Question

构造以下查询以向我提供源等于MLP的所有记录。在某些情况下，每个YPID可能有两个或更多记录。我只需要拉出YPID不止一次显示的记录而排除所有其他记录。如何使用下面的查询完成此操作？

这是我的疑问：

select
mr.ypid,
mr.source_record_id,
mr.business_name,
mr.street_address,
mr.city,
mr.state,
mr.zip5
from MDM2.records mr 
join redirects r on r.source_record_id = mr.source_record_id
where mr.source_code = 'MLP'
and mr.status = 'A'
and mr.listing_advertiser_ind = 'A'
and r.is_ad = '1'
and r.source_customer_id is not null
and r.deleted = '0'
order by mr.ypid;

Answer 1

您可以使用cte获取重复记录，并进一步加入查询以获取这些值的数据

with cte
as
(
SELECT mr.ypid
FROM MDM2.records mr
where mr.source_code = 'MLP'
and mr.status = 'A'
and mr.listing_advertiser_ind = 'A'
group by mr.ypid
having count(*) >1
)
select
mr.ypid,
mr.source_record_id,
mr.business_name,
mr.street_address,
mr.city,
mr.state,
mr.zip5
from MDM2.records mr 
join cte on 
mr.ypid = cte.ypid
join redirects r on r.source_record_id = mr.source_record_id
where  r.is_ad = '1'
and r.source_customer_id is not null
and r.deleted = '0'
order by mr.ypid;

Answer 2

试试这个选择 mr.ypid， mr.source_record_id， mr.business_name， mr.street_address， mr.city， mr.state， mr.zip5 来自MDM2.records先生，重定向r，（选择mr2.ypid，count（）cnt 来自MDM2.records mr2，重定向r2 其中r2.source_record_id = mr2.source_record_id 和mr2.source_code =＆＃39; MLP＆＃39; 和mr2.status =＆＃39; A＆＃39; 和mr2.listing_advertiser_ind =＆＃39; A＆＃39; 和r2.is_ad =＆＃39; 1＆＃39; 和r2.source_customer_id不为空和r2.deleted =＆＃39; 0＆＃39; 由mr2.ypid组成有计数（）＆gt; 1）作为重复其中r.source_record_id = mr.source_record_id 和mr.source_code =＆＃39; MLP＆＃39; 和mr.status =＆＃39; A＆＃39; 和mr.listing_advertiser_ind =＆＃39; A＆＃39; 和r.is_ad =＆＃39; 1＆＃39; 并且r.source_customer_id不为空和r.deleted =＆＃39; 0＆＃39; 和dups.ypid = mr.ypid 按mr.ypid排序;

注意：我改变了连接方式，使其看起来更清晰。我们的想法是制作一个虚拟的桌子＆＃34; dups＆＃34;只有重复的ypid

仅显示重复记录 - Oracle Sql

2 个答案: