将这两个SQL语句合并为一个

Question

我有两个表t1，t2和name, invoice, total列。

我正在使用此查询来获取T2中而不是T1中的行

select * 
from t2  
where name = (select source from info) 
except 
(select * from t1)

它工作正常并返回几行，我想要的是在一个语句中删除这些返回的行。

我已经尝试过了，但是它不仅删除了查询中返回的行，还删除了T2中的所有行。

delete from t2 
where exists ((select * 
               from t2  
               where name = (select source from info) 
               except 
               (select * from t1) 
             )) 

这里是我的数据示例：

T1

T2

返回的数据（名称为C2，在T2中，而不在T1中存在）

第三个表信息是获取名称，在这种情况下为C2。

提前感谢。

Answer 1

您可以通过左连接t2到t1来做到这一点：

delete t
from (
  select * from t2
  where name = (select source from info)
) t left join t1
on t1.name = t.name and t1.invoice = t.invoice and t1.total = t.total
where t1.name is null

请参见demo。如果要使用NOT EXISTS：

delete t
from (
  select * from t2
  where name = (select source from info)
) t 
where not exists (
  select 1 from t1
  where name = t.name and invoice = t.invoice and total = t.total
)

请参见demo。结果（行留在t2中）：

> name | invoice | total
> :--- | ------: | ----:
> C1   |       1 |   150
> C1   |       2 |   300
> C2   |       1 |   200
> C2   |       2 |   165

Answer 2

我正在使用此查询来获取T2中而不是T1中的行

我将使用的查询是：

select t2.*
from t2
where not exists (select 1 from t1 where t1.? = t2.name);

不清楚t1中匹配列的名称是什么。

很容易变成delete：

delete from t2
where not exists (select 1 from t1 where t1.? = t2.name);

Answer 3

您想通过两个条件从t2中删除：

• 名称=（从信息中选择来源）
• 行在t1中没有匹配项

声明：

delete from t2
where name = (select source from info)
and not exists
(
  select *
  from t1
  where t1.name = t2.name
    and t1.invoice = t2.invoice
    and t1.total = t2.total
);

或更短使用IN子句，仅在允许IN和元组的DBMS中可用：

delete from t2
where name = (select source from info)
and (name, invoice, total) not in (select name, invoice, total from t1);