MySQL 如何计算具有特定值的列的数量

以下是语法 –

选择
sum(yourColumnName1+yourColumnName2+yourColumnName3...N) as `anyAliasName1`,
sum(yourColumnName1 and yourColumnName2 and yourColumnName3….N) as anyAliasName
从yourTableName;

让我们创建一个表 –

mysql> create table demo36
-> (
-> id int not null auto_increment primary key,
-> value1 int,
-> value2 int,
-> value3 int
-> );
Query OK, 0 rows affected (1.68 sec)

使用插入命令将一些记录插入表中 –

mysql> insert into demo36(value1,value2,value3) values(1,0,0);
Query OK, 1 row affected (0.14 sec)

mysql> insert into demo36(value1,value2,value3) values(1,0,1);
Query OK, 1 row affected (0.11 sec)

mysql> insert into demo36(value1,value2,value3) values(1,1,1);
Query OK, 1 row affected (0.22 sec)

mysql> insert into demo36(value1,value2,value3) values(0,1,0);
Query OK, 1 row affected (0.12 sec)

mysql> insert into demo36(value1,value2,value3) values(1,1,0);
Query OK, 1 row affected (0.10 sec)

使用select语句从表中显示记录 –

mysql> select *from demo36;

这将产生以下输出 –

+----+--------+--------+--------+
| id | value1 | value2 | value3 |
+----+--------+--------+--------+
| 1  | 1      | 0      | 0      |
| 2  | 1      | 0      | 1      |
| 3  | 1      | 1      | 1      |
| 4  | 0      | 1      | 0      |
| 5  | 1      | 1      | 0      | 
+----+--------+--------+--------+
5 rows in set (0.00 sec)

以下是在MySQL中计算具有特定值的列的数量的查询 –

mysql> select
-> sum(value1+value2+value3) as `Total_Number_Of_1`,
-> sum(value1 and value2 and value3) as `Total_Number_Of_Rows_With_1`
-> from demo36;

这将产生以下输出 –

+-------------------+-----------------------------+
| Total_Number_Of_1 | Total_Number_Of_Rows_With_1 |
+-------------------+-----------------------------+
|                 9 |                           1 |
+-------------------+-----------------------------+
1 row in set (0.00 sec)

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