with t
as(
SELECT distinct c.cInvAddCode as 货号,   
  d.dDate as 日期,   
  a.cWhCode as 仓库,  
  a.iQuantity as 数量,   
  a.iNum as 件数,   
  a.cFree1 as 颜色 ,   
  a.iInvExchRate as 换算率,   
  c.cInvDefine1 as 标准条码,   
  c.cInvDefine5 as 标准名称
  FROM b,   
  a,   
  c,   
  d
  WHERE ( a.DLID = b.DLID ) and   
  ( b.cInvCode = c.cInvCode ) and   
  ( a.cDLCode = d.cDLCode )
)
select [货号],[颜色],SUM([件数]) as [件数] from t
group by [货号],[颜色]
/*
货号    颜色    件数
b-0239    150    5
g-0226    CS彩色柜93    10
g-0227    CS彩色柜93    15
g-0226    黑白柜    40
g-0227    黑白柜    60
g-0226    蓝白柜96    70
g-0227    蓝白柜96    15
*/

--结果是要这样的吗

------解决方案--------------------
SQL code

with cte as
(
    SELECT distinct 
      c.cInvAddCode ,   --as 货号
      d.dDate ,            --as 日期
      a.cWhCode ,        --as 仓库
      a.iQuantity ,        --as 数量
      a.iNum ,            --as 件数
      a.cFree1  ,        --as 颜色
      a.iInvExchRate ,  --as 换算率
      c.cInvDefine1 ,    --as 标准条码   
      c.cInvDefine5        --as 标准名称
      FROM b,   
      a,   
      c,   
      d
      WHERE ( a.DLID = b.DLID ) and   
      ( b.cInvCode = c.cInvCode ) and   
      ( a.cDLCode = d.cDLCode )
  )
select 
    cInvAddCode,
    cFree1,
    SUM(iQuantity) as [总数量],
    SUM(iNum) as [总件数]
from cte 
group by cInvAddCode,cFree1