Question

这是我的问题：

NAME    DATE           PRESENTS
bob     march 3 2011      1
bob     june 6 2008       2
bob     jan 3 2012        3
mary    feb 14 1986       4
mary    april 10 2001     5
mary    jan 3 2012        6
kate    march 3 2011      7
kate    jan 3 2012        8
kate    oct 9 2013        9
celia   march 3 2011     10
celia   feb 14 1986      11
celia   july 4 2011      12
celia   jan 3 2012       13
celia   feb 14 1991      14

因此，我们的目标是，如果他们在那些日子里收到任何礼物，我们会在同一天添加凯特和西莉亚的礼物数量。

我需要做的就是这样，除了更大的数据集。数据集的大小可能约为100,000个条目。

我需要SQL中的答案，或者作为Access 2003查询。

Answer 1

如果presents是礼物凯特或西莉亚或其他人在特定日期的礼物数量：

select name, date, sum(presents) total_on_date
from table
where name in ('kate','celia')
group by name, date

_{我绝不是一个sql天才}

Answer 2

select name, date, count(presents) total_on_date
from table
where name in ('kate','celia')
group by name, date

假设您要计算礼物ID

Answer 3

你必须加入回归，就像这样：

select DATE, SUM(PRESENTS) as TOTAL_JOINT_PRESENTS
from mytable t1
join mytable t2 on t1.DATE = t2.DATE
where t1.NAME = 'kate'
and t2.NAME = 'celia';

Answer 4

与波希米亚人给出的答案有关： -

我认为他想要计算这两个人来的日子......

select count(*)
from mytable t1 join mytable t2 
     on t1.DATE = t2.DATE 
where t1.NAME = 'kate' and t2.NAME = 'celia';

Answer 5

如果你想要

礼物凯特和西莉亚在同一天得到的数量

然后：

SELECT
    presents1.date AS presents_date,
    (COUNT(*) + 1) AS presents_count
FROM
    presents presents1,
    presents presents2
WHERE presents1.date = presents2.date
AND presents1.name = 'kate'
AND presents2.name = 'celia'
GROUP BY presents1.date

假设表名为presents。

如何在另一列的分组中对一列进行求和？

5 个答案: