这个来源很好,但太慢了。功能:如果 SC 和 %%5 和 2013.07.11 < date < 2013.07.18 和一些较旧的行代表行,则选择所有行方法:查找 X 计数行。一一看28天是否有一致性
select efi_name, efi_id, count(*) as dupes, id, mlap_date
from address m
where
mlap_date > "2013.07.11"
and mlap_date < "2013.07.18"
and mlap_type = "SC"
and calendar_id not like "%%5"
and concat(efi_id,irsz,ucase(city), ucase(address)) in (
select concat(k.efi_id,k.irsz,ucase(k.city), ucase(k.address)) as dupe
from address k
where k.mlap_date > adddate(m.`mlap_date`,-28)
and k.mlap_date < m.mlap_date
and k.mlap_type = "SC"
and k.calendar_id not like "%%5"
and k.status = 'Befejezett'
group by concat(k.efi_id,k.irsz,ucase(k.city), ucase(k.address))
having (count(*) > 1)
)
group by concat(efi_id,irsz,ucase(city), ucase(address))
感谢您的帮助!