I dont think a count will work here, can someone help me get an sql that identifies which account numbers have multiple agents, more than two agents in the where condition.
AGENT_NAME ACCOUNT_NUMBER
Clemons, Tony 123
Cipollo, Michael 123
Jepsen, Sarah 567
Joanos, James 567
McMahon, Brian 890
Novak, Jason 437
Ralph, Melissa 197
Reitwiesner, John 221
Roman, Marlo 123
Rosenzweig, Marcie 890
Results should be something like this.
ACCOUNT_NUMBER AGENT_NAME
123 Cipollo, Michael
123 Roman, Marlo
123 Clemons, Tony
890 Rosenzweig, Marcie
890 McMahon, Brian
567 Joanos, James
567 Jepsen, Sarah
You can do this using window functions:
select t.account_number, t.agent_name
from (select t.*, min(agent_name) over (partition by account_number) as minan,
max(agent_name) over (partition by account_number) as maxan
from table t
) t
where minan <> maxan;
If you know the agent names are never duplicated, you could just do:
select t.account_number, t.agent_name
from (select t.*, count(*) over (partition by account_number) as cnt
from table t
) t
where cnt > 1;
Assuming your table name is test, this should pull all the records with duplicate ACCOUNT_NUMBER:
select * from test where ACCOUNT_NUMBER in
(select ACCOUNT_NUMBER from test
group by ACCOUNT_NUMBER having
count(ACCOUNT_NUMBER)>1)
order by ACCOUNT_NUMBER
Using count function u can get the result
CREATE TABLE #TEMP
(
AGENT_NAME VARCHAR(100),
ACCOUNT_NUMBER INT
)
INSERT INTO #TEMP
VALUES ('CLEMONS, TONY',123),
('CIPOLLO, MICHAEL',123),
('JEPSEN, SARAH',567),
('JOANOS, JAMES',567),
('MCMAHON, BRIAN',890),
('NOVAK, JASON',437),
('RALPH, MELISSA',197),
('REITWIESNER, JOHN',221),
('ROMAN, MARLO',123),
('ROSENZWEIG, MARCIE',890)
SELECT a.ACCOUNT_NUMBER,a.AGENT_NAME
FROM #TEMP A
JOIN(SELECT COUNT(1) CNT,
ACCOUNT_NUMBER
FROM #TEMP
GROUP BY ACCOUNT_NUMBER) B
ON A.ACCOUNT_NUMBER = B.ACCOUNT_NUMBER
WHERE B.CNT != 1
Related
I have a table Applications
The user can submit more than one application. The user can also update an existing application, but instead of updating the record itself, we will insert a new record with the same ApplicationNumber
Id ApplicationNum ApplicantId ApplicantName CreateDate
1 101 789 John May-20-2021
2 101 789 John May-21-2021
3 102 789 John May-22-2021
4 103 123 Maria May-31-2021
I want to return the list of applications based on the ApplicantId, but I don’t want to display both records of the same ApplicationNumber
If I use this select statement
Select * from Applications where ApplicantId = 789
This is the result I currently get
1 101 789 John May-20-2021
2 101 789 John May-21-2021
3 102 789 John May-22-2021
This is the result I want to get
2 101 789 John May-21-2021
3 102 789 John May-22-2021
Notice that record Id = 1 is not displayed because it is an old version of record Id = 2
How can I achieve this?
I like using ROW_NUMBER along with a TIES trick here:
SELECT TOP 1 WITH TIES *
FROM Applications
WHERE ApplicantId = 789
ORDER BY ROW_NUMBER() OVER (PARTITION BY ApplicantId, ApplicationNum ORDER BY Id DESC);
Might be easier to just use:
select max(Id) as Id, ApplicationNum, ApplicantId, ApplicantName, max(CreateDate) as CreateDate
from Applications
where ApplicantId = 789
group by ApplicationNum, ApplicantId, ApplicantName
The traditional way which is usually the most performant is to use row_number and select the desired row from each group of Applicants
select Id, ApplicationNum, ApplicantId, ApplicantName, CreateDate
from (
select *, Row_Number() over(partition by ApplicantId, ApplicationNum order by Id desc) rn
from Applications
where ApplicantId=789
)a
where rn=1
I have the following table:
dbo.split
Name Time
Alex 120
John 80
John 300
Mary 500
Bob 900
And then another table dbo.travel
Name Time
Alex 150
Alex 160
Alex 170
John 90
John 100
John 310
Mary 550
Mary 600
Mary 499
Bob 800
Bob 700
For each value in table split I need to find the next value in table travel. I tried to do it with CTE a with ROW_NUMBER() to get next by group, but there's no way I can group by correct value, since dbo.split can containt multiple values for the same name.
I'm looking for the following output:
Name Time TravelTime
Alex 120 150
John 80 90
John 300 310
Mary 500 550
Bob 900 NULL
Here's what I have so far but it fails because split table can have multiple records per person:
;with result as (
select t.*,
ROW_NUMBER() OVER (Partition BY t.Name order by t.Time) as rn
from travel t join split s
on t.Name = s.Name and t.TIME>s.Time
)
I would use apply:
select s.*, t.time
from split s outer apply
(select top (1) t.*
from travel t
where t.name = s.name and t.time > s.time
order by t.time asc
) t;
In this case, apply is doing essentially the same thing as a correlated subquery, so you could phrase it that way as well.
You can try as below
Select * from(Select
Name,t.time,t1.time,
Row_number() over (partition by
Name,t.time order by t1.time) rn
from split t
Join travel t1 on t.time <t1.time and
t.name =t1.name)
where
rn=1;
I have 36 columns in a table but one of the columns have data multiple times like below
ID Name Ref
abcd john doe 123
1234 martina 100
123x brittany 123
ab12 joe 101
and i want results like
ID Name Ref cnt
abcd john doe 123 2
1234 martina 100 1
123x brittany 123 2
ab12 joe 101 1
as 123 has appeared twice i want it to show 2 in cnt column and so on
select ID, Name, Ref, (select count(ID) from [table] where Ref = A.Ref)
from [table] A
Edit:
As mentioned in comments below, this approach may not be the most efficient in all cases, but should be sufficient on reasonably small tables.
In my testing:
a table of 5,460 records and 976 distinct 'Ref' values returned in less than 1 second.
a table of 600,831 records and 8,335 distinct 'Ref' values returned in 6 seconds.
a table of 845,218 records and 15,147 distinct 'Ref' values returned in 13 seconds.
You should provide SQL brand to know capabilities:
1) If your DB supports window functions:
Select
*,
count(*) over ( partition by ref ) as cnt
from your_table
2) If not:
Select
T.*, G.cnt
from
( select * from your_table ) T inner join
( select count(*) as cnt from your_table group by ref ) G
on T.ref = G.ref
You can use COUNT with OVERin following:
QUERY
select ID,
Name,
ref,
count(ref) over (partition by ref) cnt
from #t t
SAMPLE DATA
create table #t
(
ID NVARCHAR(400),
Name NVARCHAR(400),
Ref INT
)
insert into #t values
('abcd','john doe', 123),
('1234','martina', 100),
('123x','brittany', 123),
('ab12','joe', 101)
I have a table which fills up with lots of transactions monthly, like below.
Name ID Date OtherColumn
_________________________________________________
John Smith 11111 2012-11-29 Somevalue
John Smith 11111 2012-11-30 Somevalue
Adam Gray 22222 2012-12-11 Somevalue
Tim Blue 33333 2012-12-15 Somevalue
John NewName 11111 2013-01-01 Somevalue
Adam Gray 22222 2013-01-02 Somevalue
From this table i want to create a dimension table with the unique names and id's. The problem is that a person can change his/her name, like "John" in the example above. The Id's are otherwise always unique. In those cases I want to only use the newest name (the one with the latest date).
So that I end up with a table like this:
Name ID
______________________
John NewName 11111
Adam Gray 22222
Tim Blue 33333
How do I go about achieving this?
Can I do it in a single query?
Use a CTE for this. It simplifies ranking and window functions.
;WITH CTE as
(SELECT
RN = ROW_NUMBER() OVER (PARTITION BY ID ORDER BY [Date] DESC),
ID,
Name
FROM
YourTable)
SELECT
Name,
ID
FROM
CTE
WHERE
RN = 1
I think creating a table is a bad idea, but this is how you get the most recent name.
select name
from yourtable yt join
(select id, max(date) maxdate
from yourtable
group by id ) temp on temp.id = yt.id and yt.date = maxdate
JNK's CTE solution is an equivalent of the following.
SELECT
Name,
ID
FROM (
SELECT
RN = ROW_NUMBER() OVER (PARTITION BY ID ORDER BY [Date] DESC),
Name,
ID
FROM theTable
)
WHERE RN = 1
Trying to think a way to get rid of the partition function without introducing the possible duplicates.
If I have the table "members" (shown below), how would I go about getting the record of the first occurrence of a membership_id (Oracle).
Expected results
123 John Doe A P
313 Michael Casey A A
113 Luke Skywalker A P
Table - members
membership_id first_name last_name status type
123 John Doe A P
313 Michael Casey A A
113 Luke Skywalker A P
123 Bob Dole A A
313 Lucas Smith A A
SELECT membership_id,
first_name,
last_name,
status,
type
FROM( SELECT membership_id,
first_name,
last_name,
status,
type,
rank() over (partition by membership_id
order by type desc) rnk
FROM members )
WHERE rnk = 1
will work for your sample data set. If you can have ties-- that is, multiple rows with the same membership_id and the same maximum type-- this query will return all those rows. If you only want to return one of the rows where there is a tie, you would either need to add additional criteria to the order by to ensure that all ties are broken or you would need to use the row_number function rather than rank which will arbitrarily break ties.
Select A.*
FROM Members AS A inner join
(Select membership_id, first(first_name) AS FN, first(last_name) AS LN
From Members
Group by membership_id) AS B
ON A.membership_id=B.membership_id and A.first_name=B.FN and A.last_name=B.LN
Hope that helps!
select *
from members
where rowid in (
select min(rowid)
from members
group by membership_id
)