Counting consecutive days for all items in SQL

Question

I'm trying to count consecutive days for all items. I've found how to count consecutive days just for one item , sample data

bob 2014-08-10 00:35:22
sue 2014-08-10 00:35:22
bob 2014-08-11 00:35:22
mike 2014-08-11 00:35:22
bob 2014-08-12 00:35:22
mike 2014-08-12 00:35:22

would give me result for Bob like this:

date_created         | streak|
2014-08-10 00:35:22  |      3|

But I'd need to get it for all users like this:

date_created         | streak|username
2014-08-10 00:35:22  |      3|     Bob
2014-08-11 00:35:22  |      2|     Mike

I've been trying to modify sql from this response , but I just can not get it work. I'd be thankful for any advice.

Answer 1

one thing you could do is self join the table to itself on consecutive days and count it. note I add one to the count because it wont count the first day

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

Sql Fiddle

Answer 2

you can solve this with CTE as follows and here is the sqlfiddle

with ranks as
(
  select
    date,
    name,
    dateadd(day, -row_number() OVER (PARTITION BY name ORDER BY date), date) as rr
  from orders
),
cnt as
(
  select
    name,
    count(*) as ttl
  from ranks
  group by
    name,
    rr
)

select 
  min(date) as date,
  r.name,
  ttl
FROM ranks r
join cnt c
on r.name = c.name
where ttl > 1
group by
  r.name,
  ttl