【发布时间】:2014-07-02 01:03:50
【问题描述】:
你好 sql 巫师,
我正在尝试根据特定参数(案例类型)计算 SLA 天数。对于每种案例类型,SLA 天数的计算方式略有不同。我的问题是:我是否使用 CASE STATEMENT 来实现这一目标?我会记下我目前所拥有的,这是行不通的:
fifthlevel --with statement declaring sla days
AS (SELECT
CASE sla_days
WHEN fourthlevel.case_type = 'Complaint'
THEN sla_days =
( SELECT COUNT (*)
FROM (SELECT business_date
FROM ( SELECT TO_DATE ('01-01-2011', 'dd-mm-yyyy') + LEVEL - 1
business_date
FROM DUAL
CONNECT BY LEVEL <=
TO_DATE ('31-12-2099', 'dd-mm-yyyy')
- TO_DATE ('01-01-2011', 'dd-mm-yyyy')
+ 1) date_tab1
WHERE TO_CHAR (business_date, 'DY') NOT IN ('SAT', 'SUN')
AND business_date NOT IN (SELECT holiday_dt
FROM cisadm.ci_cal_hol
WHERE calendar_cd = 'WAW01'
)
) work_days1
WHERE work_days1.business_date > fourthlevel.correspondence_date
AND work_days1.business_date <= fourthlevel.close_date
)
WHEN fourthlevel.case_type = 'Enquiry'
THEN sla_days = (SELECT COUNT (*)
FROM (SELECT business_date
FROM (SELECT TO_DATE ('01-01-2011',
'dd-mm-yyyy')
+ LEVEL
- 1
business_date
FROM DUAL
CONNECT BY LEVEL <=TO_DATE('31-12-2099',
'dd-mm-yyyy')
TO_DATE('01-01-2011',
'dd-mm-yyyy')+ 1
) date_tab1
WHERE TO_CHAR (business_date, 'DY')
NOT IN ('SAT', 'SUN')
AND business_date
NOT IN (SELECT holiday_dt
FROM cisadm.ci_cal_hol
WHERE calendar_cd = 'WAW01'
)
) work_days1
WHERE work_days1.business_date > fourthlevel.agreed_Date
AND work_days1.business_date <=
fourthlevel.close_date
)
END
FROM fourthlevel,
fourthlevel.* --also wanting to select * from preceding WITH statement
FROM fourthlevel)
好的,看起来有点乱,但基本上我试图根据“投诉”和“查询”的 case_types 进行两种不同的计算。
谁能指出我正确的方向?
如果我能提供更多信息,请告诉我。
传奇!
【问题讨论】:
-
大家好,感谢您的回复,很抱歉我还不能投票。还想知道如果我想要一个指定 SLA 天数是否在阈值内的字段,我是否会使用 CASE。即,一个字段表示 sla_days 是否小于 15,另一个字段表示 sla_days 是否小于 20。它类似于:
case when sla_days <=15 then 'Under SLA" WHEN sla_days > 15 then 'Exceeds SLA'