mysql-从联接表中写入两个计数的JOIN查询

提问

我花了几个小时来解决这个问题,但是我无法胜任工作.希望有人可以提供帮助？

我有一个项目表和任务表,链接在project_id上.我可以通过以下查询获取project_id,project_name和status_id：

SELECT 
    a.project_id, 
    a.project_name,
    b.status_id
FROM project_list as a
INNER JOIN task_list as b
ON a.project_id=b.project_id

我想为每个项目选择一条记录,并根据status_id添加两个计数字段.用伪代码：

SELECT 
    a.project_id, 
    a.project_name,
    (SELECT COUNT(*) FROM task_list WHERE status_id < 3) as not_completed,
    (SELECT COUNT(*) FROM task_list WHERE status_id = 3) as completed
FROM project_list as a
INNER JOIN task_list as b
ON a.project_id=b.project_id
GROUP BY project_id

我的创建表脚本如下：

CREATE TABLE `project_list` (
  `project_id` int(11) NOT NULL AUTO_INCREMENT,
  `topic_id` int(11) DEFAULT NULL,
  `project_name` varchar(45) DEFAULT NULL,
  PRIMARY KEY (`project_id`)
)

CREATE TABLE `task_list` (
  `task_id` int(11) NOT NULL AUTO_INCREMENT,
  `project_id` int(11) DEFAULT NULL,
  `task_name` varchar(45) DEFAULT NULL,
  `status_id` int(11) DEFAULT '0',
  PRIMARY KEY (`task_id`)
)

任何帮助深表感谢.谢谢！

编辑：答案：

SELECT 
    a.project_id, 
    project_name,
    SUM(status_id != 3) AS not_completed,
    SUM(status_id = 3) AS completed,
    SUM(status_id IS NOT NULL) as total
FROM tasks.project_list as a
INNER JOIN tasks.task_list as b
ON a.project_id=b.project_id
GROUP BY a.project_id

最佳答案

问题在于,在子查询中,您正在计算整个表中的所有行,而不仅仅是具有正确project_id的行.您可以通过修改每个子查询中的WHERE子句来解决此问题.

(SELECT COUNT(*)
 FROM task_list AS c
 WHERE c.status_id < 3
 AND a.project_id = c.project_id)

但是,一种更简单的方法是对布尔条件使用SUM而不是COUNT来计算与条件匹配的行：

SELECT 
    a.project_id, 
    a.project_name,
    SUM(b.status_id < 3) AS not_completed,
    SUM(b.status_id = 3) AS completed,
FROM project_list as a
INNER JOIN task_list as b
ON a.project_id = b.project_id
GROUP BY project_id

这是可行的,因为TRUE的值为1,而FALSE的值为0.