Question

我有一个设备表（dev）和device_date表（dev_data）。关系1：M

dev表：

| id  |name   |status |
|-----|-------|-------|
| 1   |a      |111    |
|-----|-------|-------|
| 2   |b      |123    |
|-----|-------|-------|
| ....|.....  |....   |

dev_data表：

|id |dev_id |status  |date                    |
|---|-------|--------|------------------------|
|1  |1      | 123    |2019-04-16T18:53:07.908Z|
|---|-------|--------|------------------------|
|2  |1      | 120    |2019-04-16T18:54:07.908Z|
|---|-------|--------|------------------------|
|3  |1      | 1207   |2019-04-16T18:55:07.908Z|
|---|-------|--------|------------------------|
|4  |2      | 123    |2019-04-16T18:53:08.908Z|
|---|-------|--------|------------------------|
|5  |2      | 121    |2019-04-16T18:54:08.908Z|
|---|-------|--------|------------------------|
|6  |2      | 127    |2019-04-16T18:55:08.908Z|
|...|.......|........|........................|

我需要选择所有dev并加入dev_data，但仅添加2条最后记录（按日期）

最终答复应如下所示：

status_calc_1和status_calc_2是dev和dev_data中status之间的区别

status_calc_1 => status与dev_data和dev的最后一行的差异

status_calc_2 => status与dev_data和dev的最后一行的差异

|id  |name  |status_calc_1  | status_calc_2 |
|----|------|---------------|---------------|
|1   |a     |1207-111       |120-111        |
|----|------|---------------|---------------|
|2   |b     |127-123        |121-123        |

我尝试了这个：

select id, "name", status, max(dd.date) as last,
       (select date from device_data p where p.dev_id = device.id and date < dd.date limit 1) as prelast
from device
  inner join device_data dd on device.id = dd.dev_id
group by id, "name", status;

但出现错误：

ERROR: subquery uses ungrouped column "device.id" from outer query

和这个：

select id, "name", status, max(dd.date) as last, max(dd2.date) as prelast,
from device
  inner join device_data dd on device.id = dd.dev_id
  inner join device_data dd2 on device.id = dd2.dev_id and dd2.date < dd.date
group by id, "name", status;

我得到正确的最后2个dev_data，但是仍然不知道如何制作2列status_calc_1和status_calc_2

status_calc_1 =最后一行dev_data.status-dev.status

status_calc_2 =前一行dev_data.status-dev.status

Answer 1

您可以使用条件聚合：

select d.id, d.name, d.status,
       max(dd.date) as last,
       max(case when dd.seqnum = 2 then dd.date end) as prelast,
       (max(case when dd.seqnum = 1 then dd.status end) - d.status) as status_calc_1,
       (max(case when dd.seqnum = 2 then dd.status end) - d.status) as status_calc_2
from device d join
     (select dd.*,
             row_number() over (partition by dd.dev_id order by dd.date desc) as seqnum
      from device_data dd
     ) dd
     on d.id = dd.dev_id
where seqnum <= 2
group by d.id, d.name, d.status;

从联接表中选择两个条目（1：m）

1 个答案: