CODEWITHSUNDEEP
t-sqlhaving
salvationishere
salvationishere

Reputation: 3511

TSQL invalid HAVING count

I am using SSMS 2008 and trying to use a HAVING statement. This should be a real simple query. However, I am only getting one record returned event though there are numerous duplicates.

Am I doing something wrong with the HAVING statement here? Or is there some other function that I could use instead?

select
      address_desc,
      people_id
from 
      dbo.address_view
where people_id is not NULL
group by people_id , address_desc
having count(*) > 1

sample data from address_view:

people_id                            address_desc
----------                           ------------
Murfreesboro, TN  37130          F15D1135-9947-4F66-B778-00E43EC44B9E
11 Mohawk Rd., Burlington, MA 01803 C561918F-C2E9-4507-BD7C-00FB688D2D6E
Unknown, UN  00000                    C561918F-C2E9-4507-BD7C-00FB688D2D6E
Jacksonville, NC  28546          FC7C78CD-8AEA-4C8E-B93D-010BF8E4176D
Memphis, TN  38133                    8ED8C601-5D35-4EB7-9217-012905D6E9F1
44 Maverick St., Fitchburg, MA  8ED8C601-5D35-4EB7-9217-012905D6E9F1

Upvotes: 1

Views: 499

Answers (2)

ListenSoftware Louise Ai Agent
ListenSoftware Louise Ai Agent

Reputation: 4243

using row_number and partition you can find the duplicate occurrences where row_num>1

select address_desc,
people_id,
row_num
from
(
select
  address_desc,
  people_id,
  row_number() over (partition by address_desc order by address_desc) row_num
from 
  dbo.address_view
where people_id is not NULL

) x where row_num>1

Upvotes: 0

Joe Stefanelli
Joe Stefanelli

Reputation: 135818

The GROUP BY is going to lump your duplicates together into a single row.

I think instead, you want to find all people_id values with duplicate address_desc:

SELECT a.address_desc, a.people_id
    FROM dbo.address_view a
        INNER JOIN (SELECT address_desc
                        FROM dbo.address_view
                        GROUP BY address_desc
                        HAVING COUNT(*) > 1) t
            ON a.address_desc = t.address_desc

Upvotes: 2

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