在 MySQL 中,如何通过连接维护数据驱动的表关系?


事实上,有时我们可以避免表中数据驱动的关系,需要将它们连接起来。借助 SELECT 列表中的 CASE 语句可以实现此目的,以处理连接可能性。为了理解这一点,我们举一个三个数据驱动的表的示例,它们分别是具有以下数据的“Student_Detail”——

mysql> Select * from student_detail;
+----+---------+
| Id | Name    |
+----+---------+
| 1  | Harshit |
| 2  | Rahul   |
| 3  | Aarav   |
+----+---------+
3 rows in set (0.00 sec)

现在,我们有三个表,分别是“Student_Harshit”、“Student_Rahul”、“Student_Aarav”,它们分别对学生 Harshit、Rahul 和 Aarav 具有说明。它们具有以下数据——

mysql> Select * from Student_Harshit;
+----+-----------+
| Id | Remarks   |
+----+-----------+
| 1  | Excellent |
+----+-----------+
1 row in set (0.00 sec)

mysql> Select * from Student_Rahul;
+----+---------+
| Id | Remarks |
+----+---------+
| 2  | Average |
+----+---------+
1 row in set (0.00 sec)

mysql> Select * from Student_Aarav;
+----+-------------+
| Id | Remarks     |
+----+-------------+
| 3  | Intelligent |
+----+-------------+
1 row in set (0.00 sec)

现在,以下查询将处理这些数据驱动的表——

mysql> Select sd.id, sd.name, CASE name WHEN 'Harshit' THEN H1.Remarks WHEN 'Rahul' THEN R1.Remarks WHEN 'Aarav' THEN A1.Remarks ELSE 'Error' END as REMARKS FROM Student_detail AS sd LEFT JOIN Student_Harshit AS H1 ON sd.id = H1.id LEFT JOIN Student_Rahul AS R1 ON sd.id = R1.id LEFT JOIN Student_Aarav AS A1 on sd.id = A1.id;
+----+---------+-------------+
| id | name    | REMARKS     |
+----+---------+-------------+
| 1  | Harshit | Excellent   |
| 2  | Rahul   | Average     |
| 3  | Aarav   | Intelligent |
+----+---------+-------------+
3 rows in set (0.00 sec)

更新时间:20-Jun-2020

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