SQL。 如何处理/比较/查找同一表中不同行内的差异

Question

我有一张桌子，看起来像这样：

ID   Date      Size   Marked
1    2010-02-02    2        X
2    2002-02-02    1
1    2010-02-03    2        X
2    2010-02-03    3        
3    2010-02-03    4        X

我有一个执行以下操作的代码（PHP）：a）计算每天的大小之和b）查找当天与最后一天的总数之差。 c）查找当天已标记的行的大小总和（昨天未标记相同ID的行）。

作为示例，我将得到以下结果：

Date        Total  DiffWithYesterday  MarkedThisDay
2010-02-02    3      0                  0
2010-02-03    9      6                  4

我有一种方法可以用SQL编写。 但是我在SQL方面相当虚弱，因此在一天内围绕内部联接，分组依据和嵌入式选择进行测试后，我放弃了。

如果您能给我一些提示，我将不胜感激。

哦..我正在使用MySQL。

问候，维克多

Answer 1

这样的事情在SQL Server中起作用。 我没有要测试的MySQL，但是一旦看到逻辑就可以转换。

create table so (sodate datetime, sosize int, somarked varchar(1))

insert into so (sodate,sosize,somarked) values ('1-jan-2010',3,'X')
insert into so (sodate,sosize,somarked) values ('2-jan-2010',1,'X')
insert into so (sodate,sosize,somarked) values ('3-jan-2010',2,'X')
insert into so (sodate,sosize,somarked) values ('4-jan-2010',0,null)
insert into so (sodate,sosize,somarked) values ('5-jan-2010',2,null)
insert into so (sodate,sosize,somarked) values ('6-jan-2010',1,null)
insert into so (sodate,sosize,somarked) values ('6-jan-2010',4,null)
insert into so (sodate,sosize,somarked) values ('6-jan-2010',1,null)
insert into so (sodate,sosize,somarked) values ('7-jan-2010',3,'X')
insert into so (sodate,sosize,somarked) values ('8-jan-2010',3,'X')
insert into so (sodate,sosize,somarked) values ('9-jan-2010',2,null)
insert into so (sodate,sosize,somarked) values ('10-jan-2010',2,'X')
insert into so (sodate,sosize,somarked) values ('11-jan-2010',1,'X')
insert into so (sodate,sosize,somarked) values ('12-jan-2010',2,null)
insert into so (sodate,sosize,somarked) values ('13-jan-2010',3,'X')

select so.sodate
    ,sum(so.sosize) as Total
    ,isnull(sum(so.sosize),0) - isnull(min(so2.sosize),0) as DiffFromYesterday
    ,sum(case when so.somarked = 'X' then so.sosize end) as MarkedThisDay
from so
    left join (select so.sodate,sum(so.sosize) sosize from so group by sodate) so2 on dateadd(dd,1,so2.sodate) = so.sodate
group by so.sodate  

..并在安装mysql之后似乎可以在其中工作...

select so.sodate
    ,sum(so.sosize) as Total
    ,ifnull(sum(so.sosize),0) - ifnull(min(so2.sosize),0) as DiffFromYesterday
    ,sum(case when so.somarked = 'X' then so.sosize end) as MarkedThisDay
from so
    left join (select so.sodate,sum(so.sosize) sosize from so group by sodate) so2 on (so2.sodate + INTERVAL 1 day )= so.sodate
group by so.sodate  ;

Answer 2

玩得很开心。

SELECT 
  today.date as Date, 
  today.total as Total, 
  (today.total - yesterday.total) as DiffWithYesterday  , 
  marked.total as MarkedThisDay
FROM
  (SELECT date, sum(size) as total
   FROM table_name
   GROUP BY date) today
LEFT JOIN 
  (SELECT date, sum(size) as total
   FROM table_name
   WHERE marked = 'X'
   GROUP BY date) marked ON today.date = marked.date
LEFT JOIN
  (SELECT (date + INTERVAL 1 day) as date, sum(size) as total
   FROM table_name
   GROUP BY date) yesterday ON today.date=yesterday.date

显然，您需要将“ table_name”替换为表名