Here is the dummy data ，这是一个通话记录数据表。

这是它的一瞥:

|  call_id  |   customer   |   company   |     call_start      | 
|-----------|--------------|-------------|---------------------|
|1411482360 | 001143792042 | 08444599175 | 2014-07-31 13:55:03 |
|1476992122 | 001143792042 | 08441713191 | 2014-07-31 14:05:10 |

customer 和 company 字段代表他们的电话号码。

• 要求根据以下逻辑计算总“ yield ”和总“损失”值:
  

编辑:

-客户A调用A公司。
- 如果客户 A 调用公司 B，则公司 B 将获得 +1 yield ，而公司 A 将损失 +1。
-如果客户 A 调用 C 公司，那么 C 公司将获得 +1 yield ，而 B 公司将损失 +1。
-如果客户A再次调用C公司，则溢出/ yield 不会受到影响。
-只有在客户 A 打出第二个电话后， yield /损失才会发挥作用。

- 如果客户按以下顺序调用公司:A、B、B、C、A、A、C、B、D，则流程应如下所示:

A ->  
B ->  B +1 gain,  A +1 lost
B ->  
C ->  C +1 gain,  B +1 lost
A ->  A +1 gain,  C +1 lost
A ->  
C ->  C +1 gain,  A +1 lost
B ->  B +1 gain,  C +1 lost
D ->  D +1 gain,  B +1 lost

经过上述过程，我们应该得到总值:

Company    Total gain    Total lost
  A            1             2            
  B            2             2       
  C            2             2         
  D            1             0     

我开始研究这个，但这是错误的，它只是一个想法，它没有根据上述条件给我单独的增量增益和损失值:

DROP TABLE IF EXISTS GetTotalGainAndLost;

CREATE TEMPORARY TABLE IF NOT EXISTS GetTotalGainAndLost
    AS 
        (
        SELECT SUM(count) as 'TotalGainAndLost', `date`, DAY(`date`) as 'DAY' 
        FROM (SELECT count(*) as 'count', customer, `date` 
            FROM (SELECT customer, company, count(*) AS 'count', DATE_FORMAT(`call_end`,'%Y-%m-%d') as 'date' 
                FROM calls 
                WHERE `call_end` LIKE CONCAT(2014, '-', RIGHT(CAST(concat('0', 01) AS CHAR),2),'-%')
                GROUP BY customer, company, DAY(`call_end`) ORDER BY `call_end` ASC)
            as tbl1 group by customer, `date` having count(*) > 1) 
        as tbl2 GROUP by `date`
        );

Select * from GetTotalGainAndLost;

DROP TABLE GetTotalGainAndLost;

此查询未显示任何结果。

• 所需的输出如下所示:
  

每个公司和日期应该一行(例如 1 月，按天计算的总 yield 和损失调用)

|  company    |  totalGain |  totalLost  |     date     |  DAY  | 
|-------------|------------|-------------|--------------|-------|
| 08444599175 |     17     |       6     | 2014-07-01   |  1    |
| 08444599175 |     12     |      10     | 2014-07-02   |  2    |
| 08444599175 |      3     |       6     | 2014-07-02   |  3    |
| 08444599175 |   ....     |      ...    |     ...      | ...   |
| 08444599175 |      7     |       6     | 2014-07-31   | 31    |

最佳答案

简化

让 N 表示公司出现的次数。让我们尝试用三个简单的规则来简化公式。

1. 第一个出现的公司将有N - 1个 yield ，N个损失。
2. 中型公司将有 N 次 yield ，N 次损失。
3. 最后一家公司将有 N 次 yield ，N - 1 次损失

测试

在你的例子中:

• 从A公司开始，出现3次。
• B公司出现3次
• C公司出现2次
• 以出现 1 次的公司 D 结尾。

结果

Company      Gain           Lost  
  A            2             3            
  B            3             3       
  C            2             2         
  D            1             0    

转换为 SQL

首先我们从统计每家公司的出现次数开始。

SELECT
    company, COUNT(*) AS gain, COUNT(*) AS lost, DATE(call_start) AS date
FROM calls 
GROUP BY DATE(call_start), company

然后，我们开始为每个客户选择每个公司第一次出现的编号。

SELECT company, -COUNT(*) AS gain, 0 AS lost, DATE(call_start) AS `date`
FROM calls INNER JOIN (
    SELECT MIN(call_id) AS call_id FROM calls GROUP BY DATE(call_start), customer
) AS t ON (calls.call_id = t.call_id)
GROUP BY DATE(call_start), calls.company

最后出现的公司数量。

SELECT company, 0 AS gain, -COUNT(*) AS lost, DATE(call_start) AS `date`
FROM calls INNER JOIN (
    SELECT MAX (call_id) AS call_id FROM calls GROUP BY DATE(call_start), customer
) AS t ON (calls.call_id = t.call_id)
GROUP BY DATE(call_start), calls.company

结合SQL

最后，我们可以使用 UNION ALL 将整个 SQL 组合在一起，然后进行另一个分组。

SELECT company, SUM(gain) AS gain, SUM(lost) AS lost, `date` FROM (
    (
        SELECT
            company, COUNT(*) AS gain, COUNT(*) AS lost, DATE(call_start) AS `date`
        FROM calls 
        GROUP BY DATE(call_start), company
    ) UNION ALL (
        SELECT company, -COUNT(*) AS gain, 0 AS lost, DATE(call_start) AS `date`
        FROM calls INNER JOIN (
            SELECT MIN(call_id) AS call_id FROM calls GROUP BY DATE(call_start), customer
        ) AS t ON (calls.call_id = t.call_id)
        GROUP BY DATE(call_start), calls.company
    ) UNION ALL (
        SELECT company, 0 AS gain, -COUNT(*) AS lost, DATE(call_start) AS `date`
        FROM calls INNER JOIN (
            SELECT MAX(call_id) AS call_id FROM calls GROUP BY DATE(call_start), customer
        ) AS t ON (calls.call_id = t.call_id)
        GROUP BY DATE(call_start), calls.company
    )
) AS t
GROUP BY `date`, company

澄清

上面的查询假设每个新的一天都是独立的。例如，

• 客户 A 调用公司 A(第 1 天)
• 客户 A 调用 B 公司(第 1 天)B 获得 1，A 损失 1
• 客户 A 调用 C 公司(第 1 天)C 获得 1，B 损失 1
• 客户 A 调用公司 D(第 2 天)
• 客户 A 调用 E 公司(第 2 天)E 获得 1，D 损失 1

结果是

COM   G     L   DAY
 ----------------
A     0     1    1
B     1     1    1
C     1     0    1
D     0     1    2
E     1     0    2

关于MySql:通过多个条件获取递增项目的计数，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/28046327/