1、如图，绑定了所有id的有A,B两个name

2、第一种Sql及效率

explain SELECT name,count(id) as count from test GROUP BY name 
HAVING count(id)=(SELECT count(DISTINCT id) from test);

3、第二种sql及效率

explain select * from (SELECT name,count(id) as count from test GROUP BY name )t1 
WHERE t1.count=(SELECT count(DISTINCT id) from test);