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

lc8prwob  于 2021-06-20  发布在  Mysql
关注(0)|答案(1)|浏览(511)

我有table:

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');

我只需要一张唱片

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

但我得到的结果是:这个查询是错误的,但它显示了我所需要的

record un_ter  u_t_a
1   2
1   1

我想得到结果:

record un_ter  u_t_a
2   3

你能给我一些建议吗?我该怎么做?谢谢。

eivgtgni

eivgtgni1#

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;

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