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 -
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)
以上是我们如何模拟返回多个表达式的 MySQL INTERSECT 查询?的详细内容。更多信息请关注PHP中文网其他相关文章!