我有一张看起来像这样的桌子
| 事件_d | event_lcl_ts | 位置_id | 事件代码 | 容器 ID |
|---|---|---|---|---|
| 4/6/24 | 2024-04-06T10:19:32.133+00:00 | 1 | 上架 | asdjhdf-323 |
| 4/6/24 | 2024-04-06T21:52:35.019+00:00 | 1 | 接收 | asdjhdf-323 |
| 4/7/24 | 2024-04-07T00:13:17.496+00:00 | 1 | 上架 | asdjhdf-323 |
| 4/7/24 | 2024-04-07T12:35:54.766+00:00 | 1 | 接收 | asdjhdf-323 |
| 4/7/24 | 2024-04-07T16:27:13.245+00:00 | 1 | 上架 | asdjhdf-323 |
| 4/8/24 | 2024-04-08T22:56:19.038+00:00 | 1 | 接收 | asdjhdf-323 |
| 4/9/24 | 2024-04-09T00:19:47.575+00:00 | 1 | 上架 | asdjhdf-323 |
| 4/10/24 | 2024-04-10T20:44:12.190+00:00 | 1 | 接收 | asdjhdf-323 |
| 4/11/24 | 2024-04-11T01:14:45.466+00:00 | 1 | 上架 | asdjhdf-323 |
| 4/11/24 | 2024-04-11T10:14:12.709+00:00 | 1 | 接收 | asdjhdf-323 |
| 4/11/24 | 2024-04-11T12:57:11.640+00:00 | 1 | 上架 | asdjhdf-323 |
我正在尝试创建一行，其中显示收到容器然后存放的时间戳，该时间戳基于
event_codePUTAWAYRECEIVEPUTAWAYRECEIVE在这种情况下，理想的输出将是这样的：
| 容器 ID | 接收时间戳 | 放置_时间戳 | 位置_id |
|---|---|---|---|
| asdjhdf-323 | 2024-04-06T21:52:35.019+00:00 | 2024-04-07T00:13:17.496+00:00 | 1 |
| asdjhdf-323 | 2024-04-07T12:35:54.766+00:00 | 2024-04-07T16:27:13.245+00:00 | 1 |
| asdjhdf-323 | 2024-04-08T22:56:19.038+00:00 | 2024-04-09T00:19:47.575+00:00 | 1 |
| asdjhdf-323 | 2024-04-10T20:44:12.190+00:00 | 2024-04-11T01:14:45.466+00:00 | 1 |
| asdjhdf-323 | 2024-04-11T10:14:12.709+00:00 | 2024-04-11T12:57:11.640+00:00 | 1 |
如何在忽略或过滤
PUTAWAYRECEIVE如果您确定行始终交错，您可以使用
LAGLEADSELECT
t.container_id,
t.event_lcl_ts AS receive_timestamp,
t.putaway_timestamp,
t.location_id
FROM (
SELECT t.*,
LEAD(CASE WHEN t.event_code = 'PUTAWAY' THEN t.event_lcl_ts END)
OVER (PARTITION BY t.location_id, t.container_id ORDER BY t.event_lcl_ts) AS putaway_timestamp
FROM YourTable AS t
) AS t
WHERE t.event_code = 'RECEIVE';