Longest Consecutive Days Count for BigQuery - mysql

Question

Right now I just have an aggregate of how many days a user has worked. I'm trying to change this query to most continuous days worked.
Where u12345 would be 4 and u1 would be 2.
Is this possible to do with a BigQuery statement?
EDIT I am Kind of close with the following query but my u1 is getting 3 instead of 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;
Schema (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')
;
Query #1
SELECT uid, COUNT(DISTINCT timestamp) Total
FROM OnSite
GROUP BY uid;
| uid | Total |
| ------ | ----- |
| u1 | 4 |
| u12345 | 5 |
View on DB Fiddle

Answer 1

Below is for BigQuery Standard SQL
In case if you are interested in max consecutive days of the users on the same worksite:
#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
In case if worksite does not matter and you are looking just for max consecutive days:
#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
You can test, play any of above with he sample data from your question as in below example
#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
with result:
Row uid max_consecuitive_days
1 u12345 4
2 u1 2

Answer 2

does this fit for you?
set #gr=1;
select uid, max(cnt) max_cnt from (
select uid, grp, count(*) cnt from (
select uid,
case when ifnull(DATE_ADD(oldDate, INTERVAL 1 DAY), timestamp)= timestamp then
#gr
else
#gr := #gr +1
end grp
from
(
SELECT
uid,
timestamp,
lag(timestamp) over (partition by uid order by timestamp asc) as oldDate
FROM OnSite
) t
)t2
group by uid, grp
)t3
group by uid
Result
| uid | max_cnt |
| ------ | ------- |
| u1 | 2 |
| u12345 | 4 |
DB Fiddle