Question

我有一个活动策划者mysql数据库，并且我试图获取一份城市表的报告，以按每次活动的天数显示所有城市排序。

例如一些记录是：

event_id - city_id - start - end
1 - 88 - 2019-01-01 - 2019-01-05
2 - 89 - 2019-02-01 2019-02-03
3 - 90 - 2019-03-01 2019-03-04
4 - 88 - 2019-01-06 - 2019-01-07

该报告可能是：

city_id - days
  88    -  7
  90    -  4
  89    -  3

Answer 1

使用以下查询-

select 
   event_id, city_id, DATEDIFF(end, start) as days
 from 
   table_name
 order by DATEDIFF(end, start) desc

Answer 2

您可以通过以下查询来做到这一点：

SELECT event_id , city_id, SUM( DATEDIFF( `end`,`start`)+1 )  as days
FROM sample
GROUP BY city_id
order by SUM( DATEDIFF( `end`,`start`) +1  ) DESC;

样本

MariaDB [test]> select * from sample;

+----------+---------+------------+------------+
| event_id | city_id | start      | end        |
+----------+---------+------------+------------+
|        1 |      88 | 2019-01-01 | 2019-01-05 |
|        2 |      89 | 2019-02-01 | 2019-02-03 |
|        3 |      90 | 2019-03-01 | 2019-03-04 |
|        4 |      88 | 2019-01-06 | 2019-01-07 |
+----------+---------+------------+------------+
4 rows in set (0.229 sec)

MariaDB [test]> 
MariaDB [test]> SELECT event_id , city_id, SUM( DATEDIFF( `end`,`start`)+1 )  as days
    -> FROM sample
    -> GROUP BY city_id
    -> order by SUM( DATEDIFF( `end`,`start`) +1  ) DESC;
+----------+---------+------+
| event_id | city_id | days |
+----------+---------+------+
|        1 |      88 |    7 |
|        3 |      90 |    4 |
|        2 |      89 |    3 |
+----------+---------+------+
3 rows in set (0.001 sec)

MariaDB [test]>

查询可按两个日期之间的天数排序

2 个答案: