我们如何模拟返回多个表达式的 MySQL INTERSECT 查询?
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由于我们无法在MySQL中使用INTERSECT查询,因此我们将使用EXIST运算符来模拟INTERSECT查询。可以通过以下示例来理解 -
示例
在此示例中,我们有两个表,即 Student_detail 和 Student_info,具有以下数据 -
mysql> Select * from Student_detail;
+-----------+---------+------------+------------+
| studentid | Name | Address | Subject |
+-----------+---------+------------+------------+
| 101 | YashPal | Amritsar | History |
| 105 | Gaurav | Chandigarh | Literature |
| 130 | Ram | Jhansi | Computers |
| 132 | Shyam | Chandigarh | Economics |
| 133 | Mohan | Delhi | Computers |
| 150 | Rajesh | Jaipur | Yoga |
| 160 | Pradeep | Kochi | Hindi |
+-----------+---------+------------+------------+
7 rows in set (0.00 sec)
mysql> Select * from Student_info;
+-----------+-----------+------------+-------------+
| studentid | Name | Address | Subject |
+-----------+-----------+------------+-------------+
| 101 | YashPal | Amritsar | History |
| 105 | Gaurav | Chandigarh | Literature |
| 130 | Ram | Jhansi | Computers |
| 132 | Shyam | Chandigarh | Economics |
| 133 | Mohan | Delhi | Computers |
| 165 | Abhimanyu | Calcutta | Electronics |
+-----------+-----------+------------+-------------+
6 rows in set (0.00 sec)
现在,使用带有 WHERE 子句的 EXIST 运算符的以下查询将模拟 INTERSECT 以返回两个表中都存在的“studentid”、姓名、地址(其中名称不是“Yashpal”) -
mysql>Select Student_detail.studentid,Student_detail.name, student_detail.address FROM student_detail WHERE Student_detail.studentid >100 AND EXISTS (SELECT * FROM Student_info WHERE Student_info.Name <> 'Yashpal' AND Student_info.studentid = Student_detail.studentid AND Student_info.name = Student_detail.name);
+-----------+--------+------------+
| studentid | name | address |
+-----------+--------+------------+
| 105 | Gaurav | Chandigarh |
| 130 | Ram | Jhansi |
| 132 | Shyam | Chandigarh |
| 133 | Mohan | Delhi |
+-----------+--------+------------+
4 rows in set (0.00 sec)
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