Question

我正在努力获得正确的数据返回，我有一个包含以下数据结构的表：

# order_id, name
'10434', 'docs_sent'
'10433', 'placed'
'10433', 'docs_sent'
'10433', 'agreement_received'
'10432', 'placed'
'10431', 'placed'
'10431', 'docs_sent'
'10430', 'placed'
'10430', 'docs_sent'
'10430', 'agreement_received'
'10429', 'placed'
'10429', 'docs_sent'
'10429', 'agreement_received'
'10428', 'placed'
'10428', 'docs_sent'
'10427', 'placed'

我要做的是返回一个独特的＆＃34; order_id＆＃34;在哪里＆＃34;名称＆＃34;不包括agreement_received。所以澄清我正在寻找一个订单号列表＆＃34; order_id＆＃34;其中订单号与agreement_received＆＃34;没有任何记录。问题是每个订单都有多个条目，我似乎无法做到正确。请帮助！

输出应类似于：

# order_id
'10434'
'10432'
'10431'
'10428'
'10427'

Answer 1

您希望按 order_id分组，并且只保留那些没有＆＃39; agreement_received＆＃39;记录。

select order_id
from mytable
group by order_id
having sum(name = 'agreement_received') = 0;

（这使用MySQL的特殊布尔处理.False = 0，True = 1.在其他dbms中你必须使用sum(case when name = 'agreement_received' then 1 else 0 end)。）

Answer 2

select distinct order_id
from orders
where name != 'agreement_received';

这对你有用吗？

Answer 3

尝试类似：

SELECT order_id
FROM MyTable
WHERE name <> 'agreement_received'
GROUP BY order_id
HAVING COUNT(order_id) = 1

Answer 4

如果您的表名是t：

select distinct t1.order_id
from t as t1
where t1.order_id not in
   (select t2.order_id
    from t as t2
    where t2.name = 'agreement_received')

输出：

+----------+
| order_id |
+----------+
|    10434 |
|    10432 |
|    10431 |
|    10428 |
|    10427 |
+----------+

MySQL SELECT与多个相似的行不同

4 个答案: