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[英]MySQL select all distinct values from column a for each distinct column b, where count distinct a >1
[英]MySQL - Display null column from child table if all values are not distinct
我有以下表格,例如:
发票
ID Name
1 A
2 B
3 C
4 D
5 E
交易
ID Invoice_ID User_ID
1 1 10
2 1 10
3 1 10
4 2 30
5 3 20
6 3 40
7 2 30
8 2 30
9 4 40
10 3 50
现在,我想进行选择,以从相关交易中提取发票和user_id,但是,如果这样做,我将不会获得所有ID,因为它们可能是不同的,但该列只有一列。 我想做的是,如果有不同的User_id,我将在该列中显示预定义的文本,而不是实际结果。
select invoices.id, invoices.name, transactions.user_id(if there are distinct user_ids -> return null)
from invoices
left join transactions on invoices.id = transactions.invoice_id
然后这就是结果
ID Name User_ID
1 A 10
2 B 30
3 C null
4 D 40
5 E null
这可能吗?
您可以执行以下操作:
select
invoices.id,
invoices.name,
IF (
(SELECT COUNT(DISTINCT user_id) FROM transactions WHERE transactions.invoice_id = invoices.id) = 1,
(SELECT MAX(user_id) FROM transactions WHERE transactions.invoice_id = invoices.id),
null
) AS user_id
from invoices
或者,您也可以使用GROUP_CONCAT函数为每个发票输出用逗号分隔的用户列表。 这并不是您所要求的,但实际上可能会更有用:
select
invoices.id,
invoices.name,
GROUP_CONCAT(DISTINCT transactions.user_id SEPARATOR ',') AS user_ids
from invoices
left join transactions on invoices.id = transactions.invoice_id
group by invoices.id
尝试类似的东西:
select i.id, i.name, t.user_id
from invoices i left join
(
select invoice_ID, User_ID
from transactions
group by invoice_ID
having count(invoice_ID)=1
) t on i.id=t.invoice_id
您可以列出具有多个用户标识的所有事务,如下所示:
select invoices.id, invoices.name, null
from invoices
left join transactions on invoices.id = transactions.invoice_id having count(distinct transactions.user_id) > 1
另外,我认为这种CASE
可能适合您的需求:
select invoices.id, invoices.name,
case when count(distinct transactions.user_id) > 1 then null else transactions.user_id end
from invoices
left join transactions on invoices.id = transactions.invoice_id
group by invoices.id
虽然,我不确定这在语法上是否正确
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