我有一个名为客户的表格,用于保存客户的数据
id | fname | lname
--- | ------ | ------
1 | John | Smith
2 | Mike | Bolton
3 | Liz | John
4 | Mark | Jobs
我还有一个名为来电的表,可以保持每个客户的每次通话。
id | timestamp | customer_id | campaign | answered |
1 |2016-09-05 15:24:08| 1 | 2016-09 | 1 |
2 |2016-09-05 15:20:08| 2 | 2016-09 | 1 |
3 |2016-08-05 15:20:08| 2 | 2016-08 | 1 |
4 |2016-08-05 13:20:08| 3 | 2016-08 | 1 |
5 |2016-08-01 15:20:08| 3 | 2016-08 | 0 |
5 |2016-08-01 12:20:08| 4 | General | 1 |
广告系列常规不计入计算。
我需要根据每个客户通话记录获得通话质量排名的客户列表。
此列表用于呼叫客户,以便:
看起来应该是这样的:
| id | fname | lname | %ans | called actual campaign | total calls | rank |
|----|--------|-------|------|------------------------|-------------|------|
| 4 | Mark | Jobs | N/A | no | 0 | 1 |
| 3 | Liz | John | 50 | no | 2 | 2 |
| 1 | John | Smith | 100 | yes | 1 | 3 | No Show
| 2 | Mike | Bolton| 100 | yes | 2 | 4 | No Show
请帮助我!
答案 0 :(得分:1)
针对指定广告系列的每个客户总呼叫和已接听电话的查询
select
c.id,
count(*) as total_calls,
sum(case when answered=1 then 1 else 0 end) as answered_calls
from customer c
join calls cs on c.id=cs.customer_id
where cs.campaign='2016-09'
group by c.id
然后,您可以使用上面的查询作为子查询来订购
select sub.id, (@rank:=@rank+1) as rank
from (the subquery above) sub, (select @rank:=1)
order by
case when sub.total_calls=0 then 0 else 1,
sub.total_calls,
sub.answered_calls*100/sub.total_calls
您可以在结果查询中包含任何所需的列