Return Results Of Consecutive Days Per User - sql

Question

I have a table such as the following:
+----+----------------+-------------------------+
| id | employeeNumber | transactionTime |
+----+----------------+-------------------------+
| 1 | 1234 | 2016-02-23 15:11:00.000 |
+----+----------------+-------------------------+
| 2 | 1234 | 2016-02-22 11:01:00.000 |
+----+----------------+-------------------------+
| 3 | 1235 | 2016-02-22 07:22:00.000 |
+----+----------------+-------------------------+
| 4 | 1236 | 2016-02-20 09:16:00.000 |
+----+----------------+-------------------------+
| 5 | 1236 | 2016-02-19 11:01:00.000 |
+----+----------------+-------------------------+
| 6 | 1236 | 2016-02-18 11:44:00.000 |
+----+----------------+-------------------------+
| 7 | 1236 | 2016-02-17 12:12:00.000 |
+----+----------------+-------------------------+
| 8 | 1236 | 2016-02-16 11:09:00.000 |
+----+----------------+-------------------------+
| 9 | 1236 | 2016-02-15 11:19:00.000 |
+----+----------------+-------------------------+
| 10 | 1236 | 2016-02-14 09:12:00.000 |
+----+----------------+-------------------------+
I Need to find a way to return the number of consecutive days that each employee logged a transaction over the past 2 weeks. Such as this:
+------+--------------+-------------------------+-------------------------+
| days |employeeNumber| startTime | endTime |
+------+--------------+-------------------------+-------------------------+
| 2 | 1234 | 2016-02-22 11:01:00.000 | 2016-02-23 15:11:00.000 |
+------+--------------+-------------------------+-------------------------+
| 1 | 1235 | 2016-02-22 11:01:00.000 | 2016-02-22 11:01:00.000 |
+------+--------------+-------------------------+-------------------------+
| 7 | 1236 | 2016-02-14 09:12:00.000 | 2016-02-20 09:16:00.000 |
+------+--------------+-------------------------+-------------------------+
I have been working with the following query, but It only returns a single user and doesn't take into account only the past 2 weeks.
WITH
dates(date) AS (
SELECT DISTINCT CAST(transactionTime AS DATE)
FROM Fuel.dbo.comdata
WHERE employeeNumber = 123456
),
groups AS (
SELECT ROW_NUMBER() OVER (ORDER BY date) AS rn,
DATEADD(DAY, -ROW_NUMBER() OVER (ORDER BY date), date) AS grp,
date
FROM dates
)
SELECT COUNT(*) AS consecutiveDates,
MIN(date) AS minDate, MAX(date) AS maxDate
FROM groups
GROUP BY grp
ORDER BY 1 DESC, 2 DESC
Any help is appreciated.
UPDATE
So I have found the following query very helpful thanks to Gordon Linoff's answer below. However I notice that the Min/Max Dates don't match up to the number of consecutive days. As shown Here with live data:
SELECT * FROM (
SELECT employeeNumber, COUNT(*) AS consecutiveDays,
MIN(transactionTime) AS startTime, MAX(transactionTime) AS endTime
FROM (
SELECT cd.*, DATEADD(DAY, -DENSE_RANK() OVER (PARTITION BY
employeeNumber ORDER BY transactionTime), CAST(transactionTime AS
DATE)) AS grp
FROM Fuel.dbo.comdata cd
WHERE transactionTime >= DATEADD(DAY, -14, GETDATE())
) cd
GROUP BY employeeNumber, grp
) AS tbl1
WHERE consecutiveDays >= 7
+--------------+-------------------------+------------------------+
| empNum | days| startTime | endTime |
+--------+-------------------------------+------------------------+
| 16742 | 7 | 2016-04-28 17:00:00.000 | 2016-05-07 17:04:00.000|
+--------+-------------------------------+------------------------+
| 15056 | 8 | 2016-04-27 13:03:00.000 | 2016-05-08 09:51:00.000|
+--------+-------------------------------+------------------------+
As you can see the number of consecutive days does not match the start/end time. Any ideas?

Answer 1

I would do this with the difference using row number approaches (assuming there is never more than one record per day per employee):
select employee, count(*) as numdays,
min(timestamp) as startTime, max(timestamp) as endTime
from (select cd.*,
dateadd(day,
- row_number() over (partition by employee order by transactionTime),
cast(transactionTime as date)
) as grp
from Fuel.dbo.comdata cd
) cd
group by employee, grp;
The idea is to generate a series of sequential numbers for each employee based on the transactionTime. The difference between this and the transactionTime is constant, when the transactions are on consecutive days.
If you can have multiple transactions on the same day, then you can use dense_rank().
If you have duplicates on the same day:
select employee, count(*) as numdays,
min(timestamp) as startTime, max(timestamp) as endTime
from (select cd.*,
dateadd(day,
- dense_rank() over (partition by employee order by cast(transactionTime as date)),
cast(transactionTime as date)
) as grp
from Fuel.dbo.comdata cd
) cd
group by employee, grp;