派生表查询不能在mysql中工作

Question

以下查询导致错误。

select monstart, 
     sum(datediff(least(m.monend, t.end_date) + interval 1 day, 
        greatest(m.monstart, t.start_date) 
        ) 
      ) as days_worked 
from travel t join 
    (select date('2016-01-01') as monstart, date('2016-01-31') as monend union all 
     select date('2016-02-01') as monstart, date('2016-02-29') as monend union all 
select date('2016-03-01') as monstart, date('2016-03-31') as monend union all 
select date('2016-04-01') as monstart, date('2016-04-30') as monend union all 
select date('2016-05-01') as monstart, date('2016-05-31') as monend union all 
select date('2016-06-01') as monstart, date('2016-06-30') as monend union all 
select date('2016-07-01') as monstart, date('2016-07-31') as monend union all 
select date('2016-08-01') as monstart, date('2016-08-31') as monend union all 
select date('2016-09-01') as monstart, date('2016-09-30') as monend union all 
select date('2016-10-01') as monstart, date('2016-10-31') as monend union all 
select date('2016-11-01') as monstart, date('2016-11-30') as monend union all 
select date('2016-12-01') as monstart, date('2016-12-31') as monend union all 
    ) m 
    on t.end_date >= m.monstart and t.start_date <= m.monend 
group by m.monstart; 

错误如下。

ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ') m 
    on t.end_date >= m.monstart and t.start_date <= m.monend 
group by m.mon' at line 19 

Answer 1

您最后还有一个额外的union all。只要删除它。

select monstart, 
     sum(datediff(least(m.monend, t.end_date) + interval 1 day, 
        greatest(m.monstart, t.start_date) 
        ) 
      ) as days_worked 
from travel t join 
    (select date('2016-01-01') as monstart, date('2016-01-31') as monend union all 
     select date('2016-02-01') as monstart, date('2016-02-29') as monend union all 
select date('2016-03-01') as monstart, date('2016-03-31') as monend union all 
select date('2016-04-01') as monstart, date('2016-04-30') as monend union all 
select date('2016-05-01') as monstart, date('2016-05-31') as monend union all 
select date('2016-06-01') as monstart, date('2016-06-30') as monend union all 
select date('2016-07-01') as monstart, date('2016-07-31') as monend union all 
select date('2016-08-01') as monstart, date('2016-08-31') as monend union all 
select date('2016-09-01') as monstart, date('2016-09-30') as monend union all 
select date('2016-10-01') as monstart, date('2016-10-31') as monend union all 
select date('2016-11-01') as monstart, date('2016-11-30') as monend union all 
select date('2016-12-01') as monstart, date('2016-12-31') as monend -- removed a union all from here 
    ) m 
    on t.end_date >= m.monstart and t.start_date <= m.monend 
group by m.monstart; 

Answer 2

你应该这样做。如下：

select monstart, 
      sum(datediff(least(m.monend, t.end_date) + interval 1 day, 
         greatest(m.monstart, t.start_date) 
         ) 
       ) as days_worked 
    from travel t join 
     (select date('2016-01-01') as monstart, date('2016-01-31') as monend union all 
      select date('2016-02-01') as monstart, date('2016-02-29') as monend union all 
    select date('2016-03-01') as monstart, date('2016-03-31') as monend union all 
    select date('2016-04-01') as monstart, date('2016-04-30') as monend union all 
    select date('2016-05-01') as monstart, date('2016-05-31') as monend union all 
    select date('2016-06-01') as monstart, date('2016-06-30') as monend union all 
    select date('2016-07-01') as monstart, date('2016-07-31') as monend union all 
    select date('2016-08-01') as monstart, date('2016-08-31') as monend union all 
    select date('2016-09-01') as monstart, date('2016-09-30') as monend union all 
    select date('2016-10-01') as monstart, date('2016-10-31') as monend union all 
    select date('2016-11-01') as monstart, date('2016-11-30') as monend union all 
    select date('2016-12-01') as monstart, date('2016-12-31') as monend 
     ) m 
     on t.end_date >= m.monstart and t.start_date <= m.monend 
    group by m.monstart; 

Answer 3

您简化查询和像这样的东西删除加入伪表：不必更新查询

select 
    sum(datediff(least(monend, end_date) + interval 1 day,greatest(monstart, start_date))) as `days_worked`, 
    (adddate(last_day(start_date), 1) - interval 1 month) as `monstart`, 
    last_day(start_date) as `monend` 
from travel 
group by `monstart`,`monend` 

这将在任何一年的工作。

编辑 - 试试这个：

select 
    (adddate(last_day(start_date), 1) - interval 1 month) as `monstart`, 
    last_day(start_date) as `monend`, 
    sum(
     datediff(
      least(last_day(start_date), end_date) + interval 1 day, 
      greatest((adddate(last_day(start_date), 1) - interval 1 month), start_date)) 
    ) as `days_worked` 
from travel 
group by `monstart`,`monend`