我们如何模拟MySQL INTERSECT查询?
由于我们无法在MySQL中使用INTERSECT查询,因此我们将使用IN运算符来模拟INTERSECT查询。通过以下示例可以理解-
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,以下使用IN运算符的查询将模拟INTERSECT以返回两个表中都存在的所有“ studentid”值-
mysql> Select Student_detail.studentid FROM Student_detail WHERE student_detail.studentid IN(SELECT Student_info.studentid FROM Student_info);+-----------+
| studentid |
+-----------+
| 101 |
| 105 |
| 130 |
| 132 |
| 133 |
+-----------+
5 rows in set (0.06 sec)
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