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The next problem we ll look at requires you to combine, or merge, all overlapping sessions for the same user into one session group, returning the user, start time, and end time of the session group . The purpose of such a request is to determine the amount of time a user was connected, regardless of the number of simultaneous active sessions the user had . The solution to this problem would be especially helpful to service providers that allow multiple sessions at no extra charge . You might want to tackle the problem in steps: identify starting times of session groups, identify ending times of session groups, and then match each ending time to its corresponding starting time .
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To isolate starting times of session groups, you first need to come up with a logical way of identifying them . A start time S starts a group if no session (for the same user) starts before S and continues until S or later . With this definition of a session group start time, if you have multiple identical start times, you will get them all . By applying DISTINCT, you will get only one occurrence of each unique start time . Here s the query that translates this logic to T-SQL:
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SELECT DISTINCT username, starttime FROM dbo.Sessions AS O WHERE NOT EXISTS (SELECT * FROM dbo.Sessions AS I WHERE I.username = O.username AND O.starttime > I.starttime AND O.starttime <= I.endtime);
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This generates the following output:
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username ---------User1 User1 User2 User2 User2 User2 User3 User3 starttime ---------------------2009-12-01 08:00:00.00 2009-12-01 10:00:00.00 2009-12-01 08:00:00.00 2009-12-01 11:00:00.00 2009-12-01 11:32:00.00 2009-12-01 12:04:00.00 2009-12-01 08:00:00.00 2009-12-01 09:30:00.00
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To identify end times of session groups, you essentially use the inverse of the previous logic . An end time E ends a group if there is no session (for the same user) that had already begun by time E but that ends after E . Here s the query returning the ending times of session groups:
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SELECT DISTINCT username, endtime FROM dbo.Sessions AS O WHERE NOT EXISTS (SELECT * FROM dbo.Sessions AS I WHERE I.username = O.username AND O.endtime >= I.starttime AND O.endtime < I.endtime);
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username ---------User1 User1 User2 User2 User2 User2 User3 User3 endtime ---------------------2009-12-01 09:30:00.00 2009-12-01 12:30:00.00 2009-12-01 10:30:00.00 2009-12-01 11:30:00.00 2009-12-01 12:00:00.00 2009-12-01 12:30:00.00 2009-12-01 09:00:00.00 2009-12-01 09:30:00.00
10 Working with Date and Time
Next, you need to match a session group ending time to each session group starting time . You can achieve this by adding a row number calculation to the queries that return start and end times . Then, in the outer query, associate the right end to each start by matching the user name and row number . The row number needs to be partitioned by user name and ordered by start or end time . The tricky part is that the queries that calculate start and end times use a DISTINCT clause . Remember that in terms of logical query processing, a ROW_ NUMBER calculation that appears in the SELECT list is evaluated before the DISTINCT clause . But for our purposes, we need the row numbers to be assigned after removal of duplicates . To overcome this problem, use the DENSE_RANK function instead of ROW_NUMBER . Because DISTINCT eliminates duplicates, you will effectively get the row numbers that you need in terms of the result rows . Here s the complete solution query:
WITH StartTimes AS ( SELECT DISTINCT username, starttime, DENSE_RANK() OVER(PARTITION BY username ORDER BY starttime) AS rownum FROM dbo.Sessions AS O WHERE NOT EXISTS (SELECT * FROM dbo.Sessions AS I WHERE I.username = O.username AND O.starttime > I.starttime AND O.starttime <= I.endtime) ), EndTimes AS ( SELECT DISTINCT username, endtime, DENSE_RANK() OVER(PARTITION BY username ORDER BY endtime) AS rownum FROM dbo.Sessions AS O WHERE NOT EXISTS (SELECT * FROM dbo.Sessions AS I WHERE I.username = O.username AND O.endtime >= I.starttime AND O.endtime < I.endtime) ) SELECT S.username, S.starttime, E.endtime FROM StartTimes AS S JOIN EndTimes AS E ON S.username = E.username AND S.rownum = E.rownum;
This generates the following output:
username ---------User1 User1 User2 User2 User2 starttime ---------------------2009-12-01 08:00:00.00 2009-12-01 10:00:00.00 2009-12-01 08:00:00.00 2009-12-01 11:00:00.00 2009-12-01 11:32:00.00 endtime ---------------------2009-12-01 09:30:00.00 2009-12-01 12:30:00.00 2009-12-01 10:30:00.00 2009-12-01 11:30:00.00 2009-12-01 12:00:00.00
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