比較同一表格的行和列

Question

+--------------+--------------+------------+
| company_name | address_type |  address   |
+--------------+--------------+------------+
| Company A    | Billing      | 111 Street |
| Company A    | Shipping     | 111 Street |
| Company B    | Billing      | 222 Street |
| Company B    | Shipping     | 333 street |
| Company B    | Shipping     | 444 street |
+--------------+--------------+------------+

我有一張與此類似的桌子。
我需要的是帳單地址和送貨地址不同的所有公司 。

注意-每個公司只有一個帳單地址。 但是可以有多個收貨地址

這似乎是一個非常簡單的查詢，但我不僅能得到它。

我的嘗試-我嘗試從帳單中“減去”所有收貨地址，但沒有任何輸出。 區別也無濟於事

查詢：

select company_name 
from tableA 
where address_type='Billing' 
and company_name not in (select to_char(company_name) from tableA where address_type='Shipping');

輸出應為公司B （因為帳單和送貨地址不同）

編輯1：嘗試了Indra的查詢，但它會永遠運行。 沒有反應

select A.* from company A inner join company B on A.company_Name = B.company_Name
and (A.address_type = 'Billing' and B.address_type = 'Shipping') 
AND A.address <> B.address 

Answer 1

如何使用join ？ 下面顯示了所有不同的對：

select tb.*, ts.*
from company tb join
     company ts
     on tb.company_name = ts.company_name and
        ts.address_type = 'shipping' and
        tb.address_type = 'billing' and
        ts.address <> tb.address;

如果您只想要與眾不同的公司：

select company_name
from company t
group by company_name
having count(distinct case when t.address_type = 'billing' then address end) = 1 and
       count(distinct case when t.address_type = 'shipping' then address end) = 1 and
       (max(case when t.address_type = 'billing' then address end) <>
        max(case when t.address_type = 'shipping' then address end)
       );

注意：這還會檢查是否只有一個不同的帳單和送貨地址。

Answer 2

使用以下查詢

select A.* from company A inner join company B on A.company_Name = B.company_Name
and (A.address_type = 'Billing' and B.address_type = 'Shipping') 
AND A.address <> B.address   

Answer 3

在這個SQL Fiddle示例中 ，您可以找到另一家公司的其他方法：

+--------------+--------------+------------+
| company_name | address_type |  address   |
+--------------+--------------+------------+
| Company A    | Billing      | 111 Street |
| Company A    | Shipping     | 111 Street |
| Company B    | Billing      | 222 Street |
| Company B    | Shipping     | 333 street |
| Company B    | Shipping     | 444 street |
| Company C    | Shipping     | 555 street |
| Company C    | Shipping     | 666 street |
| Company C    | Billing      | 555 street |
| Company C    | Billing      | 666 street |
+--------------+--------------+------------+

POSTDATA ：SQL Fiddle對我來說很好。 這是所有代碼：

架構：

CREATE TABLE companies
( 
  company_name  VARCHAR2(40),
  address_type  VARCHAR2(40),
  address    VARCHAR2(40)
);

insert into companies values ('Company A','Billing','111 Street' );
insert into companies values ('Company A','Shipping','111 Street' );
insert into companies values ('Company B','Billing','222 Street' );
insert into companies values ('Company B','Shipping','333 street' );
insert into companies values ('Company B','Shipping','444 street');
insert into companies values ('Company C','Billing','555 Street' );
insert into companies values ('Company C','Billing','666 Street' );
insert into companies values ('Company C','Shipping','555 Street' );
insert into companies values ('Company C','Shipping','666 Street' );

Sentece：

SELECT DISTINCT(comp.company_name) FROM
    (select company_name, address, count(*) regs
    from companies
    group by company_name, address) comp
WHERE MOD(comp.regs,2) = 1;

結果：

Company B