如何查看特定列中是否有多行具有相同的值？

Question

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.

Answer 1

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

Answer 2

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

Answer 3

SELECT * 
FROM #Temp
WHERE SSN NOT IN (SELECT ssn FROM #Temp GROUP BY ssn HAVING COUNT(ssn) > 1)

Answer 4

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