Question

这是我为了获取用户选择的计划而编写的查询。但是这将返回 usersubscription 表中的记录，即使用户未订阅（如果表中没有与用户对应的记录）。

$userid=$_POST['userid'];
$videoid=$_POST['videoid'];
$subscribedquery=$this->db->query("select id from usersubscription where plan_id IN 
        (SELECT DISTINCT plan_id FROM subscribed_videos sv where sv.videoid = $videoid)
        OR id IN (SELECT DISTINCT assosiated_plan_id 
        FROM subscription_groups sg
        JOIN subscribed_videos sv ON sv.plan_id = sg.plan_id
        WHERE sv.videoid = $videoid) and user_id=$userid");

下面我分享所有表格的结构。

CREATE TABLE IF NOT EXISTS `subscribed_videos` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `plan_id` int(11) NOT NULL,
  `videoid` int(11) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=26 ;

INSERT INTO `subscribed_videos` (`id`, `plan_id`, `videoid`) VALUES
(7, 2, 1),
(8, 2, 2),
(14, 1, 3),
(15, 1, 4),
(16, 1, 5),
(17, 1, 21),
(18, 1, 28),
(19, 1, 2),
(20, 3, 4),
(21, 3, 6),
(24, 5, 25),
(25, 6, 5);

CREATE TABLE IF NOT EXISTS `subscription_groups` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `plan_id` int(11) NOT NULL,
  `assosiated_plan_id` int(11) NOT NULL,
  `added_on` int(11) NOT NULL,
  PRIMARY KEY (`id`)
);


INSERT INTO `subscription_groups` (`id`, `plan_id`, `assosiated_plan_id`, `added_on`) VALUES
(1, 1, 1, 0),
(2, 2, 2, 0),
(3, 3, 3, 0),
(4, 4, 1, 0),
(5, 4, 2, 0),
(6, 4, 3, 0),
(12, 5, 5, 0),
(13, 5, 1, 0),
(14, 5, 2, 0),
(15, 6, 1, 0);

CREATE TABLE IF NOT EXISTS `subscription_plans` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `plan` varchar(256) NOT NULL,
  `days_limit` int(11) NOT NULL,
  `added_on` int(11) NOT NULL,
  `status` int(11) NOT NULL,
  `rate` decimal(6,2) NOT NULL,
  PRIMARY KEY (`id`)
);



INSERT INTO `subscription_plans` (`id`, `plan`, `days_limit`, `added_on`, `status`, `rate`) VALUES
(1, 'PlanA', 15, 1398249706, 1, 150.00),
(2, 'PlanB', 15, 1398249679, 1, 100.00),
(3, 'PlanC', 15, 1398249747, 1, 100.00),
(4, 'PlanD', 10, 1398249771, 1, 500.00),
(5, 'PlanE', 15, 1398250104, 1, 200.00),
(6, 'Plan R1', 20, 1398250104, 1, 200.00);


CREATE TABLE IF NOT EXISTS `usersubscription` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `user_id` int(11) NOT NULL,
  `plan_id` int(11) NOT NULL,
  `subscribed_on` int(11) NOT NULL,
  PRIMARY KEY (`id`)
);



INSERT INTO `usersubscription` (`id`, `user_id`, `plan_id`, `subscribed_on`) VALUES
(1, 1, 1, 1399091458);

内容：

SELECT * FROM subscribed_videos; 
+----+---------+---------+
| id | plan_id | videoid |
+----+---------+---------+
|  7 |       2 |       1 |
|  8 |       2 |       2 |
| 14 |       1 |       3 |
| 15 |       1 |       4 |
| 16 |       1 |       5 |
| 17 |       1 |      21 |
| 18 |       1 |      28 |
| 19 |       1 |       2 |
| 20 |       3 |       4 |
| 21 |       3 |       6 |
| 24 |       5 |      25 |
| 25 |       6 |       5 |
+----+---------+---------+

SELECT * FROM subscription_groups; 
+----+---------+--------------------+----------+
| id | plan_id | assosiated_plan_id | added_on |
+----+---------+--------------------+----------+
|  1 |       1 |                  1 |        0 |
|  2 |       2 |                  2 |        0 |
|  3 |       3 |                  3 |        0 |
|  4 |       4 |                  1 |        0 |
|  5 |       4 |                  2 |        0 |
|  6 |       4 |                  3 |        0 |
| 12 |       5 |                  5 |        0 |
| 13 |       5 |                  1 |        0 |
| 14 |       5 |                  2 |        0 |
| 15 |       6 |                  1 |        0 |
+----+---------+--------------------+----------+

SELECT * FROM subscription_plans;
+----+---------+------------+------------+--------+--------+
| id | plan    | days_limit | added_on   | status | rate   |
+----+---------+------------+------------+--------+--------+
|  1 | PlanA   |         15 | 1398249706 |      1 | 150.00 |
|  2 | PlanB   |         15 | 1398249679 |      1 | 100.00 |
|  3 | PlanC   |         15 | 1398249747 |      1 | 100.00 |
|  4 | PlanD   |         10 | 1398249771 |      1 | 500.00 |
|  5 | PlanE   |         15 | 1398250104 |      1 | 200.00 |
|  6 | Plan R1 |         20 | 1398250104 |      1 | 200.00 |
+----+---------+------------+------------+--------+--------+

 SELECT * FROM usersubscription
+----+---------+---------+---------------+
| id | user_id | plan_id | subscribed_on |
+----+---------+---------+---------------+
|  1 |       1 |       1 |    1399091458 |
+----+---------+---------+---------------+

小提琴：http://sqlfiddle.com/#!2/90b2e/1

如果用户已经订阅了所选视频的计划，我希望结果是这样的，否则查询应返回空记录：

id
---
1

另外，如果使用查询本身的用户的计划未过期，我该如何返回记录。即，当用户购买计划时，它将被输入 usersubscription 表中。 subscribed_on 字段将包含购买它的php unix time()。因此，我想在此查询中仅获取与用户和视频相对应的计划，该计划未过期。到期日在 subscription_plans 表的 days_limit 字段中存储为天数（例如：15）。

任何人都可以帮我找到合适的解决方案。

提前致谢。

Answer 1

这是你在看什么？

select 
id 
from 
usersubscription 
where 
(
  plan_id 
    IN 
    (
      (
        SELECT DISTINCT plan_id 
        FROM 
        subscribed_videos sv where sv.videoid = 5
      )
    )
   OR id IN 
  (
       SELECT DISTINCT assosiated_plan_id AS plan_id
       FROM subscription_groups sg
       JOIN subscribed_videos sv ON sv.plan_id = sg.plan_id
       WHERE sv.videoid = 5
   )
) 

and user_id=1;

DEMO

Answer 2

我会说你应该尝试使用连接

SELECT DISTINCT s.id ,
FROM_UNIXTIME(p.`added_on`),
DATE_ADD(FROM_UNIXTIME(s.`subscribed_on`), INTERVAL p.`days_limit` DAY) `expiry_date`
FROM usersubscription s
LEFT JOIN subscribed_videos v ON (s.plan_id = v.plan_id)
LEFT JOIN subscription_groups g ON(s.id = assosiated_plan_id )
LEFT JOIN subscribed_videos sv ON sv.plan_id = g.plan_id
LEFT JOIN `subscription_plans` p ON (p.id = s.plan_id)
WHERE s.user_id=1 AND  sv.videoid = 5
AND  v.videoid = 5
AND  DATE_ADD(FROM_UNIXTIME(s.`subscribed_on`), INTERVAL p.`days_limit` DAY) > CURRENT_DATE()

在上面的查询中，我有一个额外的联接subscription_plans来检查您的到期日期条件，还要注意您在查询中直接使用帖子变量即$userid=$_POST['userid'];$videoid=$_POST['videoid'];这是不安全的，当您使用codeigniter时您应该使用活动记录库来构建您的查询，该查询将在其自身结束时进行所有转义

Fiddle Demo

以下是上述查询的有效记录版本

$query = $this->db
    ->select('s.id')
    ->distinct()
    ->from('usersubscription s')
    ->join('subscribed_videos v ','s.plan_id = v.plan_id','LEFT')
    ->join('subscription_groups g ','s.id = assosiated_plan_id','LEFT')
    ->join('subscribed_videos sv','sv.plan_id = g.plan_id','LEFT')
    ->join('`subscription_plans` p','p.id = s.plan_id','LEFT')
    ->where('s.user_id',$userid)
    ->where('sv.videoid',$videoid)
    ->where('v.videoid',$videoid)
    ->where('DATE_ADD(FROM_UNIXTIME(s.`subscribed_on`), INTERVAL p.`days_limit` DAY) > CURRENT_DATE()',null,FALSE)
    ->get();

Answer 3

尝试将查询更改为：

$subscribedquery=$this->db->query("select id from usersubscription where plan_id IN 
    ((SELECT DISTINCT plan_id FROM subscribed_videos sv where sv.videoid = $videoid)
    OR id IN (SELECT DISTINCT assosiated_plan_id 
    FROM subscription_groups sg
    JOIN subscribed_videos sv ON sv.plan_id = sg.plan_id
    WHERE sv.videoid = $videoid)) and user_id=$userid");

我添加了括号来更改运算符优先级

子查询未返回预期结果

3 个答案:

Fiddle Demo