集团通过总结变量

Question

SELECT deposit.numberSuccessfulDeposits, count(distinct userid) 
FROM deposit WHERE deposit.asOfDate between '2016-04-01 00:00:00' and '2016-04-03 23:59:59' 
AND deposit.licenseeId = 1306 
GROUP BY deposit.numberSuccessfulDeposits 

样本输出

numberSuccessfulDeposits  count(distinct userid) 
0 228 
1 878 
2 90 
3 37 
4 17 

然而，如果Bob在周二做出周一1个存3个存款，那么这将计入这两个“1”和“3”号的成功存款。

numberSuccessfulDeposits  count(distinct userid) 
0 ## 
1 1 
2 ## 
3 1 
4 ## 

理想的情况下，它应该只对 “4”

numberSuccessfulDeposits  count(distinct userid) 
0 ## 
1 ## 
2 ## 
3 ## 
4 1 

思考算什么？

Answer 1

计算子查询中的每用户总和，然后计算主查询中的总计数。

SELECT totalDeposits, COUNT(*) 
FROM (SELECT userid, SUM(numberOfSuccessfulDeposits) AS totalDeposits 
     FROM deposit 
     WHERE deposit.asOfDate between '2016-04-01 00:00:00' and '2016-04-03 23:59:59' 
     AND deposit.licenseeId = 1306 
     GROUP BY userid) AS subquery 
GROUP BY totalDeposits 

Answer 2

将分组更改为基于用户，并对所有存款进行求和。再算上用户对这些存款的总和每个：

SELECT 
    numberSuccessfulDeposits, 
    COUNT(userid) AS users_count 
FROM (
    SELECT 
    sum(numberSuccessfulDeposits) AS numberSuccessfulDeposits, 
    userid 
    FROM deposit 
    WHERE asOfDate between '2016-04-01 00:00:00' and '2016-04-03 23:59:59' 
    AND licenseeId = 1306 
    GROUP BY userid 
) t 
GROUP BY numberSuccessfulDeposits 

编辑：分组deposints为0，1，2，3 +类看起来像：

SELECT 
    numberSuccessfulDeposits, 
    COUNT(userid) AS user_count 
FROM (
    SELECT 
    CASE WHEN numberSuccessfulDeposits >= 3 THEN '3+' ELSE numberSuccessfulDeposits::TEXT END AS numberSuccessfulDeposits, 
    userid 
    FROM (
    SELECT 
     sum(numberSuccessfulDeposits) AS numberSuccessfulDeposits, 
     userid 
    FROM deposit 
    WHERE asOfDate between '2016-04-01 00:00:00' and '2016-04-03 23:59:59' 
     AND licenseeId = 1306 
    GROUP BY userid 
    ) t 
) f 
GROUP BY numberSuccessfulDeposits