Get Month columns from datetime column and count entries - sql

Question

I have the following table:
| ID | Name | DateA | TimeToWork | TimeWorked |
|:--:|:----:|:----------:|:----------:|:----------:|
| 1 |Frank | 2013-01-01 | 8 | 5 |
| 2 |Frank | 2013-01-02 | 8 | NULL |
| 3 |Frank | 2013-01-03 | 8 | 7 |
| 4 |Jules | 2013-01-01 | 4 | 9 |
| 5 |Jules | 2013-01-02 | 4 | NULL |
| 6 |Jules | 2013-01-03 | 4 | 3 |
The table is very long, every person has an entry for every day in a year. For each person I have the Date he worked (DateA), the hours he has to work according to contract (TimeToWork) and the hours he worked (TimeWorked). As you can see some days a person didnt work on a day he had to. This is when a person took a full day overtime.
What I try to accomplish is to get the following table out of the first one above.
| Name | January | Feburary | March | ... | Sum |
|:----:|:----------:|:--------:|:-----:|:---:|:---:|
|Frank | 2 | 0 | 1 | ... | 12 |
|Jules | 5 | 1 | 3 | ... | 10 |
For each month I want to count all days where a person took A FULL day off and sum all up in the Sum column.
I tried something like Select (case when Datetime(month, DateA = 1 then count(case when timetowork - (case when timeworked then 0 end) = timetowork then 1 else 0 end) end) as 'January' but my TSQL is just not that good and the code doent work at all. Btw using this my select command would be about 40 lines.
I really would appreciate if anyone could help me or give me a link to a good source so I can read myself into it.

Answer 1

If I understand the question right, than Gordon Linoff's answer is a good beginning, but doesn't deal with "full day off".
select Name,
sum(case when month(DateA) = 01 and TimeWorked is null then 1 else 0 end) as Jan,
sum(case when month(DateA) = 02 and TimeWorked is null then 1 else 0 end) as Feb,
...
sum(case when month(DeteA) = 12 and TimeWorked is null then 1 else 0 end) as Dec,
sum(case when TimeWorked is null then 1 else 0 end) as Sum
from table T
where year(DateA) = 2013
group by name
This method solves the problem?

Answer 2

The correct syntax is conditional aggregation:
select name,
sum(case when month(datea) = 1 then timeworked else 0 end) as Jan,
sum(case when month(datea) = 2 then timeworked else 0 end) as Feb,
. . .
sum(case when month(datea) = 12 then timeworked else 0 end) as Dec,
sum(timeworked)
from table t
where year(datea) = 2013
group by name;

Answer 3

The CASE can be removed using bit logic
SELECT name
, January = SUM((1 - CAST(MONTH(DateA) - 1 as bit))
* (1 - CAST(COALESCE(TimeWorked, 0) as bit)))
, February = SUM((1 - CAST(MONTH(DateA) - 2 as bit))
* (1 - CAST(COALESCE(TimeWorked, 0) as bit)))
...
, December = SUM((1 - CAST(MONTH(DateA) - 12 as bit))
* (1 - CAST(COALESCE(TimeWorked, 0) as bit)))
, Total = SUM((1 - CAST(COALESCE(TimeWorked, 0) as bit)))
FROM table1
GROUP BY name;
To check if there is a dayoff the formula is:
(1 - CAST(COALESCE(TimeWorked, 0) as bit))
that is equivalent to TimeWorked IS NULL: the CAST to BIT return 1 for every value different from 0, 1 - BIT invert those values.
The month filter is:
(1 - CAST(MONTH(DateA) - %month% as bit))
using the same idea as before this formula return 1 only for the given month (the cast give 1 for every other month, the 1 - BIT invert that result)
Multipling the two formulas we have the days off only for the given month

Answer 4

You can get your required result by using pivot also. You can get more information about pivot here http://technet.microsoft.com/en-in/library/ms177410(v=sql.105).aspx
Also you can get your output using the following query. I did it for up to April only. You can extend it up to December.
Select [Name], [January], [February], [March], [April]
From
(
Select Name, MName, DaysOff from
(
select Name, DATENAME(MM, dateA) MName,
count(case isnull(timeworked,0) when 0 then 1 else null end) DaysOff
from tblPivot
Where Year(DateA) = 2013
group by Name, DATENAME(MM, dateA)
) A ) As B
pivot(Count(DaysOff)
For MName in ([January], [February],[March],[April])
) As Pivottable;