Question

如果我的信用额度为12和14，注销为0和-4且可打印= 1，则在新列中输入qty * paidamt。

我使用了case语句，没有运气；代码和下面的查询。

ALTER VIEW [dbo].[GEMTQAF]
AS
select tdate,job,qty as units, paidamt as unitAmount, qty*paidamt as ExtendedAmount,credit as credit, -123 as Charge, credits.writeoff as credittype,
woseq as seq, postwho, credits.defaultsystemid as syscode, postdate, printable, cmpy, payor,'-123' as HCPCS, -123 as uniqueID,

    Case
            When forCharge=0 then '1'
            Else
            forCharge
    end as forCharge,

    Case    
            When credits.code in (12,14) and printable=1 and credits.writeOff in (0,-4)
                then qty*paidamt
            Else
            0
    end as pprimary

from trip_credits,credits where trip_credits.credit=credits.code
union all
select ta.tdate,ta.job,units, costperunit as unitAmount, convert(int, convert(int,round(units * costperunit,0))) as ExtendedAmount,-123 as credit,
ta.addon as Charge, -99 as credittype, ta.seq, postwho, '-1' as syscode, postdate, printable, cmpy, -123 as payor, HCPCS, ta.uniqueID, -123 as forCharge,-123 as pprimary


from trip_addons ta,HCPCSandModifiers hm where ta.tdate=hm.tdate and ta.job=hm.job and ta.addon=hm.addon

这是输出，其中tdate ='2017-07-30'和job ='1240-A'：

tdate   job units   unitAmount  ExtendedAmount  credit  Charge  credittype  seq postwho syscode postdate    printable   cmpy    payor   HCPCS   uniqueID    forCharge   pprimary
2017-07-30  1240-A  1.00    74092   74092   1   -123    -2  1   -201804 7   2018-05-04  1   250 11262   -123    -123    1   0
2017-07-30  1240-A  1.00    20944   20944   1   -123    -2  1   -201804 7   2018-05-04  1   250 11262   -123    -123    1   0
2017-07-30  1240-A  1.00    108180  108180  2   -123    -2  1   -201711     2017-12-21  0   250 11265   -123    -123    1   0
2017-07-30  1240-A  1.00    34580   34580   2   -123    -2  1   -201711     2017-12-21  0   250 11265   -123    -123    1   0
2017-07-30  1240-A  -1.00   108180  -108180 2   -123    -2  1   -201804     2018-05-04  0   250 11265   -123    -123    1   0
2017-07-30  1240-A  -1.00   34580   -34580  2   -123    -2  1   -201804     2018-05-04  0   250 11265   -123    -123    1   0
2017-07-30  1240-A  1.00    35991   35991   12  -123    0   0   201805      2018-05-29  1   250 11262   -123    -123    1   35991
2017-07-30  1240-A  1.00    12118   12118   12  -123    0   0   201805      2018-05-29  1   250 11262   -123    -123    2   12118
2017-07-30  1240-A  1.00    12273   12273   14  -123    -4  0   201805      2018-05-29  1   250 11265   -123    -123    1   12273
2017-07-30  1240-A  1.00    735 735 23  -123    -4  0   201805      2018-05-29  1   250 11262   -123    -123    1   0
2017-07-30  1240-A  1.00    247 247 23  -123    -4  0   201805      2018-05-29  1   250 11262   -123    -123    2   0
2017-07-30  1240-A  1.00    120000  120000  -123    4991    -99 1   201707  -1  2017-08-23  1   250 -123    A0427   1   -123    -123
2017-07-30  1240-A  14.00   2600    36400   -123    5000    -99 1   201707  -1  2017-08-23  1   250 -123    A0425   2   -123    -123

输出仅将CreditAlt列中12和14中的行中的ExtendedAmount放在pprimary中，并将credittype列中的0和-4放在pprimary中。当tdate和job匹配的credit列和credittype列都存在两个值时，我只需要看到ExtendedAmount。换句话说，对于每项工作，我要检查医疗保险是否是tdate +职位的主要付款人，而medicaid是否是次要付款人，则仅添加这些工资。

如何在日期和工作相等的列中测试2个值

0 个答案: