如何将这两个查询合并为一个？ （针对同一个表的多个连接）

Question

给出两个表，一个用于工人，一个用于工人完成的任务，

CREATE TABLE IF NOT EXISTS `workers` (
  `id` int(11) NOT NULL,
  PRIMARY KEY  (`id`)
);

INSERT INTO `workers` (`id`) VALUES
(1);


CREATE TABLE IF NOT EXISTS `tasks` (
  `id` int(11) NOT NULL,
  `worker_id` int(11) NOT NULL,
  `status` int(11) NOT NULL,
  PRIMARY KEY  (`id`)
);

INSERT INTO `tasks` (`id`, `worker_id`, `status`) VALUES
(1, 1, 1),
(2, 1, 1),
(3, 1, 2),
(4, 1, 2),
(5, 1, 2);

我正在尝试获取每个工作人员拥有的每个状态代码的任务数量。

我可以说

SELECT w.*
,COUNT(t1.worker_id) as status_1_count
FROM workers w
LEFT JOIN tasks t1 ON w.id = t1.worker_id AND t1.status = 1 
WHERE 1 
GROUP BY 
t1.worker_id
ORDER BY w.id

或

SELECT w.*
,COUNT(t2.worker_id) as status_2_count
FROM workers w
LEFT JOIN tasks t2 ON w.id = t2.worker_id AND t2.status = 2
WHERE 1 
GROUP BY 
t2.worker_id
ORDER BY w.id

并获取具有单个给定状态代码的任务数量，但是当我尝试在单个查询中获取多个任务状态的计数时，它不起作用！

SELECT w.*
,COUNT(t1.worker_id) as status_1_count
,COUNT(t2.worker_id) as status_2_count
FROM workers w
LEFT JOIN tasks t1 ON w.id = t1.worker_id AND t1.status = 1 
LEFT JOIN tasks t2 ON w.id = t2.worker_id AND t2.status = 2
WHERE 1 
GROUP BY t1.worker_id
,t2.worker_id
ORDER BY w.id

如果我不愿意，那么任务表就会交叉加入自身！

有没有办法将这两个查询合并为一个，以便我们可以在一个查询中检索多个任务状态的计数？

谢谢！

Answer 1

SELECT w.*,
  SUM(t1.status = 1) AS status_1_count,
  SUM(t1.status = 2) AS status_2_count
FROM workers w
  LEFT JOIN tasks t1 ON w.id = t1.worker_id AND t1.status IN (1, 2) 
GROUP BY w.id
ORDER BY w.id;

Answer 2

  

我正在尝试获取每个工作人员拥有的每个状态代码的任务数量。

SELECT worker_id, status, COUNT(*)
    FROM tasks
    GROUP BY worker_id, status;

就是这样。

Answer 3

我这里没有MySQL的实例，但是我在t-sql框上测试了它并且它有效。

select distinct(worker_id), 
       (select count(*) from tasks where status = 1) as Status1, 
       (select count(*) from tasks where status = 2) as Status2 
   from tasks;