MySQL 简化这个查询？

Question

我已经编写 MySQL 查询很长时间了，但是在涉及到这个问题时我完全觉得自己像个新手，而且我似乎无法弄清楚如何简化这个查询。基本上我只是想从我们的数据库中生成一个以逗号分隔的收入列表，最终插入谷歌图表。我还必须为“过去 31 天”制作一个，所以我有点搞砸了，除非我想出了如何简化它。

我的表中的数据库列是...

• 时间：订单进来的时间 int()（UNIX 时间戳格式）
• 价格：十进制（10,2）的订单是多少

SELECT
 CONCAT(
 (SELECT SUM(price) FROM orders WHERE time < (UNIX_TIMESTAMP() - (6 * 86400)) AND time > (UNIX_TIMESTAMP() - (7 * 86400))),
 ',',
 (SELECT SUM(price) FROM orders WHERE time < (UNIX_TIMESTAMP() - (5 * 86400)) AND time > (UNIX_TIMESTAMP() - (6 * 86400))),
 ',',
 (SELECT SUM(price) FROM orders WHERE time < (UNIX_TIMESTAMP() - (4 * 86400)) AND time > (UNIX_TIMESTAMP() - (5 * 86400))),
 ',',
 (SELECT SUM(price) FROM orders WHERE time < (UNIX_TIMESTAMP() - (3 * 86400)) AND time > (UNIX_TIMESTAMP() - (4 * 86400))),
 ',',
 (SELECT SUM(price) FROM orders WHERE time < (UNIX_TIMESTAMP() - (2 * 86400)) AND time > (UNIX_TIMESTAMP() - (3 * 86400))),
 ',',
 (SELECT SUM(price) FROM orders WHERE time < (UNIX_TIMESTAMP() - 86400) AND time > (UNIX_TIMESTAMP() - (2* 86400)) ),
 ',',
 (SELECT SUM(price) FROM orders WHERE time > (UNIX_TIMESTAMP() - 86400))
 ) as chart_rev 
FROM orders_basic
LIMIT 0,1

如果可能的话，我需要像这样以 24 小时为增量。

一如既往，我们非常感谢任何帮助。谢谢！

解决方案

请注意：仔细查看您的结果，它们可能存在，只是顺序错误，如果是，请反转您的排序。

SELECT GROUP_CONCAT(total_sum) AS sum_list FROM
(
    SELECT
        FLOOR((UNIX_TIMESTAMP() - time) / 86400) AS date,
        SUM(price) AS total_sum
    FROM
        orders_basic
    WHERE
        is_testorder < 1
          AND FROM_UNIXTIME(time) > DATE_SUB(NOW(), INTERVAL 1 WEEK)
    GROUP BY
        date
) AS s

Answer 1

可能是这样的吗？

SELECT GROUP_CONCAT(total_sum) AS sum_list FROM
(
    SELECT
        DATE(FROM_UNIXTIME(time)) AS date,
        SUM(price) AS total_sum
    FROM
        orders
    WHERE
        FROM_UNIXTIME(time) > DATE_SUB(NOW(), INTERVAL 1 MONTH)
    GROUP BY
        date
) AS s

这应该为您提供上个月销售总额的逗号分隔列表。但它会按日期分组，而不是 24 小时增量（根据time 列，截止日期为午夜）。