Question

我需要所有相关儿童记录状态已关闭的所有门票。因此，如果其中一个孩子是其他孩子，我不希望结果集中的票证。

我尝试使用antijoin patern，但我的问题是孩子们住在不同的桌子里。

t1：票 t2：relatedrecord t3：child1 t4：child1

select ticket.ticketid, rr.relatedrecordkey, rr.relatedrecordclass, c1.id, c1.status, c2.id, c2.status
from ticket
inner join relatedrecord rr on rr.recordkey = ticket.ticketid and rr.class = ticket.class
left join child1 c1 on c1.id = rr.relatedreckey and c1.class = rr.relatedrecclass
left join child2 c2 on c2.id = rr.relatedreckey and c2.class = rr.relatedrecclass

结果：

+----------+--------+-----------+-------+-----------------+---------------+--------+--------+--------+--------+
| ticketid | status | RECORDKEY | class | RELATEDRECCLASS | relatedreckey |   id   | status |   id   | status |
+----------+--------+-----------+-------+-----------------+---------------+--------+--------+--------+--------+
|     1183 | NEW    |      1183 | SR    | WORKORDER       |          1238 | 1238   | NEW    | (null) | (null) |
|     1183 | NEW    |      1183 | SR    | SR              |          1184 | (null) | (null) | 1184   | NEW    |
|     1185 | NEW    |      1185 | SR    | WORKORDER       |          1239 | 1239   | CLOSE  | (null) | (null) |
|     1185 | NEW    |      1185 | SR    | SR              |          1186 | (null) | (null) | 1186   | CLOSE  |
|     1187 | NEW    |      1187 | SR    | WORKORDER       |          1240 | 1240   | CLOSE  | (null) | (null) |
|     1187 | NEW    |      1187 | SR    | SR              |          1188 | (null) | (null) | 1188   | NEW    |
|     1190 | NEW    |      1190 | SR    | SR              |          1191 | (null) | (null) | 1191   | CLOSE  |
|     1192 | NEW    |      1192 | SR    | WORKORDER       |          1241 | 1241   | CLOSE  | (null) | (null) |
+----------+--------+-----------+-------+-----------------+---------------+--------+--------+--------+--------+

所以

ticket 1183有两条相关记录，两条记录均尚未结束。（拒绝）
票证1185有两条相关记录，均已关闭（接受）
票证1187有两个相关记录，一个是新的，一个是关闭的。（拒绝）
票证1190有一个关闭（接受）的相关记录
票证1192有一个关闭（接受）的相关记录（不同的表格）
票证1189没有相关记录（拒绝）

从这个集合中，我只想在结果集中看到票证1185,1190,1192。它应该看起来像：

+----------+--------+-----------+-------+-----------------+---------------+--------+--------+--------+--------+
| ticketid | status | RECORDKEY | class | RELATEDRECCLASS | relatedreckey |   id   | status |   id   | status |
+----------+--------+-----------+-------+-----------------+---------------+--------+--------+--------+--------+
|     1185 | NEW    |      1185 | SR    | WORKORDER       |          1239 | 1239   | CLOSE  | (null) | (null) |
|     1185 | NEW    |      1185 | SR    | SR              |          1186 | (null) | (null) | 1186   | CLOSE  |
|     1190 | NEW    |      1190 | SR    | SR              |          1191 | (null) | (null) | 1191   | CLOSE  |
|     1192 | NEW    |      1192 | SR    | WORKORDER       |          1241 | 1241   | CLOSE  | (null) | (null) |
+----------+--------+-----------+-------+-----------------+---------------+--------+--------+--------+--------+

我尝试过类似的事情：

select ticket.ticketid, rr.relatedrecordkey, rr.relatedrecordclass, c1.id, c1.status, c2.id, c2.status
from ticket
inner join relatedrecord rr on rr.recordkey = ticket.ticketid and rr.class = ticket.class
where not exists (
    select 1 from child1 c1
    where c1.id = rr.relatedreckey and c1.class = rr.relatedrecclass
    and c1.status <> 'CLOSE'
) and not exists (
    select 1 from child2 c2
    where c2.id = rr.relatedreckey and c2.class = rr.relatedrecclass
    and c2.status <> 'CLOSE'
)

这导致ticket2的两行和来自ticket 3的一行（因为它关闭了一个子）我有点困惑如何正确解决这个问题。

Answer 1

仍然不能完全确定你想要的结果，但我认为这可能是它。

select *
from
(
    select t.ticketid
    from ticket t
    join relatedrecord rr on rr.recordkey = t.ticketid and rr.class = t.class
    join child1 c1 on c1.id = rr.relatedreckey 
        and c1.class = rr.relatedrecclass
        and c1.status = 'CLOSE'
    group by t.ticketid
    having MAX(c1.status) = MIN(c1.status)
    UNION ALL
    select t.ticketid
    from ticket t
    join relatedrecord rr on rr.recordkey = t.ticketid and rr.class = t.class
    join child2 c2 on c2.id = rr.relatedreckey 
        and c2.class = rr.relatedrecclass
        and c2.status = 'CLOSE'
    group by t.ticketid
    having MAX(c2.status) = MIN(c2.status)
) x
group by ticketid

Answer 2

我找到了合适的答案。

请参阅此sqlfiddle：http://sqlfiddle.com/#!3/febde/24

SELECT ticket.ticketid,
   count(rr.relatedreckey) AS children,
   count(wo.id) AS wo,
   count(tt.id) AS sr
FROM ticket
INNER JOIN relatedrecord rr ON rr.recordkey = ticket.ticketid
AND rr.class = ticket.class
LEFT JOIN child1 wo ON wo.id = rr.relatedreckey
AND wo.class = rr.relatedrecclass
AND wo.status = 'CLOSE'
LEFT JOIN child2 tt ON tt.id = rr.relatedreckey
AND tt.class = rr.relatedrecclass
AND tt.status = 'CLOSE'
GROUP BY ticket.ticketid 
HAVING (count(wo.id) + count(tt.id)) = count(rr.relatedreckey);

基本上它为每个相关类型计算已关闭记录的数量，并将其与相关记录的总数进行比较。如果它们相同，则必须关闭所有相关记录。

在多个表中仅查找具有特定状态子项的父记录

2 个答案: