惊人的混乱（不可能）查询

Question

我有一个查询，我甚至不知道它是否可能。

我有3张桌子。一个父表和两个孩子

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

像这样（将在一秒内完成99％）

因此查询需要让epirica_score和mfin_score确定网格上的展示位置（例如99％的位置）

并获取密钥匹配的distinct的{​​{1}}值 或多或少像这样

[NAEDO]startdate

我对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))

一无所知

1. 因此，您可以获得2个分数来确定展示位置。
2. 金额（paysoft_result / NAEDO）* 100 = JOINS
3. 需要按客户计算。并且总计投入了网格。

我不知道如何处理这样的查询。它是一份碧玉报道。

任何想法？

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次付款和3次NAEDO记录（start_date DISTINCT）
即3/3 * 100 = 100％
他已经在网格上代表（虽然它的99％）

安德鲁有4笔付款和4封NAEDO记录（start_date DISTINCT）
即4/4 * 100 = 100％
他应该在网格上的同一个地方。

也许在后台我可以收集所有网格位置分数并点按百分比计算。

一个客户的代表

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（row和col组）。我计算每个帐户成功的paysoft结果和每个帐户的不同的naedo开始日期。然后我加入并聚合。然后由Jasper Reports决定是否在网格中显示和显示数据。

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;

Answer 2

我可以提供此请求以显示您需要的结果

如果有任何错误告诉我纠正它

如果请求正常，jasper很容易

祝你好运

SELECT paysoft_results.COUNT(custno),
  account_details.impirica_score,
  account_details.mfin_score ,
  naedo.start_date
FROM account_details
INNER JOIN paysoft_results pr
ON (account_details.account_no = paysoft_results.custno)
INNER JOIN naedo
ON (account_details.account_no =naedo.customer_code)
WHERE pr.result_date BETWEEN '2014/08/01' AND '2014/09/01'
AND paysoft_results.result = 'successful'