如何在层次结构树的每个节点中进行 SQL 分组查询？

Question

有一个 table 具有通常的“子到父”结构，其中每个节点都有权重。

TREE_TABLE
----------------------------------------
   id  | parent_id  |  name    | weight
----------------------------------------
   1   |  NULL      |   N1     |   51
   2   |  1         |   N12    |   62
   3   |  1         |   N13    |   73
   4   |  2         |   N124   |   84
   5   |  2         |   N125   |   95

// for convenience the "name" column shows path from a node to the root node

如何构建一个 SQL 查询以生成一组行，其中每行代表每个特定节点及其子节点的分组？

请根据您的选择使用任何 SQL 方言。只需要一个大概的想法或解决方案的原型。

为了解决这个问题，我正在试验 GROUPING SETS 和 ROLLUP 报告，但无法弄清楚处理“动态”数量的分组级别的方法。

预期查询结果示例：

RESULT
--------------------------
 name  |  summary_weight
--------------------------
  N1   |   (51+62+73+84+95)
  N12  |   (62+84+95)    
  N124 |   (84)
  N125 |   (95)
  N13  |   (73)

Answer 1

例如：

with 
 datum(id,parent_id,name,weight)
 as
 (
 select 1,NULL,'N1',51 from dual union all
 select 2 ,  1 , 'N12' , 62  from dual union all
 select 3 ,  1 , 'N13' , 73 from dual union all
 select 4 ,  2 , 'N124' , 84 from dual union all
 select 5 ,  2 , 'N125' , 95 from dual
 ),
 step1 as
 (
 select id,parent_id,name,weight, connect_by_root name  root,connect_by_isleaf is_parent 
 from datum
 connect by prior id = parent_id 
 )
select root,sum(weight)  sum_w,
 '('||listagg(weight,'+') within group(order by null) ||')' str_w,
 '('||listagg(name,'+') within group(order by null) ||')' str_n
from step1
group by root
order by 1;

link: Hierarchical Queries

Answer 2

如果你想SUM起来，你可以使用这个选项：

SELECT 
  name, 
  (SELECT 
     sum(t2.weight) 
   FROM tree t2 
   WHERE t2.name LIKE t1.name || '%' )
FROM tree t1
ORDER BY rpad(name,10,'0');

如果你想连接文本，你可以使用这个：

SELECT 
  name, 
  (SELECT 
    '(' || string_agg(t2.weight || '','+') || ')' 
   FROM tree t2 
   WHERE t2.name LIKE t1.name || '%' )
FROM tree t1
ORDER BY rpad(name,10,'0');

如何在层次结构树的每个节点中进行 SQL 分组查询？

How to make SQL query with grouping in each node of an hierarchy tree?

sql

oracle

grouping-sets