Question

表 payment_transaction

+----+---------+--------+------------+
| ID | user_id | status | time_stamp |
+----+---------+--------+------------+
|  1 |    1    |    1   | 1414541884 |
|  2 |    2    |    0   | 1414576722 |
|  3 |    2    |    0   | 1414577273 |
|  4 |    3    |    0   | 1414782966 |
|  5 |    3    |    1   | 1414785691 |
|  6 |    4    |    0   | 1415112933 |
+----+---------+--------+------------+

此表存储所有付款交易。 status = 0表示付款失败，无论出于何种原因，status = 1表示付款成功。

我现在想设置一个cronjob，所有用户都会在过去7天内收到无法购买会员资格的消息。与上面的示例表中一样，我只需要2行，其中包含最近付款的user_id，transaction_id和最近付款的time_stampe。由于user_id = 3在首次付款失败后成功获得会员资格，因此应将其排除在外：

+---------+----------------+------------+
| user_id | transaction_id | time_stamp |
+---------+----------------+------------+
|    2    |       3        | 1414577273 |
|    4    |       6        | 1415112933 |
+---------+----------------+------------+

是否可以在一个查询中完成所有这些操作？如果是，它比使用两个查询更有效吗？

到目前为止我得到了什么：

SELECT DISTINCT
    `t`.`user_id`,
    `t`.`id` AS `transaction_id`,
    `t`.`time_stamp`
FROM `payment_transaction` AS `t`
WHERE
    `t`.`status` = 0
    AND `t`.`time_stamp` < UNIX_TIMESTAMP() - 60*60*24*7
ORDER BY
    `t`.`id`

但是，它不会将用户user_id = 3排除在外，而且还会向我提供首次付款失败，而不是最近的付款。

----编辑----

感谢RST在下面的评论中提供解决方案。经过一些额外的更改后，最终查询如下所示：

SELECT
    `t`.`user_id`,
    MAX(`t`.`id`) AS `transaction_id`,
    MAX(`t`.`time_stamp`) AS `time_stamp`
FROM `payment_transaction` AS `t`
WHERE
    `t`.`status` = 0
    AND `t`.`time_stamp` < UNIX_TIMESTAMP() - 60*60*24*7
    AND `t`.`user_id` NOT IN (
        SELECT `user_id`
        FROM `payment_transaction`
        WHERE `status` = 1
        AND UNIX_TIMESTAMP(`time_stamp`) < UNIX_TIMESTAMP() - 60*60*24*7
    )
GROUP BY
    `t`.`user_id`
ORDER BY
    `t`.`id`

Answer 1

稍微偏离您的请求，但仍应获得相同的结果，具体取决于您对应用程序进行编码的方式。假设最新交易与最准确的交易最相关吗？

如果是这样，您可以查看以下内容（未经测试）：

SELECT DISTINCT
    `t`.`user_id`,
    `t`.`id` AS `transaction_id`,
    `t`.`time_stamp`
FROM `payment_transaction` AS `t`
WHERE
    `t`.`time_stamp` < UNIX_TIMESTAMP() - 60*60*24*7
GROUP BY
    `t`.`user_id`
ORDER BY
    `t`.`id`

请注意我没有测试过这个SQL，因为我离开了我的开发机器。我建议您仔细查看GROUP BY命令的复杂性，以确保您不会遇到一些未捕获的边缘情况。

Answer 2

 DROP TABLE IF EXISTS payment_transaction;

 CREATE TABLE payment_transaction
 (ID INT NOT NULL AUTO_INCREMENT PRIMARY KEY
 ,user_id INT NOT NULL
 ,status INT NOT NULL
 ,time_stamp BIGINT NOT NULL
 );

 INSERT INTO payment_transaction VALUES
 (1 ,1 ,1 ,1414541884),
 (2 ,2 ,0 ,1414576722),
 (3 ,2 ,0 ,1414577273),
 (4 ,3 ,0 ,1414782966),
 (5 ,3 ,1 ,1414785691),
 (6 ,4 ,0 ,1415112933);

 SELECT x.* 
   FROM payment_transaction x
   JOIN 
      ( SELECT a.user_id
             , MAX(a.id) max_id
          FROM payment_transaction a 
          LEFT 
          JOIN payment_transaction b 
            ON b.user_id = a.user_id 
           AND b.status = 1 
           AND b.time_stamp > a.time_stamp 
         WHERE a.time_stamp > UNIX_TIMESTAMP(NOW())-604800
           AND a.status = 0 
           AND b.id IS NULL
      ) y
     ON y.user_id = x.user_id
    AND y.max_id = x.id;
 +----+---------+--------+------------+
 | ID | user_id | status | time_stamp |
 +----+---------+--------+------------+
 |  6 |       4 |      0 | 1415112933 |
 +----+---------+--------+------------+

MySQL：获取值= 0的特定时间范围内的所有条目，除非在同一时间范围内输入值为value = 1的条目

2 个答案: