Question

Here is the dummy data，这是一个电话记录数据表。

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'增益'和总'丢失'值：

修改

- 客户A致电公司A.
- 如果客户A致电B公司，那么B公司将获得+1收益，而A公司将失去+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发出第二个电话后，才会发挥收益/损失。

- 如果客户按此顺序呼叫公司：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

在上述过程之后，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

简化

将 N 表示为公司出现的次数。让我们尝试用三个简单的规则简化公式。

出现的第一家公司将获得N - 1收益，N亏损。
中间公司将有N收益，N亏。
最后一家公司将有N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。
公司B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的每个客户的号码。

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，然后按照。

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致电公司B（第1天）B获得1，A丢失1
客户A致电公司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致电公司E（第2天）E获得1，D失去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

Answer 2

这应该有效 -

CTEGains 了解公司每个客户每个日期出现的次数。

CTEFirst 查明该公司当天是否是该客户的第一次联系。

CTELast 查明该公司是否是当天该客户的最后一次联系。

然后代码应遵循您指出的逻辑。

CREATE TEMPORARY TABLE CTEGains (RNo int, customer varchar(14), company varchar(16), startdate date, gains int)
CREATE TEMPORARY TABLE CTEFirst (customer varchar(14), call_start date, company varchar(16))
CREATE TEMPORARY TABLE CTELast (customer varchar(14), call_start date, company varchar(16))
Insert into CTEGains
Select ROW_NUMBER() over (partition by customer order by Customer) Rno, customer, company, Convert(date,call_start) startdate, count(company) gains 
from calls
group by customer, company, Convert(date,call_start), call_start

Insert into CTEFirst
Select customer, min(Convert(date,call_start)) call_start, min(company) company
from calls
group by customer, Convert(date,call_start)

Insert into CTELast
Select customer, max(Convert(date,call_start)) call_start, max(company) company
from #calls
group by customer, Convert(date,call_start)

Select c1.company, 
SUM(gains) - case when exists (Select * from CTEGains c2 where c2.customer = max(c1.customer) and max(c1.Rno) = c2.Rno - 1 and c1.company = c2.company and c1.startdate = c2.startdate) then 1 else 0 end --Didn't gain as same company called
           - case when exists (select * from CTEFirst c2 where c2.company = c1.company and c2.call_start = c1.startdate) then 1 else 0 end TotalGain -- Didn't gain as first company
, SUM(gains) - case when exists (Select * from CTEGains c2 where c2.customer = max(c1.customer) and max(c1.Rno) = c2.Rno - 1 and c1.company = c2.company and c1.startdate = c2.startdate) then 1 else 0 end --Didn't lose as same company as last called
             - case when exists (select * from CTELast c2 where c2.company = c1.company and c2.call_start = c1.startdate) then 1 else 0 end TotalLost -- didn't lose as last company
, startdate [date], DatePart(DAY, startdate) [Day]
from CTEGains c1
group by c1.company, c1.startdate

Drop Table CTEFirst
Drop Table CTEGains
Drop Table CTELast

Answer 3

我认为最简单的方法是使用两个查询。首先，我们可以获得总收益，计算每个客户对不同公司的每次通话：

select g.company company, count(g.call_id) gain
from calls c
join calls g on c.customer = g.customer and c.company <> g.company and c.call_start < g.call_start
left join calls m on g.customer = m.customer and g.company <> m.company and g.call_start > m.call_start and m.call_start > c.call_start
where m.call_id is null
group by g.company;

如果客户向各个公司拨打各种电话，则需要左连接才能计算额外的收益（即，如果客户电话公司a，b和c公司c只有一个获得，而不是两个）。

采用相同方法的总损失：

select l.company company, count(l.call_id) lost
from calls c
join calls l on c.customer = l.customer and c.company <> l.company and c.call_start > l.call_start
left join calls m on l.customer = m.customer and l.company <> m.company and c.call_start > m.call_start and l.call_start < m.call_start
where m.call_id is null
group by l.company;

这里有一个小小的演示解决方案：http://sqlfiddle.com/#!2/3236ab/7

Answer 4

让我们先做一些定义：

非首次通话：任何不是第一次致电客户的电话。
非最后一次通话：任何不是客户最后一次通话的电话。

我们已经介绍了 first 和 last 的概念，这意味着我们需要在我们的调用集上定义一个总订单。我们可以遵循我们想要的任何规则，但出于解释的目的，我假设呼叫按开始时间排序，并且在相同的开始时间由id排序。换句话说：

如果callA.sartTime < callB.startTime，则callA < callB
如果callA.startTime = callB.startTime和callA.id = callB.id，则callA < callB

请注意我们如何使用以下查询获取集合的所有非首次调用：

SELECT *
FROM calls AS non_first_calls
    RIGHT JOIN calls
    ON non_first_calls.customer = calls.customer
    AND non_first_calls.call_start >= calls.call_start
    AND non_first_calls.call_id > calls.call_id
WHERE non_first_calls.call_id IS NOT NULL

（查询输出有重复，即调用可以出现多次）

同样，我们可以按如下方式获取所有非最后一次调用：

SELECT *
FROM calls AS non_last_calls
    RIGHT JOIN calls
    ON non_last_calls.customer = calls.customer
    AND non_last_calls.call_start <= calls.call_start
    AND non_last_calls.call_id < calls.call_id
WHERE non_last_calls.call_id IS NOT NULL

业务逻辑

每次客户拨打任何其他电话后，公司都会获得+1。这意味着，对于任何给定的公司，其收益等于它收到的非首次呼叫的数量。同样，公司的损失等于它收到的非最后一次通话的数量。

强大的查询

因此，对于每家公司，我们只需要计算它收到的非首次呼叫和非最后呼叫的数量。

每个公司的部分意味着我们需要获得完整的公司列表。我们可以使用此查询执行此操作：

SELECT DISTINCT company FROM calls

全部放在一起：

SELECT

    -- The company
    companies.company

    -- How many non-first calls (gains) it has received
    ,(SELECT COUNT(DISTINCT non_first_calls.call_id) gains
        FROM calls AS non_first_calls
        RIGHT JOIN calls
            ON non_first_calls.customer = calls.customer
            AND non_first_calls.call_start >= calls.call_start
            AND non_first_calls.call_id > calls.call_id
        WHERE non_first_calls.company = companies.company
    ) gains

    -- How many non-last calls (losses) it has received    
    ,(SELECT COUNT(DISTINCT non_last_calls.call_id) gains
        FROM calls AS non_last_calls
        RIGHT JOIN calls
            ON non_last_calls.customer = calls.customer
            AND non_last_calls.call_start <= calls.call_start
            AND non_last_calls.call_id < calls.call_id
    WHERE non_last_calls.company = companies.company
    ) losses

-- From the set of all companies
FROM (SELECT DISTINCT company FROM calls) companies

关于效果

我不确定在处理大量数据时此查询的效率是否可以接受。

至少你需要（customer，call_start）（按此顺序）和（company）上的另一个索引的组合索引。这是我在此查询上运行EXPLAIN后获得的输出，他提到了索引和您提供的示例数据。

Output of EXPLAIN

MySql：按多个条件获取递增项目的计数

4 个答案:

简化

测试

转换为SQL

结合SQL

澄清