Counting Consecutive Dates Using SQL

• Download source code - 334 B

Recently, I answered one of the SQL based questions on CodeProject. Thanks to the participation, that question inspired me to write this blog post.

Question

He has a table (tblLeave) with data like below:

PAYCODE		LV_TYPE	FROM_DATE		TO_DATE	        	LVALUE
5023		SL    	14/12/2012 0:00		14/12/2012 0:00		1
5023		SL    	15/12/2012 0:00		15/12/2012 0:00		1
5023		COF   	16/12/2012 0:00		16/12/2012 0:00		1
5023		SL    	19/12/2012 0:00		19/12/2012 0:00		1
5023		SL    	22/12/2012 0:00		22/12/2012 0:00		1
5023		SL    	23/12/2012 0:00		23/12/2012 0:00		1
5023		SL    	24/12/2012 0:00		24/12/2012 0:00		1
5023		PL    	28/12/2012 0:00		28/12/2012 0:00		1
5023		PL    	29/12/2012 0:00		29/12/2012 0:00		1
5023		PL    	30/12/2012 0:00		30/12/2012 0:00		1
5023		PL    	31/12/2012 0:00		31/12/2012 0:00		1

And he wants to output the data as below:

PAYCODE LV_TYPE FROM_DATE       TO_DATE         LVALUE
5023    SL      14/12/2012 0:00 15/12/2012 0:00 2
5023    COF     16/12/2012 0:00 16/12/2012 0:00 1
5023    SL      19/12/2012 0:00 19/12/2012 0:00 1
5023    SL      22/12/2012 0:00 24/12/2012 0:00 3
5023    PL      28/12/2012 0:00 31/12/2012 0:00 4

Condition: If same type of leave is taken continuously, it should be merged in one row mentioning from_date to to_date.

Answer

When I saw that question for the first time, I didn’t go through in detail and just thought that was an easy grouping query. And I just gave the following answer:

SELECT LV_TYPE,LV_TYPE,MIN(FROM_DATE) AS FROM_DATE ,MAX(TO_DATE) AS TO_DATE, COUNT(LVALUEP) AS LVALUE
FROM tblLeave
GROUP BY PAYCODE,LV_TYPE

But that’s wrong, he made comments saying it doesn’t make sense and highlighted the condition he wants. (Thanks to him; he didn’t down vote my answer). Again, I read the question… Oh… that was a tricky question. He needs to group the leave by consecutive date. Isn’t that tricky?

To answer that, I use the DATEDIFF SQL function:

SQL

DATEDIFF ( datepart , startdate , enddate )

http://msdn.microsoft.com/en-us/library/ms189794.aspx

Following is my answer and the output:

SELECT PAYCODE,LV_TYPE, MIN(FROM_DATE) AS FROM_DATE,
       MAX(FROM_DATE) AS TO_DATE, COUNT('A') AS LVALUE
FROM (
SELECT PAYCODE,LV_TYPE,FROM_DATE,
    DATEDIFF(D, ROW_NUMBER() OVER(ORDER BY FROM_DATE), FROM_DATE) AS Diff
FROM tblLeave) AS dt
GROUP BY PAYCODE,LV_TYPE, Diff
ORDER BY FROM_DATE

PAYCODE     LV_TYPE FROM_DATE               TO_DATE                 LVALUE
----------- ------- ----------------------- ----------------------- -----------
5023        SL      2012-12-14 00:00:00.000 2012-12-15 00:00:00.000 2
5023        COF     2012-12-16 00:00:00.000 2012-12-16 00:00:00.000 1
5023        SL      2012-12-19 00:00:00.000 2012-12-19 00:00:00.000 1
5023        SL      2012-12-22 00:00:00.000 2012-12-24 00:00:00.000 3
5023        PL      2012-12-28 00:00:00.000 2012-12-31 00:00:00.000 4

Query Explanation

Before explaining the logic, take a look at the following query and the output.

SELECT PAYCODE,LV_TYPE,FROM_DATE,
	ROW_NUMBER() OVER(ORDER BY FROM_DATE) AS ROW_NUMBER,
    DATEDIFF(D, ROW_NUMBER() OVER(ORDER BY FROM_DATE), FROM_DATE) AS Diff
FROM tblLeave

PAYCODE     LV_TYPE FROM_DATE               ROW_NUMBER           Diff
----------- ------- ----------------------- -------------------- -----------
5023        SL      2012-12-14 00:00:00.000 1                    41254
5023        SL      2012-12-15 00:00:00.000 2                    41254
5023        COF     2012-12-16 00:00:00.000 3                    41254
5023        SL      2012-12-19 00:00:00.000 4                    41256
5023        SL      2012-12-22 00:00:00.000 5                    41258
5023        SL      2012-12-23 00:00:00.000 6                    41258
5023        SL      2012-12-24 00:00:00.000 7                    41258
5023        PL      2012-12-28 00:00:00.000 8                    41261
5023        PL      2012-12-29 00:00:00.000 9                    41261
5023        PL      2012-12-30 00:00:00.000 10                   41261
5023        PL      2012-12-31 00:00:00.000 11                   41261

By seeing this, you will realize that above query generates the same Diff value for all the consecutive dates. Now, you can easily group this and get the counts as you like.