我們如何藉助 LEFT JOIN 來減去 MySQL 表中的值?
可以透過一個示例來理解,其中兩個表具有一些值,並且藉助 LEFT JOIN 減去值。這裡我們取兩個包含以下資料的表 −
mysql> Select * from value_curdate; +----+----------+-------+ | Id | Product | Price | +----+----------+-------+ | 1 | Notebook | 100 | | 2 | Pen | 40 | | 3 | Pencil | 65 | +----+----------+-------+ 3 rows in set (0.00 sec) mysql> Select * from value_prevdate; +----+-----------+-------+ | Id | Product | Price | +----+-----------+-------+ | 1 | Notebook | 85 | | 2 | Pen | 34 | | 3 | Pencil | 56 | | 4 | Colors | 65 | | 5 | Fevistick | 25 | +----+-----------+-------+ 5 rows in set (0.00 sec)
以上兩張表分別具有產品的當前價格和先前價格。現在,以下查詢透過使用 LEFT JOIN 將找出儲存在兩個表中的相同產品的價格差異。
mysql> Select value_curdate.id, value_curdate.product, value_curdate.price as Curprice,value_prevdate.price as 'prevprice', value_curdate.price-value_prevdate.price as 'Difference' from value_curdate LEFT JOIN value_prevdate ON value_curdate.id = value_prevdate.id ; +----+----------+----------+-----------+------------+ | id | product | Curprice | prevprice | Difference | +----+----------+----------+-----------+------------+ | 1 | Notebook | 100 | 85 | 15 | | 2 | Pen | 40 | 34 | 6 | | 3 | Pencil | 65 | 56 | 9 | +----+----------+----------+-----------+------------+ 3 rows in set (0.00 sec)
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