すべての異常を一覧表示するには:
SELECT name, count(*) FROM TableA GROUP BY name HAVING count(*) > 1;
重複の削除に取り組む方法はいくつかあり、パスは重複の数に大きく依存します。
これ を参照してください テーブルからそれらを削除する方法について質問してください。
そこで提供したソリューションは次のとおりです。
-- Setup for example
create table people (fname varchar(10), lname varchar(10));
insert into people values ('Bob', 'Newhart');
insert into people values ('Bob', 'Newhart');
insert into people values ('Bill', 'Cosby');
insert into people values ('Jim', 'Gaffigan');
insert into people values ('Jim', 'Gaffigan');
insert into people values ('Adam', 'Sandler');
-- Show table with duplicates
select * from people;
-- Create table with one version of each duplicate record
create table dups as
select distinct fname, lname, count(*)
from people group by fname, lname
having count(*) > 1;
-- Delete all matching duplicate records
delete people from people inner join dups
on people.fname = dups.fname AND
people.lname = dups.lname;
-- Insert single record of each dup back into table
insert into people select fname, lname from dups;
-- Show Fixed table
select * from people;