sql - 令人惊讶的困惑(不可能的)查询

Question

我有一个问题，我不知道它是否可行。

我有 3 个表。一张父表和两个 child

数据将显示在网格上。与
[account_detail]empirica_score 作为行标题 和
[account_detail]mfin_score 作为列标题

像这样(一秒就能飞99%)

因此查询需要获取 epirica_score 和 mfin_score 以确定在网格上的位置(例如，99% 的位置)

并获取 [NAEDO]startdate 键匹配的 distinct 值。
大致是这样的

SELECT * 
FROM paysoft_result 
WHERE result_date >= "2014-08-01" and <= "2014-09-01"
AND ( SELECT COUNT( custno ) from Paysoft_result where result = 'successful' )
AND (.. some join to get the scores from account_detail .. )
AND (.. some join to get to NAEDO..  WHERE DISTINCT(start_date))

我对JOINS一无所知

1. 因此您将获得 2 个分数来确定排名。
2. 数量 (paysoft_result/NAEDO) * 100 = 网格值
3. 需要按客户计算。并将总计放入网格中。

我不知道如何处理这样的查询。这是一份碧 Jade 报告。

有什么想法吗？

这里有一些(希望)有用的数据

account_details 表

account_no    title    surname    firstname    empirica_score    mfin_score   
1000002222    "Mr"    "Petersen"    "Andrew peter"    602    97
1000002828    "Mr"    "Mars"    "Abraham"    599    97

NAEDO TBALE

customer_code    name                        start_date
1000002828       "Abraham Mars"              "2014-08-01"
1000002828       "Abraham Mars"              "2014-09-03"
1000002828       "Abraham Mars"              "2014-08-01"
1000002828       "Abraham Mars"              "2014-07-03"
1000002222       "Andrew peter Petersen"     "2014-05-23"
1000002222       "Andrew peter Petersen"     "2014-08-22"
1000002222       "Andrew peter Petersen"     "2014-07-23"
1000002222       "Andrew peter Petersen"     "2014-07-23"
1000002222       "Andrew peter Petersen"     "2014-06-23"

paysoft_results 表

custno         name                        result           result_date
1000002828     "Abraham Mars"              "Successful"     "2014-07-05"
1000002828     "Abraham Mars"              "Tracking"       "2014-08-01"
1000002828     "Abraham Mars"              "Successful"     "2014-08-01"
1000002828     "Abraham Mars"              "Tracking"       "2014-09-03"
1000002828     "Abraham Mars"              "Successful"     "2014-09-06"
1000002222     "Andrew peter Petersen"     "Successful"     "2014-08-25"
1000002222     "Andrew peter Petersen"     "Successful"     "2014-06-25"
1000002222     "Andrew peter Petersen"     "Tracking"       "2014-08-22"
1000002222     "Andrew peter Petersen"     "Successful"     "2014-05-23"
1000002222     "Andrew peter Petersen"     "Successful"     "2014-07-25"
1000002222     "Andrew peter Petersen"     "Tracking"       "2014-07-23"

亚伯拉罕有 3 笔付款和 3 条 NAEDO 记录(开始日期不同)
即 3/3 * 100 = 100%
他已经出现在网格上(尽管是 99%)

安德鲁有 4 笔付款和 4 条 NAEDO 记录(开始日期不同)
即 4/4 * 100 = 100%
他应该在发车区的同一个位置。

也许在后台我可以收集所有的网格位置分数并打点百分比计算。

一位客户的代表

Answer 1

所以这应该是关于所需的查询。我从 account_detail 中选择帐户及其分组字符串(行和列组)。我计算每个帐户的成功 paysoft 结果和每个帐户不同的 naedo 开始日期。然后我加入并聚合。然后由 Jasper Reports 在网格中旋转和显示数据。

select 
  ad.escore,
  ad.mscore,
  sum(ps.cnt) / sum(n.cnt) * 100 as percent
from 
(
  select 
    account_no,
    -- 602 becomes '595-604'
    to_char(trunc(empirica_score - 5, -1) + 5, , '9999') || '-' || to_char(trunc(empirica_score - 5, -1) + 14, '9999') as escore,
    -- 97 becomes '76-100'. Change the expression to group differently.
    to_char(((mfin_score - 1) / 25) * 25 + 1) || '-' || to_char(((mfin_score - 1) / 25) * 25 + 25) as mscore
  from account_detail
) ad
join 
(
  select custno, count(*) as cnt
  from paysoft_result 
  where result = 'Successful'
  group by custno
) ps on ps.custno = ad.account_no
join 
(
  select customer_code, count(distinct start_date) as cnt
  from naedo 
  group by customer_code
) n on n.customer_code =  = ad.account_no
group by ad.escore, ad.mscore;