mysql - 2 count(*)+group by+having+join

Question

我有表:

CREATE TABLE IF NOT EXISTS `un_pl` (
  `coo` double DEFAULT NULL,
  `pl` varchar(255) DEFAULT NULL,
  `gal` varchar(255) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;


INSERT INTO `un_pl` (`coo`, `pl`, `gal`) VALUES
    (21.33333, 'One', 'Zero'),
    (22.33333, 'Mars', 'Sofar');



CREATE TABLE IF NOT EXISTS `un_tr` (
 `ut_id` int(11) NOT NULL AUTO_INCREMENT,
 `date` date DEFAULT NULL,
 `s_p_c` double NOT NULL,
 `d_c` double NOT NULL,
 `tr_id` int(11) DEFAULT NULL,
 `ev_t` varchar(50) DEFAULT NULL,
 PRIMARY KEY (`ut_id`)
) ENGINE=InnoDB AUTO_INCREMENT=8 DEFAULT CHARSET=utf8;



INSERT INTO `un_tr` (`ut_id`, `date`, `s_p_c`, `d_c`, `tr_id`, `ev_t`) 
VALUES
(1, '2018-06-01', 20.33333, 21.33333, 1, 'accident'),
(2, '2018-07-02', 21.33333, 23.33333, 1, 'accident'),
(3, '2018-06-03', 21.33333, 24.33333, 1, 'accident'),
(4, '2018-06-04', 25.33333, 26.33333, 1, 'travel'),
(5, '2018-06-04', 21.33333, 26.33333, 2, 'travel'),
(6, '2018-06-04', 21.33333, 26.33333, 2, 'accident'),
(7, '2018-06-04', 21.33333, 26.33333, 2, 'travel'),
(8, '2018-06-04', 21.33333, 26.33333, 3, 'travel'),
(9, '2018-08-04', 19.33333, 26.33333, 4, 'travel');

我只需要得到一条记录 un_ter 和 u_t_a

select 
count(distinct ut.tr_id) as un_ter, 
count(case when ut.ev_t = 'accident' then 1 end) as u_t_a
from un_tr ut
left join un_pl up ON  (ut.s_p_c = up.coo) or  (ut.d_c = up.coo) 
where up.gal = 'Zero' and date between '2018-06-01' and '2018-06-31'
group by ut.tr_id
having count(ut.ut_id)>1

但我得到了结果:这个查询是错误的，但它显示了我需要的东西

record un_ter  u_t_a
1   2
1   1

我想得到结果:

record un_ter  u_t_a
2   3

你能给我建议吗，我该怎么做？谢谢。

Answer 1

这个查询不会提供想要的结果吗？

select sum(un_ter), sum(u_t_a) from (
select 
count(distinct ut.tr_id) as un_ter, 
count(case when ut.ev_t = 'accident' then 1 end) as u_t_a
from un_tr ut
left join un_pl up ON  (ut.s_p_c = up.coo) or  (ut.d_c = up.coo) 
where up.gal = 'Zero' and date between '2018-06-01' and '2018-06-31'
group by ut.tr_id
having count(ut.ut_id)>1
) t1;