Question

我在mysql数据库中有一个包含货币交易的表。假设tabled被称为 transactions ：

id int,
to_user int,
from_user int,
created datetime,
amount double

我正在尝试创建一个查询，该查询返回按每天显示的交易数量分组的平均交易金额。适用的列是金额并创建。

所以我试图回答这个问题：在0到2个交易，2到4个交易，4到6个交易之间等的日子里发生的平均交易量是多少。

有什么想法吗？

Answer 1

经过几次误解后，我终于得到了你所需要的东西：）

SELECT  cnt_range * 2 AS days_range,
        CASE WHEN SUM(trans_cnt) > 0 THEN
             SUM(trans_sum) / SUM(trans_cnt)
             ELSE 0
        END AS average_amount
FROM    (
        SELECT  SUM(amount) AS trans_sum,
                COUNT(*) AS trans_cnt,
                FLOOR(COUNT(*) / 2) AS cnt_range
        FROM    transactions
        GROUP BY
                TO_DATE(created)
        ) ao
GROUP BY
        cnt_range

Answer 2

这是一个尝试（但没有经过测试）：

SELECT CASE WHEN txn_per_day BETWEEN 0 AND 2 THEN 2
            WHEN txn_per_day BETWEEN 2 AND 4 THEN 4
            WHEN txn_per_day BETWEEN 4 AND 6 THEN 6
            ELSE 8 -- or more
       END CASE AS num_txns, AVG(t3.amount)
FROM (
  SELECT t1.*, COUNT(*) AS txn_per_day
  FROM transactions t1 JOIN transactions t2 
    ON (TO_DAYS(t1.created) = TO_DAYS(t2.created))
  GROUP BY t1.id
) t3
GROUP BY num_txns;

这不是特别快。生成报告是可以的，但是如果你需要它在PHP请求期间运行，那就不太好了。

Answer 3

SELECT AVG(`num`), ((`num` - 1) DIV 2) * 2 AS `tier`
FROM (
    SELECT DATE_FORMAT(`created`, '%Y-%m-%d') AS `day`, COUNT(*) AS `num`
    FROM `yourtable`
    GROUP BY 1
) AS `src`
GROUP BY `tier`

Answer 4

尝试类似

的内容

SELECT AVG(`amount`) as `amount`,
    DATE_FORMAT(`created`, '%Y-%m-%d') as `date`
FROM `transactions`
GROUP BY DATE_FORMAT(`created`, '%Y-%m-%d')

在Date-and-time functions处查看更多内容，并查看BETWEEN function

每日实例摘要的SQL查询

4 个答案: