN
(
SELECT ssn
FROM accounts b
GROUP BY ssn
HAVING COUNT(1) = 1
) b ON a.ssn = b.ssn
SELECT *
FROM #Temp
WHERE SSN NOT IN (SELECT ssn FROM #Temp GROUP BY ssn HAVING COUNT(ssn) > 1)
Thank you all for your detailed suggestions. When it was all said and done, I needed to use a correlated subquery. Essentially, this is what I had to do:
SELECT acn, ssn, [date] FROM Account a
WHERE NOT EXISTS (SELECT 1 FROM Account WHERE ssn = a.ssn AND [date] < a.[date])
Hope this helps someone.
I never updated this... In my final submission, I achieved this through a left join to increase efficiency (the correlated subquery was not acceptable as it took a significant amount of time to run, checking each record against over 150K others).
Here is what had to be done to solve my problem:
SELECT acn, ssn
FROM Account a
LEFT JOIN (SELECT ssn, COUNT(1) AS counter FROM Account
GROUP BY ssn) AS counters
ON a.ssn = counters.ssn
WHERE counter IS NULL OR counter = 0