如何在 LEFT JOIN 幫助下從 MySQL 表中減去值?
可以藉助一個包含一些值的兩個表的示例,然後藉助 LEFT JOIN 減去值來理解。在此,我們將採用包含以下資料的兩個表 -
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)
以上兩個表分別具有產品的當前價格和之前的價格。現在,以下查詢將使用 LEFT JOIN 找出儲存在兩個表中相同產品之間的價格差。
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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