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计算 MySQL 中 NOT NULL 值的存在

2023-06-16 17:15:03 444

要计算NOTNULL值的存在，请使用聚合函数COUNT(yourColumnName)。让我们首先创建一个表-

mysql> create table DemoTable
   (
   Id int NOT NULL AUTO_INCREMENT PRIMARY KEY,
   NumberOfQuestion int,
   NumberOfSolution int
   );

使用插入命令在表中插入一些记录。在这里，一些值是NULL-

mysql> insert into DemoTable(NumberOfQuestion,NumberOfSolution) values(20,10);

mysql> insert into DemoTable(NumberOfQuestion,NumberOfSolution) values(20,2);

mysql> insert into DemoTable(NumberOfQuestion,NumberOfSolution) values(20,NULL);

mysql> insert into DemoTable(NumberOfQuestion,NumberOfSolution) values(20,NULL);

mysql> insert into DemoTable(NumberOfQuestion,NumberOfSolution) values(30,19);

mysql> insert into DemoTable(NumberOfQuestion,NumberOfSolution) values(30,1);

使用select语句显示表中的所有记录-

mysql> select *from DemoTable;

这将产生以下输出-

+----+------------------+------------------+
| Id | NumberOfQuestion | NumberOfSolution |
+----+------------------+------------------+
| 1  | 20               | 10               |
| 2  | 20               | 2                |
| 3  | 20               | NULL             |
| 4  | 20               | NULL             |
| 5  | 30               | 19               |
| 6  | 30               | 1                |
+----+------------------+------------------+
6 rows in set (0.00 sec)

以下是计算值存在的查询。相同的值应该不是NULL，即不会计算NULL值-

mysql> SELECT NumberOfQuestion, COUNT(NumberOfSolution) as NumberOfRows FROM DemoTable GROUP BY NumberOfQuestion;

这将产生以下输出-

+------------------+--------------+
| NumberOfQuestion | NumberOfRows |
+------------------+--------------+
| 20               | 2            |
| 30               | 2            |
+------------------+--------------+
2 rows in set (0.00 sec)

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计算 MySQL 中 NOT NULL 值的存在

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