Question

我有以下数据：

case "$(whoami)" in
    <USER1>|<USER2>)
        alias python="/usr/bin/python2.7"
        ;;

我事先知道Key1 | Key2 | KValue =====+======+======= A | X | 100 B | Y | 200 C | X | 220 B | X | 300 B | Y | 50可以包含以下值Key1和['A','B','C','D']值Key2。

我需要SQL，它将返回所有['X','Y']组合的矩阵及其值的总和（也是零和的组合）。因此，之前的结果应该返回：

Key1 x Key2

Key1 | Key2 | KSUM =====+======+====== A | X | 100 A | Y | 0 B | X | 300 B | Y | 250 C | X | 220 C | Y | 0 D | X | 0 D | Y | 0不会返回零和行！

Answer 1

with
     table1 ( key1 ) as (
       select 'A' from dual union all
       select 'B' from dual union all
       select 'C' from dual union all
       select 'D' from dual
     ),
     table2 ( key2 ) as (
       select 'X' from dual union all
       select 'Y' from dual
     ),
     test_data ( key1, key2, kvalue ) as (
       select 'A', 'X', 100 from dual union all
       select 'B', 'Y', 200 from dual union all
       select 'C', 'X', 220 from dual union all
       select 'B', 'X', 300 from dual union all
       select 'B', 'Y',  50 from dual
     )
-- End of simulated inputs (for testing purposes only, not part of the solution).
-- SQL query begins BELOW THIS LINE.
select t1.key1, t2.key2, nvl(ksum, 0) as ksum
from     table1 t1
         cross join
         table2 t2
         left join
         ( select   key1, key2, sum(kvalue) as ksum
           from     test_data
           group by key1, key2
         ) t3
              on t1.key1 = t3.key1 and t2.key2 = t3.key2
order by key1, key2
;

<强>输出：

KEY1 KEY2 KSUM
---- ---- ----
A    X     100
A    Y       0
B    X     300
B    Y     250
C    X     220
C    Y       0
D    X       0
D    Y       0

Answer 2

获取Key1和Key2的所有可能组合（您可以存储它们或使用笛卡尔连接生成它们），并将其连接到由Key1和Key2分组的聚合sum（）查询，将空和值合并为零

矩阵结果的SQL查询

2 个答案: