I have an SQL query which looks at date-of-birth, last name and a soundex of first name to identify duplicates. The following query finds some 8,000 rows (which I assume means there are around 8,000 duplicate records).
select dob,last_name,soundex(first_name),count(*)
from clients
group by dob,last_name,soundex(first_name)
having count(*) >1
Almost all of the results have a count of 2, a few have a count of 3 where obviously the record existed twice in one of the two databases which were merged.
The next step I need to take is to mark one of the rows, doesn't really matter, with a duplicate flag and to mark each row with the opposite rows key. Is there a way of doing this using SQL?
Here is a query that will give you not only the duplicates, but also the first id inserted (assuming Id is the sequential primary-key column) and the newest id.
OTTOMH
select dob, last_name, soundex(first_name) firstnamesoundex, min (Id) OldestId, max (Id) NewestId, Count (*) NumRows
from clients
group by dob,last_name,soundex(first_name)
having count(*) >1
You can use this in a JOIN to do your update
UPDATE Clients
SET OppositeRowId = DuplicateRows.NewestId
FROM
(
select dob, last_name, soundex(first_name) firstnamesoundex, min (Id) OldestId, max (Id) NewestId, Count (*) NumRows
from clients
group by dob,last_name,soundex(first_name)
having count(*) >1
) DuplicateRows
WHERE
DuplicateRows.OldestId = Clients.Id
All of this assumes that you have one duplicate . If you have more than one, you are going to have to try something different.
好吧,您可以使用SELECT DISTINCT,然后将单行标记为“ notplicate”-然后搜索“ notplicate”的行以找到重复的行。
This should do what you are after, the UPDATE in one go.
UPDATE FROM clients c
INNER JOIN
(
select dob,last_name,soundex(first_name),MIN(id) as keep
from clients
group by dob,last_name,soundex(first_name)
having count(*) >1
) k
ON c.dob=k.dob AND c.last_name=k.last_name AND soundex(c.first_name)=soundex(k.first_name)
SET duplicateid = NULLIF(k.keep, c.id),
hasduplicate = (k.keep = c.id)
It assumes you have 3 columns not stated in the question
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