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 |
+--------------+--------------+------------+

我有一张类似的表。
我需要的是所有公司的账单地址和送货地址不同的公司。

注意 - 每家公司只有一个帐单邮寄地址。但它可以有多个送货地址

这似乎是一个相当简单的查询，但我不仅能够得到它。

我的尝试 - 我尝试减去＆＃39;来自Billing的所有送货地址，但没有输出。明显没有帮助

查询：

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');

输出应公司B （因为它的结算和送货地址不同）

编辑1：尝试过Indra的查询，但它会永远运行。没有回复

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？以下显示了所有不同的对：

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;

如果你只想要不同的公司：

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：这还会检查只有一个不同的结算和送货地址。

Answer 2

使用以下查询

buildscript {
    repositories {
        ...
    }
    dependencies {
        classpath 'com.android.tools.build:gradle:1.3.1'
        classpath 'com.google.gms:google-services:1.3.1'
    }
}

Answer 3

在这个SQL Fiddle example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小提琴适用于我。这就是所有代码：

架构：

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;

结果：

Company B

比较同一个表的行和列

3 个答案: