Question

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_type`) VALUES
(1, '2018-06-01', 20.33333, 21.33333, 1, 'accident', NULL),
(2, '2018-07-02', 21.33333, 23.33333, 1, 'accident', NULL),
(3, '2018-06-03', 21.33333, 24.33333, 1, 'accident', NULL),
(4, '2018-06-04', 25.33333, 26.33333, 1, 'travel', NULL),
(5, '2018-06-04', 21.33333, 26.33333, 2, 'travel', NULL),
(6, '2018-06-04', 21.33333, 26.33333, 2, 'accident', NULL),
(7, '2018-06-04', 21.33333, 26.33333, 2, 'travel', NULL),
(8, '2018-06-04', 21.33333, 26.33333, 3, 'travel', NULL),
(9, '2018-08-04', 19.33333, 26.33333, 4, 'travel', NULL);

我只需要一个记录计数（ut.tr_id）

select  count(distinct ut.tr_id) as count_tr from un_tr ut group  by ut.tr_id having count(ut.ut_id)>1

但我得到了结果：

count_tr
1
1

我想得到结果：

count_tr
2

您想给我建议吗，我该怎么做？感谢。

Answer 1

您查询

select  count(distinct ut.tr_id) as count_tr 
from un_tr ut 
group  by ut.tr_id 
having count(ut.ut_id)>1

可以简化为

select  ut.tr_id
from un_tr ut 
group  by ut.tr_id 
having count(ut.ut_id)>1

因为count(distinct ut.tr_id)没有意义：group by ut.tr_id因此此计数始终每ut.tr_id。

您现在可以通过以下方式计算不同tr_id次出现次数：

select count(*)
from
(
   select  ut.tr_id
   from un_tr ut 
   group  by ut.tr_id 
   having count(ut.ut_id)>1

) as t

Answer 2

您可以尝试此查询：

select count(*) as count_tr from (
   select distinct tr_id as dtr_id, count(ut_id)
   from un_tr 
   group by dtr_id
   having count(ut_id) > 1
  ) a;

结果：

+----------+
| count_tr |
+----------+
|        2 |
+----------+

count（*）+ group by + having

2 个答案: