MySql:按多个条件获取递增项目的计数

Question

这是虚拟数据,它是一个调用记录数据表.

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 |

在customer和company领域代表了他们的电话号码.

• 要求是根据以下逻辑计算总"增益"和总"丢失"值:
  

编辑:

- 客户A致电公司
A.-如果客户A致电公司B,则公司B将获得+1收益,而公司A将失去+1.
- 如果客户A致电公司C,则公司C将获得+1收益,而公司B将失去+1.
- 如果客户A再次致电公司C,则溢出/收益不会受到影响.
- 只有在客户A第二次拨打电话后,收益/损失才会发挥作用.

- 如果客户按此顺序呼叫公司:A,B,B,C,A,A,C,B,D过程应该是这样的:

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

经过上述过程后,我们应该将总值设为:

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

我开始研究这个但是它错了,它只是一个想法,它不会根据上述条件给我单独增加的增益和丢失值:

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;

此查询不显示任何结果.

• 所需的输出如下所示:
  

每个公司和日期应该是一行(总收益和每天丢失的电话,例如1月)

|  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    |

Answer 1

简单化

让我们将N表示为公司出现的次数.让我们尝试在三个简单的规则中简化公式.

1. 出现的第一家公司将获得N - 1收益,N亏损.
2. 中型公司将有N收益,N亏损.
3. 最后一家公司将有N个收益,N - 1损失

测试

在你的例子中:

• 从公司A开始,它出现了3次.
• B公司出现3次
• 公司C出现2次
• 以公司D结束,出现1次.

结果

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

转换为SQL

首先,我们首先计算每家公司的数量.

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

然后,我们开始选择每个公司第一次出现在每个客户的号码.

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,然后再执行另一个组.

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

上述查询假设每个新的一天都是独立的.例如,

• 客户A致电公司A(第1天)
• 客户A呼叫公司B(第1天)B获得1,A丢失1
• 客户A呼叫公司C(第1天)C获得1,B输掉1
• 客户A致电公司D(第2天)
• 客户A呼叫公司E(第2天)E获得1,D丢失1

结果将是

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