0

我有这两张MySQL表：

desc students;
+---------------------------+---------------+
| Field                     | Type          |
+---------------------------+---------------+
| student_id                | INT(11)       |
| student_firstname         | VARCHAR(255)  |
| student_lastname          | VARCHAR(255)  |
+---------------------------+---------------+

desc studentabsence;
+---------------------------+---------------+
| Field                     | Type          |
+---------------------------+---------------+
| student_id                | INT(11)       |
| student_absence_startdate | date          |
| student_absence_enddate   | date          |
+---------------------------+---------------+

如果我运行以下查询：

SELECT 
    * 
FROM 
    students s 
        INNER JOIN studentabsence sa 
            ON s.student_id = sa.student_id
WHERE s.student_id = '1'

我得到以下输出：

+------------+-------------------+-----------------+------------+---------------------------+-------------------------+
| student_id | student_firstname | student_lastname| student_id | student_absence_startdate | student_absence_enddate |
+------------+-------------------+-----------------+------------+---------------------------+-------------------------+
|       1    | John              | Doe             | 1          | 2012-08-01                | 2012-08-08              |
+------------+-------------------+-----------------+------------+---------------------------+-------------------------+

现在，我有<form>一个用户可以选择学生的位置和日期范围。我想知道是否可以将用户选择的两个日期与数据库值进行比较，如果学生拥有数据的范围之间的所有工作日都存在匹配回显。

因此，如果用户发布以下表单，则使用上面的示例：

student_id = '1'
student_absence_startdate = '2012-08-02'
student_absence_enddate = '2012-08-06'

我想echo：

2012-08-02
2012-08-03
2012-08-06

任何指向正确方向的指针都会受到高度赞赏。

4

1 回答 1

0

我仅根据查询发布此回复。此外，如果您将数据类型保留为日期，将对您有很大帮助。

SELECT * FROM students s
INNER JOIN studentabsence sa ON s.student_id = sa.student_id
AND s.student_id = '1'
AND student_absence_startdate >= 2012-08-02
AND student_absence_enddate < 2012-08-06
于 2012-08-13T09:56:04.713 回答