mysql - SQL获取同一列但同一表中不同日期的值差异

Question

如何获得特定日期的特定列的值与前一个日期的同一列的值之间的差异？

Current Table:
Date          Qty
8-Jan-17      100
9-Jan-17      120
10-Jan-17     180

Desired Output:

Date         Diff Qty
9-Jan-17      20
10-Jan-17     60

Answer 1

假设日期不重复并且日期之间也没有间隙，这将起作用。

    --Sample data as provided. This script works in SQL Server 2005+
    CREATE TABLE #Table1
        ([Date] datetime, [Qty] int)
    ;

    INSERT INTO #Table1
        ([Date], [Qty])
    VALUES
        ('2017-01-08 00:00:00', 100),
        ('2017-01-09 00:00:00', 120),
        ('2017-01-10 00:00:00', 180)
    ;

   --This script is plain SQL for any DMBS 

   select y.Date, y.Qty-x.Qty as 'Diff Qty'
   from #table1 x inner join #Table1 y 
     on x.Date+1=y.Date

结果

+-------------------------+----------+
|          Date           | Diff Qty |
+-------------------------+----------+
| 2017-01-09 00:00:00.000 |       20 |
| 2017-01-10 00:00:00.000 |       60 |
+-------------------------+----------+