Question

我有2张桌子 invm和invt

invm已

i_b_no , i_b_dt , freight , tot_amt
1001 , 04/01/2015, 805 , 5000
1002 , 04/01/2015, 205 , 300

invt已

i_b_no , item_code, item_desc , stax_rate , amount
 1001 , xyz     , adlasjdlja,  12.5, 1000
 1001  , abc     , dfjsdffsdf,  12.5, 1000
 1001  , lkm     , sfkgsfksdn,  10.5, 1000
 1001  , rmn     , yeryoueoqe,  5.00, 1000
 1001  , the     , qwelskjfsdf,  12.5, 1000
1002   , dad     , sdfkdsffsdf , 10.0, 100
1002   , dekw    , kdasdahdafa , 3.2 , 100
1002    , dasdna , kdfnasldakdf, 6.8 , 100

现在我想要sum ( (freight) * max(staxrate) /100 )

是(805* 12.5)/100 = 1006.25 (205 * 10)/100 = 20.5

ans应为1026.75

我用过

select sum (( sum (INVM.FREIGHT ) *  MAX (INVT.STAX_RATE) /100 ))
from invm inner join INVT on INVM.I_B_NO = INVT.I_B_NO 
where 
invm.i_b_dt between '04/01/2015 00:00:00' and '04/01/2015 00:00:00' 
group by INVM.FREIGHT

给我错误

消息130，级别15，状态1，行1无法执行聚合对包含聚合或子查询的表达式的函数

我被困在这请求帮助

Answer 1

使用public static class RecyclerViewFragment extends Fragment { @Override public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) { View root = inflater.inflate(R.layout.fragment_recyclerview, container, false); RecyclerView recyclerView = (RecyclerView) root.findViewById(R.id.recycler_view); recyclerView.setHasFixedSize(true); recyclerView.setItemAnimator(new DefaultItemAnimator()); recyclerView.setLayoutManager(new LinearLayoutManager(getActivity())); recyclerView.addItemDecoration(new DividerItemDecoration(getActivity(), DividerItemDecoration.VERTICAL_LIST)); RecyclerViewAdapter adapter = new RecyclerViewAdapter(getActivity(), getResources() .getStringArray(R.array.countries)); recyclerView.setAdapter(adapter); return root; } }和JOIN执行派生表格中的GROUP BY：

MAX

sql中2表上的sum和max的聚合函数

1 个答案: