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select dictcharge.no1 as no1 ,
dictcharge.name1 as name1,
patinfo.hospcode,
sum(hosp_detail_price.price) as price
from dictcharge, his_history..hosp_detail_price hosp_detail_price, dictmedi , his_history..patinfo patinfo
where dictmedi.mediid = hosp_detail_price.mediid
    and dictcharge.chargeid = dictmedi.hosptype 
    and hosp_detail_price.hospid = patinfo.hospid
    and hosp_detail_price.isprice = 1
    and hosp_detail_price.ischeck = 1
    and hosp_detail_price.isconfirm = 1
    and hosp_detail_price.ispay <> 1
    and dictcharge.name1 = 'ÊÖÊõ·Ñ'
group by dictcharge.no1, patinfo.hospcode,dictcharge.name1 ;

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25 ÊÖÊõ·Ñ 20090001 700.00
25 ÊÖÊõ·Ñ 20090003 300.00
25 ÊÖÊõ·Ñ 20090004 400.00
25 ÊÖÊõ·Ñ 20090005 600.00
25 ÊÖÊõ·Ñ 20090009 1030.00
25 ÊÖÊõ·Ñ 20090011 980.00
25 ÊÖÊõ·Ñ 20090012 780.00
25 ÊÖÊõ·Ñ 20090013 460.00
25 ÊÖÊõ·Ñ 20090016 600.00

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SQL code:
select COUNT(*) as ÐÐÊý
from (
select dictcharge.no1 as no1 ,
dictcharge.name1 as name1,
patinfo.hospcode,
sum(hosp_detail_pri


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