Question

我想找出哪个学生没有提交他的任务以及什么日期。我想检查每个学生当前与否。我不介意答案是在sql还是vba代码中。更多规格如下：

Task Table
-------------------------------
SubID  |ID | Task  | Date 
-------------------------------
1      |1  | Dance | 01-01-2014
2      |1  | Sing  | 02-01-2014
3      |1  | shout | 05-01-2014
4      |2  | try   | 02-01-2014
5      |3  | Okay  | 01-01-2014
6      |2  | random| 06-01-2014
8      |4  |Jumping| 01-01-2014
9      |4  | try   | 02-01-2014
10     |4  | Piano | 03-01-2014
11     |4  | try   | 04-01-2014
12     |4  | guitar| 05-01-2014
13     |4  | try   | 06-01-2014

Student table --the Date is in the dd-mm-yyyy format. - also it is a date/time datatype

ID | Name | Current
--------------------
1  | Ron  | YES
2  | sqlJ | YES
3  | jque | NO
4  | holy | YES
5  | htdoc| YES

期望的结果：

谁未在01-01-2014和06-01-2014

之间提交任务

ID | Date
---------------
1  | 03-01-2014
1  | 04-01-2014
1  | 06-01-2014
2  | 01-01-2014
2  | 03-01-2014
2  | 04-01-2014
2  | 05-01-2014
3  | 02-01-2014
3  | 03-01-2014
3  | 04-01-2014
3  | 05-01-2014
3  | 06-01-2014

我尝试过：

SELECT w.ID, w.Date, student.[first name], student.[last name], student.[id]
FROM tasktbl AS w 
     right join student 
           on w.id = student.[id];

//I was thinking of using a vba-for loop to iterate over the range of date and store it in an array spot every Id that doesn't have a date but it didn't work out quite well.

任何帮助，从伪代码到sql代码到vba代码（基本上任何暗示我的任务）都将不胜感激

Answer 1

我使用了一个日历表，每个日期包含一行。将该表与学生交叉加入，并将日期范围限制为6个值，这给了我30行（6个日期乘以5个学生）：

SELECT sub.*
FROM
    (
        SELECT c.the_date, s.ID
        FROM tblCalendar AS c, student AS s
        WHERE c.the_date Between #1/1/2014# And #1/6/2014#
    ) AS sub

然后我将其用作子查询，并将LEFT JOIN编辑为 tasktbl 。所以＆＃34;右边的行＃34;是空的，是那些学生在相关日期没有完成任务的地方。

SELECT sub.ID AS student_id, sub.the_date
FROM
    (
        SELECT c.the_date, s.ID
        FROM tblCalendar AS c, student AS s
        WHERE c.the_date Between #1/1/2014# And #1/6/2014#
    ) AS sub
    LEFT JOIN tasktbl AS t
    ON (sub.ID = t.ID) AND (sub.the_date = t.Date)
WHERE t.ID Is Null
ORDER BY sub.ID, sub.the_date;

Answer 2

您应该可以在没有任何VBA的情况下执行此操作。

select s.id, t.date
from 
( --this will populate every possible date, for every student
    select distinct s.id 
    from students s
) s
cross join 
(
    select distinct t.date
    from taskTable t
) t
left join taskTable tt on s.id = tt.id and t.date = tt.date
where tt.date is null
--(optional) order by clause, if needed

基本上这是为每个可能的学生（来自＆＃34;＆＃34;表）拉一个条目，为每个可能的日期（来自＆＃34; t＆＃34;表）拉一个条目进行交叉连接，生成一个结果集，其中包含每个可能学生的每个可能日期的行。通过将此连接返回到taskTable（其中taskTable的日期字段为空）将仅返回交叉连接表中的条目，这些条目不具有与之关联的任务。

我还建议您重新考虑TaskTable的设计，而不是仅跟踪已完成的任务，跟踪所有任务并添加一个字段＆＃34;已完成＆＃34;如果任务已完成（是），则存储（否）。

找到缺少的日期 - vba Sql

2 个答案: