Self Join to Subtract Customer Out Time of a Row from Customer In Time of the Next Row

The problem presented in this question is related to calculating the free time between consecutive customers for a waiter. The query provided attempts to achieve this, but it yields incorrect results. This article will delve into the issue with the original query and provide a corrected approach using self-joins.

Understanding the Problem

Given a table t containing information about waiters and their respective customer interactions (in and out times), we want to calculate the free time between consecutive customers for each waiter. The first customer of the day should have an initial free time of 0, while subsequent customers’ free times are calculated by subtracting their in-time from the next customer’s out-time.

The Original Query

The original query attempts to solve this problem using a self-join:

DECLARE @start DATE, @end DATE
SET @start = '01/01/2020'
SET @end = '03/03/2020'

SELECT Waiter, t.Customer_In_Time, Customer_Out_Time,
       COALESCE(SUM(DATEDIFF(MINUTE, prev_cot, Customer_in_time)), 0) AS free_minutes
FROM (
    SELECT t.*,
           TOP 1 t2.*
    FROM t OUTER APPLY (
        SELECT TOP (1) t3.*
        FROM t t3
        WHERE t3.Waiter = t.Waiter AND t3.date = t.date AND
              t3.Customer_in_time < t.Customer_in_time
        ORDER BY t.Waiter, t3.Customer_in_time
    ) tprev
) t
WHERE t.date BETWEEN @start AND @end
AND Waiter IS NOT NULL
AND Room_no NOT LIKE '%TH%'
GROUP BY Waiter, Customer_In_Time, Customer_Out_Time
ORDER BY t.Waiter, t.Customer_In_Time;

The query has several issues. Firstly, it only considers the first row as the previous row for the entire group. This is because the TOP 1 clause in the subquery selects the first matching record based on the order specified (Waiter and then Customer_in_time). To fix this, we need to use a numbering function to assign unique numbers to each waiter’s records within their respective groups.

Correct Approach using Self-Join

To correctly calculate the free time between consecutive customers for each waiter, we can use the ROW_NUMBER() function to assign unique numbers to each waiter’s records:

SELECT Waiter, Customer_In_Time, Customer_Out_Time,
       LAG(Customer_Out_Time) OVER (PARTITION BY Waiter ORDER BY Customer_In_Time) AS prev_cot
INTO #temp
FROM @t t;

SELECT Waiter, Customer_In_Time, Customer_Out_Time,
       DATEDIFF(MINUTE, LAG(Customer_Out_Time) OVER (PARTITION BY Waiter ORDER BY Customer_In_Time), Customer_in_time) AS free_minutes
FROM #temp;

This corrected approach uses the LAG() function to access the previous row’s customer out-time for each waiter. The OVER clause specifies that we want to partition by waiter and order by customer in time, allowing us to calculate the free minutes correctly.

Why ROW_NUMBER() is Used

The original query used ROW_NUMBER() but applied it incorrectly. It created a numbering function within groups but did not select just the last row for each waiter. By using PARTITION BY with the OVER clause, we ensure that we are partitioning by waiter and ordering by customer in time, allowing us to access the previous row’s customer out-time.

Conclusion

In conclusion, the original query had several issues that prevented it from calculating the free time between consecutive customers for each waiter correctly. By using a self-join with the ROW_NUMBER() function, we can accurately calculate the free minutes and provide a corrected solution to this problem.

Last modified on 2024-06-04