Question

我有下面提到的表：

ID    Var1      Date
A-1   5         2017-04-01 18:45:05
A-2   8         2017-04-01 18:45:05
A-3   5         2017-04-01 18:45:05
A-3   5         2017-04-02 18:45:05
A-4   8         2017-04-02 18:45:05
A-5   8         2017-04-02 18:45:05
A-6   8         2017-04-03 18:45:05

我想计算每天ID Var1值8为SELECT COUNT( ID) AS ID, ((COUNT( Var1) / (COUNT( ID) )) * 100 ) AS percentage FROM Table1 WHERE Var1=8 GROUP BY Date(Date);的百分比。我试过了，但它无法工作。

db.feed.find({
  "_id" : {
    "$in" : 
      [ObjectId("55880c251df42d0466919268"), 
       ObjectId("55bf528e69b70ae79be35006")
      ]
   }
});

Answer 1

drop table if exists t;
create table t (ID varchar(3),   Var1 int,     Dt datetime);
insert into t values
('A-1'  , 5     ,    '2017-04-01 18:45:05'),
('A-2'  , 8     ,    '2017-04-01 18:45:05'),
('A-3'  , 5     ,    '2017-04-01 18:45:05'),
('A-3'  , 5     ,    '2017-04-02 18:45:05'),
('A-4'  , 8     ,    '2017-04-02 18:45:05'),
('A-6'  , 8     ,    '2017-04-03 18:45:05');

select dt,count(id) NofID,
         sum(case when var1 = 8 then 1 else 0 end) nofeights,
         (sum(case when var1 = 8 then 1 else 0 end) / count(id)) * 100 eights
from t
group by dt;

+---------------------+-------+-----------+----------+
| dt                  | NofID | nofeights | eights   |
+---------------------+-------+-----------+----------+
| 2017-04-01 18:45:05 |     3 |         1 |  33.3333 |
| 2017-04-02 18:45:05 |     2 |         1 |  50.0000 |
| 2017-04-03 18:45:05 |     1 |         1 | 100.0000 |
+---------------------+-------+-----------+----------+
3 rows in set (0.00 sec)

Answer 2

我建议将其写成：

SELECT Date(Date), SUM(Var1 = 8) AS ID, 
       AVG(Var1 = 8) * 100 AS percentage
FROM Table1 
GROUP BY Date(Date);

这使用MySQL快捷方式，允许在数字上下文中将布尔值视为数字。

在具有特定条件的Mysql中计算

2 个答案: