Question

我有以下表格结构。我想从TAB2获得每个级别的总和。

TAB1将层次结构存储在级别列中。

softmax_w = tf.get_variable("softmax_w", [hidden_size, 1], )
softmax_b = tf.get_variable("softmax_b", [1])

init_op = tf.initialize_all_variables()

这是SQLFiddle链接：LINK TO FIDDLE

Answer 1

Oracle设置

Create table TAB1 (pKey varchar2(10),level1 varchar2(10),level2 varchar2(10),level3 varchar2(10),level4 varchar2(10));
insert into TAB1(pKey,level1) values('A','A');
insert into TAB1(pKey,level1,level2) values('B','A','B');
insert into TAB1(pKey,level1,level2,level3) values('C','A','B','C');
insert into TAB1(pKey,level1,level2,level3) values('D','A','B','D');

Create table TAB2 (pKey varchar(10), tc integer);
insert into TAB2(pKey,tc) values('A',10);
insert into TAB2(pKey,tc) values('B',11);
insert into TAB2(pKey,tc) values('C',6);
insert into TAB2(pKey,tc) values('D',12);
insert into TAB2(pKey,tc) values('X',11);

<强>查询：

SELECT t2.pKey,
       SUM( COALESCE( t4.TC, t2.tc ) ) AS tc
FROM   tab2 t2
       LEFT OUTER JOIN
       tab1 t1
       ON ( t2.pKey = t1.pKey )
       LEFT OUTER JOIN
       tab1 t3
       ON (    t1.level1 = t3.level1
           AND ( t1.level2 IS NULL OR t1.level2 = t3.level2 )
           AND ( t1.level3 IS NULL OR t1.level3 = t3.level3 )
           AND ( t1.level4 IS NULL OR t1.level4 = t3.level4 ) )
       LEFT OUTER JOIN
       tab2 t4
       ON ( t3.pKey = t4.pKey )
GROUP BY t2.pKey;

<强>输出：

PKEY               TC
---------- ----------
D                  12 
A                  39 
B                  29 
C                   6 
X                  11

Answer 2

在下面提供的解决方案中（包括输入数据作为因子子查询），首先我展示如何使用unpivot和其他操作来规范化tab1（结果是因子子查询{{1} } for＆＃34; n ormalized＆＃34;）。然后，如果你有正常形式的数据，可以通过直接应用标准分层查询获得输出，如我的代码底部所示。

在此过程中，在with tab1 (key, L1, L2, L3) as ( select 'A', 'A', null, null from dual union all select 'B', 'A', 'B' , null from dual union all select 'C', 'A', 'B' , 'C' from dual union all select 'D', 'A', 'B' , 'D' from dual ), tab2 (key, TC) as ( select 'A', 10 from dual union all select 'B', 11 from dual union all select 'C', 6 from dual union all select 'D', 12 from dual union all select 'X', 11 from dual ), unpiv (key, l, ancestor) as ( select key, to_number(substr(lv, 2)), ancestor from tab1 unpivot (ancestor for lv in (L1, L2, L3)) ), d (key, depth) as ( select key, max(l) from unpiv group by key ), n (child, parent, TC) as ( select d.key, u.ancestor, tab2.TC from unpiv u right outer join d on u.key = d.key and u.l = d.depth - 1 left outer join tab2 on d.key = tab2.key ) SELECT key, sum(TC) as sum_TC from ( select connect_by_root child as key, TC from n connect by prior child = parent ) group by key order by key;中，我已经拥有了所有父子关系，因此我可以直接与unpiv上的tab2一起加入unpiv.key = tab2.key按TC分组（类似于MT0＆＃39;解决方案）。相反，我想展示两个单独的步骤：（1）规范化ancestor和（2）在规范化表上使用分层查询是多么容易。

<强>输出：

tab1

每个级别的层次结构查询总和

2 个答案: