Question

我正在尝试应用递归逻辑。

我在tbl_appointment表中有以下数据（recur_type：1 =每周，2 =每月，0 =不递归）：

appointment_id    user_id    date        recur_type    .....
18                56         2014-06-02  1
19                56         2014-06-15  2
20                56         2014-06-20  0
21                56         2014-07-20  2

我有以下标准来获取数据：如果我获取 2014年7月（第7个月）的数据，那么

appointment_id(21)（1行）和...
appointment_id(18)每周递归，7月份重复获取（4行）：
```
appointment_id    user_id    date        recur_type    .....
18                56         2014-07-07  1
18                56         2014-07-14  1
18                56         2014-07-21  1
18                56         2014-07-28  1
```
注意：日期已更改，因为每周的约会是递归的，这意味着我会在每个日期添加7天。 2014-06-02 + 7天= 2014-06-09等。因此，对于7月，日期是2014-07-07。
appointment_id(19)是按月递归的，7月份的重复次数被提取（1行）：
```
appointment_id    user_id    date        recur_type    .....
19                56         2014-07-15  2
```
注意：日期已更改，因为每个月的约会都是递归的，这意味着我会在该日期添加一个月。

最终输出是（总共6行）：

appointment_id    user_id    date        recur_type    .....
21                56         2014-07-20  2
18                56         2014-07-04  1
18                56         2014-07-11  1
18                56         2014-07-18  1
18                56         2014-07-15  1
19                56         2014-07-15  2

我尝试了以下代码：

SELECT
    tu.email,
    ta.appointment_id,
    ta.user_id,
    ta.date,
    ta.time,
    ta.recur_type,
    0 recursive
FROM
    tbl_appointment ta
        LEFT JOIN
    tbl_user tu ON ta.user_id = tu.user_id
WHERE
    1 AND YEAR(ta.date) = '2014'
      AND MONTH(ta.date) = '06'
      AND ta.user_id = 56
UNION ALL SELECT
    tu.email,
    ta.appointment_id,
    ta.user_id,
    ta.date,
    ta.time,
    ta.recur_type,
    1 recursive
FROM
    tbl_appointment ta
        LEFT JOIN
    tbl_user tu ON ta.user_id = tu.user_id
WHERE
    1 AND recur_type = '2'
      AND ta.user_id = 56
UNION ALL SELECT
    tu.email,
    ta.appointment_id,
    ta.user_id,
    ta.date,
    ta.time,
    ta.recur_type,
    2 recursive
FROM
    tbl_appointment ta
        LEFT JOIN
    tbl_user tu ON ta.user_id = tu.user_id
WHERE
    1 AND recur_type = '1'
      AND ta.user_id = 56
ORDER BY date DESC, time

如何满足上述要求？

Answer 1

对每个递归类型使用查询，联合在一起。

每周和每月递归使用几个交叉连接查询来生成要添加到日期的一系列数字。这可以应对多达1000次重复约会，但很容易扩展到更多（如果约会要重复超过20年）。

SELECT a.appoinemnt_id, a.user_id, a.recur_type, a.date AS appoint_date
FROM tbl_appointment a
WHERE a.recur_type = 0
HAVING appoint_date BETWEEN '2014-07-01' AND '2014-07-31'
UNION
SELECT a.appoinemnt_id, a.user_id, a.recur_type, DATE_ADD(a.date, INTERVAL units.i + tens.i * 10 WEEK) AS appoint_date
FROM tbl_appointment a
CROSS JOIN (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9)units
CROSS JOIN (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9)tens
WHERE a.recur_type = 1
HAVING appoint_date BETWEEN '2014-07-01' AND '2014-07-31'
UNION
SELECT a.appoinemnt_id, a.user_id, a.recur_type, DATE_ADD(a.date, INTERVAL units.i + tens.i * 10 MONTH) AS appoint_date
FROM tbl_appointment a
CROSS JOIN (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9)units
CROSS JOIN (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9)tens
WHERE a.recur_type = 2
HAVING appoint_date BETWEEN '2014-07-01' AND '2014-07-31'

这里的SQL小提琴： -

http://www.sqlfiddle.com/#!2/6346a2/2

PHP + MySQL中的递归逻辑

1 个答案: