Question

类别

| id | tID | category |
-----------------------
| 1  | 1   | Foo      |
| 2  | 1   | Bar      |
| 3  | 2   | Fox      |
| 4  | 3   | Box      |
| 5  | 2   | Chair    |

上传

| id | category | data | act
| 1  | 1        | some |  1
| 2  | 1        | thin |  0
| 3  | 3        | fat  |  0
| 4  | 2        | key  |  0
| 5  | 1        | lad  |  1

从类别中选择类别，并计算从上传中具有相同类别ID的总上传ID，其中act等于1。

结果

|category| count|
|   1    |   2  |
|   2    |   0  |
|   3    |   0  |
|   4    |   0  |
|   5    |   1  |

到目前为止我的尝试。

SELECT COUNT(u.category), c.category FROM categories c LEFT JOIN upload u ON c.id = u.category GROUP BY u.category

查询应该是什么？

Answer 1

正确的答案可能是以下几行。如果这不对，那么你需要更仔细地考虑你的标准......

DROP TABLE IF EXISTS categories;

CREATE TABLE categories
(id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,tID INT NOT NULL
,category VARCHAR(12) NOT NULL
);

INSERT INTO categories VALUES
(1,1,'Foo'),
(2,1,'Bar'),
(3,2,'Fox'),
(4,3,'Box'),
(5,2,'Chair');

DROP TABLE IF EXISTS upload;

CREATE TABLE upload
(id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,category INT NOT NULL
,data VARCHAR(12) NOT NULL
,act TINYINT NOT NULL
);

INSERT INTO upload VALUES
(1,1,'some',1),
(2,1,'thin',0),
(3,3,'fat',0),
(4,2,'key',0),
(5,1,'lad',1);

SELECT c.category
     , COALESCE(SUM(u.act = 1),0) ttl 
  FROM categories c 
  LEFT 
  JOIN upload u 
    ON u.category = c.id 
 GROUP 
    BY c.category;
+----------+------+
| category | ttl  |
+----------+------+
| Bar      |    0 |
| Box      |    0 |
| Chair    |    0 |
| Foo      |    2 |
| Fox      |    0 |
+----------+------+

计算两个表之间的所有相关字段

1 个答案: