问题描述
限时送ChatGPT账号..+--------------+--------------+------------+
| 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 |
+--------------+--------------+------------+
我有一张类似的桌子.
我需要的是帐单地址和送货地址不同的所有公司.
I have a table similar to this.
What i need is All the companies whose billing address and shipping address are different.
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
NOTE - Each company has only ONE billing address. But it can have multiple Shipping addresses
这似乎是一个相当简单的查询,但我无法理解.
This seems like a fairly simple query but I'm not just able to get it.
我的尝试 - 我尝试从 Billing 中减去"所有送货地址,但没有任何输出.不同也无济于事
My attempt - I tried 'subtracting' all the shipping address from Billing but there's just no output. Distinct doesn't help as well
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');
输出应该是Company B(因为它的帐单和送货地址不同)
Output should be Company B (since it's billing and shipping address is different)
编辑 1:尝试了 Indra 的查询,但它永远运行.没有反应
EDIT 1 : Tried Indra's query but it runs forever. No response
select A.* from company A inner join company B on Apany_Name = Bpany_Name
and (A.address_type = 'Billing' and B.address_type = 'Shipping')
AND A.address <> B.address
推荐答案
如何使用 join
?下面显示了所有不同的对:
How about using a join
? The following shows all pairs that are different:
select tb.*, ts.*
from company tb join
company ts
on tbpany_name = tspany_name and
ts.address_type = 'shipping' and
tb.address_type = 'billing' and
ts.address <> tb.address;
如果您只想要不同的公司:
If you just want the companies that are different:
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:这还会检查是否只有一个不同的帐单邮寄地址.
Note: this also checks that there is only one distinct billing and shipping address.
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