MySQL 如何检查数据是否为NULL

您可以使用IF()函数来检查数据是否为NULL。让我们首先创建一个表 −

mysql> create table DemoTable
(
   Id int NOT NULL AUTO_INCREMENT PRIMARY KEY,
   Name varchar(200),
   Age int
);
Query OK, 0 rows affected (0.44 sec)

使用insert命令将记录插入到表中 −

mysql> insert into DemoTable(Name,Age) values('John',23);
Query OK, 1 row affected (0.12 sec)
mysql> insert into DemoTable(Name,Age) values('Sam',null);
Query OK, 1 row affected (0.12 sec)
mysql> insert into DemoTable(Name,Age) values('Mike',23);
Query OK, 1 row affected (0.20 sec)
mysql> insert into DemoTable(Name,Age) values('David',21);
Query OK, 1 row affected (0.21 sec)
mysql> insert into DemoTable(Name,Age) values('Carol',null);
Query OK, 1 row affected (0.13 sec)

使用select命令从表中显示记录 −

mysql> select *from DemoTable;

这将产生以下输出 −

+----+-------+------+
| Id | Name  | Age  |
+----+-------+------+
| 1  | John  | 23   |
| 2  | Sam   | NULL |
| 3  | Mike  | 23   |
| 4  | David | 21   |
| 5  | Carol | NULL |
+----+-------+------+
5 rows in set (0.00 sec)

这是一个检查数据是否为NULL的查询。对于记录中可见的NULL，此查询将添加一条消息 −

mysql> select if(Age IS NULL,'Age is missing',Age) from DemoTable;

这将产生以下输出 −

+--------------------------------------+
| if(Age IS NULL,'Age is missing',Age) |
+--------------------------------------+
| 23                                   |
| Age is missing                       |
| 23                                   |
| 21                                   |
| Age is missing                       |
+--------------------------------------+
5 rows in set (0.00 sec)

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