SQL - 从一个表中查找存在于另一个表中的记录，如果不存在则删除

Question

我有以下四个 SQL 表：

Table 1:
-----------------------
Product | Date_Purchase
-----------------------
abc     | 06-Jan-19
def     | 05-Jan-18
ghi     | 05-Apr-19
abc     | 06-Feb-19

Table 2:

------------------------
Product | Date_Purchase
------------------------
jkl    | 6-Feb-19
mno    | 2-Aug-18
ghi    | 9-May-19
pqr    | 1-Sep-19

Table 3:

-------------------------
Product | Date_Purchase
-------------------------
ghi    | 2-Aug-18
mno    | 9-May-19
pqr    | 2-Aug-18
abc    | 06-Jan-19

Table 4:

-------------------------
Product | Date_Purchase
-------------------------
stu    | 9-May-19
vwx    | 05-Apr-19
ghi    | 9-May-19
def    | 05-Jan-18

我有以下代码将表格与 Union 连接起来：

SELECT Product, Date_Purchase FROM Table1 UNION ALL
SELECT Product, Date_Purchase FROM Table2 UNION ALL
SELECT Product, Date_Purchase FROM Table3 UNION ALL SELECT Product, Date_Purchase FROM Table4
ORDER BY Product, Date_Purchase;

我想从表格中删除所有在所有表格中只出现一次的行，无论是哪个表格。

例如 jkl、stu 和 vwx 只出现一次，所以我想从它们出现的表中删除整行。有谁知道该怎么做？ 另外，如何删除表格中出现且购买日期相同的所有产品？

Answer 1

试一试 Scratte 的版本，仅当产品和日期出现两次时（未选中，因为写在移动设备上）：

SELECT pdo.*
FROM (SELECT pd.*, COUNT(*) OVER (PARTITION BY Product) as cnt
      FROM ((SELECT Product, Date_Purchase FROM Table1
            ) UNION ALL
            (SELECT Product, Date_Purchase FROM Table2
            ) UNION ALL
            (SELECT Product, Date_Purchase FROM Table3
            ) UNION ALL
            (SELECT Product, Date_Purchase FROM Table4
            )
           ) pd
     ) pdo
Group by pdo.Product,pdo.Date_Purchase
Having cnt=1

Answer 2

MySql 的解决方案，您可以在 1 条语句中从所有 4 个表中删除：

delete t1, t2, t3, t4
from (
  select u.product, count(*) counter 
  from (
    select * from table1 union all
    select * from table2 union all
    select * from table3 union all
    select * from table4
  ) u  
  group by u.product
) t 
left join table1 t1 on t1.product = t.product
left join table2 t2 on t2.product = t.product
left join table3 t3 on t3.product = t.product
left join table4 t4 on t4.product = t.product
where t.counter = 1; 

请参阅demo。
结果：

表1

> Product | Date_Purchase
> :------ | :------------
> abc     | 06-Jan-19    
> def     | 05-Jan-18    
> ghi     | 05-Apr-19    
> abc     | 06-Feb-19  

表2

> Product | Date_Purchase
> :------ | :------------
> mno     | 2-Aug-18     
> ghi     | 9-May-19     
> pqr     | 1-Sep-19 

表3

> Product | Date_Purchase
> :------ | :------------
> ghi     | 2-Aug-18     
> mno     | 9-May-19     
> pqr     | 2-Aug-18     
> abc     | 06-Jan-19 

表4

> Product | Date_Purchase
> :------ | :------------
> ghi     | 9-May-19     
> def     | 05-Jan-18 

Answer 3

如果“删除”意味着不返回，则将它们返回到select，那么：

SELECT pd.*
FROM (SELECT pd.*, COUNT(*) OVER (PARTITION BY Product) as cnt
      FROM ((SELECT Product, Date_Purchase FROM Table1
            ) UNION ALL
            (SELECT Product, Date_Purchase FROM Table2
            ) UNION ALL
            (SELECT Product, Date_Purchase FROM Table3
            ) UNION ALL
            (SELECT Product, Date_Purchase FROM Table4
            )
           ) pd
     ) pd
WHERE cnt = 1;

如果“删除”的意思是delete，那么您需要四个delete 语句，每个语句如下：

delete t
   from table1 t
   where not exists (select 1 from table2 where t2.product = t.product) and
         not exists (select 1 from table3 where t3.product = t.product) and
         not exists (select 1 from table4 where t4.product = t.product);

实际上，这会删除仅在桌子上的产品，即使它们多次出现。如果也有必要，可以对其进行调整以仅删除单例。