如何创建基于某些条件从表中获取值的MySQL视图？
如果要创建一个基于某些特定条件从表中获取值的视图，则在创建视图时必须使用WHERE子句。取决于WHERE子句的值将存储在视图中。使用WHERE子句创建MySQL视图的语法可以如下-
语法
Create View view_name AS Select_statements FROM table WHERE condition(s);
示例
为了说明上述概念，我们使用表'Student_info'中的以下数据-
mysql> Select * from student_info; +------+---------+------------+------------+ | id | Name | Address | Subject | +------+---------+------------+------------+ | 101 | YashPal | Amritsar | History | | 105 | Gaurav | Chandigarh | Literature | | 125 | Raman | Shimla | Computers | | 130 | Ram | Jhansi | Computers | +------+---------+------------+------------+ 4 rows in set (0.08 sec)
现在，借助以下查询，我们将创建视图名称“Info”，条件是它存储仅以计算机为主题的行。所以我们在创建视图时需要如下使用WHERE子句-
mysql> Create OR Replace VIEW Info AS Select Id, Name, Address, Subject from student_info WHERE Subject = 'Computers'; mysql> Select * from info; +------+-------+---------+-----------+ | Id | Name | Address | Subject | +------+-------+---------+-----------+ | 125 | Raman | Shimla | Computers | | 130 | Ram | Jhansi | Computers | +------+-------+---------+-----------+ 2 rows in set (0.00 sec)
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