如何从日期范围中获取当前年份的天数？

Question

In my database, I have the attributes "startingdate", "endingdate", "daterange", "numdays" and a few others that are not of any importance for my question. I am trying to build a leave application system, allowing users to apply leave.

The attributes "startingdate" and "endingdate" are rather self explanatory. "numdays" is the number of days from the "startingdate" to the "endingdate". As for "daterange" it is all the dates from the starting date to the ending date, excluding the starting date. For example, if the starting date is 2014-01-01 and the ending date is 2014-01-03, the "daterange" would be "2014-01-02 2014-01-03". I need this for a php calendar to highlight the dates that the user had already applied leave on.

The problem that I encountered is that I am not sure how to find the number of days a user had taken a leave this year. Lets say that a user applied leave on 2014-12-30 to 2015-01-02, the "daterange" would be "2014-12-31 2015-01-01 2015-01-02". How do I find the number of days that are in 2014?

Sorry for the long question.

Answer 1

Can probably be done more efficiently but I'll leave optimizing it to you.

Assumes that endingdate is after startingdate :

select startingdate,endingdate,
case 
when year(startingdate)=year(now()) 
 and year(endingdate)=year(now()) 
then datediff(endingdate,startingdate) 
when year(startingdate)=year(now()) 
 and year(endingdate)>year(now()) 
then datediff(date(concat(year(now()),'-12-31')),startingdate)
when year(startingdate)<year(now()) 
 and year(endingdate)=year(now()) 
then datediff(endingdate,date(concat(year(now()),'-01-01')))
when year(startingdate)<year(now()) 
 and year(endingdate)>year(now()) 
then datediff(date(concat(year(now()),'-12-31')),date(concat(year(now()),'-01-01')))
else 0 end as daysOffThisYear
from myTable m;