mysql – BigQuery的最长连续日数

现在我只是总计用户工作了多少天.我正在尝试将此查询更改为大多数连续工作日.

其中u12345为4,u1为2.

这可能与BigQuery语句有关吗？

编辑我是Kind of close与以下查询,但我的u1得到3而不是2.

SELECT MIN(e.timestamp) as date_created, e.uid, COUNT(e.uid) + 1 AS streak
FROM OnSite e
LEFT JOIN OnSite ee 
  ON e.uid = ee.uid 
AND DATE(e.timestamp) = DATE(DATE_ADD(ee.timestamp, INTERVAL -1 DAY))
WHERE ee.uid IS NOT NULL
GROUP BY e.uid;

架构(MySQL v5.7)

CREATE TABLE OnSite
    (`uid` varchar(55), `worksite_id`  varchar(55), `timestamp` datetime)
;

INSERT INTO OnSite
    (`uid`, `worksite_id`, `timestamp`)
VALUES
  ("u12345", "worksite_1", '2019-01-01'),
  ("u12345", "worksite_1", '2019-01-02'),
  ("u12345", "worksite_1", '2019-01-03'),
  ("u12345", "worksite_1", '2019-01-04'),
  ("u12345", "worksite_1", '2019-01-06'),
  ("u1", "worksite_1", '2019-01-01'),
  ("u1", "worksite_1", '2019-01-02'),
  ("u1", "worksite_1", '2019-01-05'),
  ("u1", "worksite_1", '2019-01-06')

;

查询#1

SELECT    uid, COUNT(DISTINCT timestamp) Total
FROM      OnSite
GROUP BY  uid;

| uid    | Total |
| ------ | ----- |
| u1     | 4     |
| u12345 | 5     |

View on DB Fiddle

解决方法:

以下是BigQuery Standard SQL

如果您对同一工作地点的用户最多连续几天感兴趣：

#standardSQL
SELECT uid, MAX(consecuitive_days) max_consecuitive_days
FROM (
  SELECT uid, grp, COUNT(1) consecuitive_days
  FROM (
    SELECT uid, 
      COUNTIF(step > 1) OVER(PARTITION BY uid, worksite_id ORDER BY ts) grp
    FROM (
      SELECT uid, worksite_id, ts, 
        DATE_DIFF(ts, LAG(ts) OVER(PARTITION BY uid, worksite_id ORDER BY ts), DAY) step 
      FROM `project.dataset.table`
    )
  ) GROUP BY uid, grp
) GROUP BY uid  

如果工地无关紧要,并且您正在寻找连续最多天数：

#standardSQL
SELECT uid, MAX(consecuitive_days) max_consecuitive_days
FROM (
  SELECT uid, grp, COUNT(1) consecuitive_days
  FROM (
    SELECT uid, 
      COUNTIF(step > 1) OVER(PARTITION BY uid ORDER BY ts) grp
    FROM (
      SELECT uid, ts, 
        DATE_DIFF(ts, LAG(ts) OVER(PARTITION BY uid ORDER BY ts), DAY) step 
      FROM `project.dataset.table`
    )
  ) GROUP BY uid, grp
) GROUP BY uid  

您可以使用问题中的示例数据测试,播放以上任何一项,如下例所示

#standardSQL
WITH `project.dataset.table` AS (
  SELECT 'u12345' uid, 'worksite_1' worksite_id, DATE '2019-01-01' ts UNION ALL
  SELECT 'u12345', 'worksite_1', '2019-01-02' UNION ALL
  SELECT 'u12345', 'worksite_1', '2019-01-03' UNION ALL
  SELECT 'u12345', 'worksite_1', '2019-01-04' UNION ALL
  SELECT 'u12345', 'worksite_1', '2019-01-06' UNION ALL
  SELECT 'u1', 'worksite_1', '2019-01-01' UNION ALL
  SELECT 'u1', 'worksite_1', '2019-01-02' UNION ALL
  SELECT 'u1', 'worksite_1', '2019-01-05' UNION ALL
  SELECT 'u1', 'worksite_1', '2019-01-06' 
)
SELECT uid, MAX(consecuitive_days) max_consecuitive_days
FROM (
  SELECT uid, grp, COUNT(1) consecuitive_days
  FROM (
    SELECT uid, 
      COUNTIF(step > 1) OVER(PARTITION BY uid ORDER BY ts) grp
    FROM (
      SELECT uid, ts, 
        DATE_DIFF(ts, LAG(ts) OVER(PARTITION BY uid ORDER BY ts), DAY) step 
      FROM `project.dataset.table`
    )
  ) GROUP BY uid, grp
) GROUP BY uid   

结果：

Row uid     max_consecuitive_days    
1   u12345  4    
2   u1      2    

标签：gaps-and-islands,google-bigquery,mysql
来源： https://codeday.me/bug/20190727/1548920.html