MySQL过去7天的计数数据

MySQL没有递归功能,所以您只需使用数字表技巧-

1. DROP TABLE IF EXISTS `example`.`numbers`;
   CREATE TABLE  `example`.`numbers` (
     `id` int(10) unsigned NOT NULL auto_increment,
      PRIMARY KEY  (`id`)
   ) ENGINE=InnoDB DEFAULT CHARSET=latin1;
   

2. 使用以下命令填充表:

   INSERT INTO NUMBERS
     (id)
   VALUES
     (NULL)
   

3. 使用 DATE_ADD

   SELECT x.dt
     FROM (SELECT DATE_SUB(CURRENT_DATE, INTERVAL (n.id - 1) DAY) AS dt
             FROM numbers n
            WHERE n.id <= 7 ) x
   

4. 基于datetime部分将左连接到数据表上。

      SELECT x.dt,
              COUNT(v.aid) AS num
        FROM (SELECT DATE_SUB(CURRENT_DATE, INTERVAL (n.id - 1) DAY) AS dt
                FROM numbers n
               WHERE n.id <= 7 ) x
   LEFT JOIN VOTES v ON DATE(FROM_UNIXTIME(v.timestamp)) = DATE(x.dt)
    GROUP BY x.dt
    ORDER BY x.dt
   

简单-日期可以根据数字生成,如我提供的示例中所示。它还意味着使用单个表,而不是说每个数据类型使用一个表。

以前:

  SELECT DATE(FROM_UNIXTIME(v.timestamp)) AS dt,
         COUNT(v.aid)
    FROM VOTES v
   WHERE DATE(FROM_UNIXTIME(v.timestamp)) BETWEEN DATE_SUB(CURRENT_DATE, INTERVAL 7 DAY)
                                              AND CURRENT_DATE
GROUP BY DATE(FROM_UNIXTIME(v.timestamp))

选择COUNT(*)作为投票,aid FROM table WHERE timestamp>@7DaysAgo GROUP BY aid