Oracle 树上的逻辑

Question

查询的输出：

STARTING_REF_ID REF_LABEL

---

           501 article 501
           502 par 502,article 501
           503 (503,par 502,article 501

预期结果：

STARTING_REF_ID REF_LABEL

---

           501 article 501
           502 article 501,par 502
           503 article 501,par 502,(503,

Answer 1

向选择添加了反向

WITH
refs (ref_id,
      ref_type_id,
      parent_ref_id,
      ref_value)
AS
    (SELECT 501, 1, NULL, 207 FROM DUAL
     UNION ALL
     SELECT 502, 2, 501, 4 FROM DUAL
     UNION ALL
     SELECT 503, 3, 502, 1 FROM DUAL)
SELECT CONNECT_BY_ROOT r.ref_id as starting_ref_id,
       TRIM (
           ',' FROM
           reverse( --reverse path order (also reverts letters in words)
               SYS_CONNECT_BY_PATH (
                 
                     reverse( --put letters in words and numbers in ids back in the right order
                          CASE r.ref_type_id
                          WHEN 1 THEN 'article '
                          WHEN 2 THEN 'par '
                          WHEN 3 THEN '('
                          WHEN 4 THEN 'point '
                          WHEN 5 THEN 'sous '
                          WHEN 6 THEN NULL
                          WHEN 8 THEN NULL
                          ELSE '/'
                      END
                   || r.ref_id),
                   ',')
                   
                   ) )   AS ref_label
  FROM refs r
 WHERE CONNECT_BY_ISLEAF = 1
CONNECT BY PRIOR r.parent_ref_id = r.ref_id

Answer 2

你正在以一种非常不自然的方式做这一切。

您应该从根（父 ID 为

null

）开始，然后从根开始“向下”，而不是相反。那么你不需要

where

子句，你需要

start with

子句；并且

connect by

条件必须以相反的方式进行。然后您不需要对查询进行任何其他更改。

像这样：

WITH
    refs (ref_id,
          ref_type_id,
          parent_ref_id,
          ref_value)
    AS
        (SELECT 501, 1, NULL, 207 FROM DUAL
         UNION ALL
         SELECT 502, 2, 501, 4 FROM DUAL
         UNION ALL
         SELECT 503, 3, 502, 1 FROM DUAL)
    SELECT CONNECT_BY_ROOT r.ref_id as starting_ref_id,
           TRIM (
               ',' FROM
                   SYS_CONNECT_BY_PATH (
                          CASE r.ref_type_id
                              WHEN 1 THEN 'article '
                              WHEN 2 THEN 'par '
                              WHEN 3 THEN '('
                              WHEN 4 THEN 'point '
                              WHEN 5 THEN 'sous '
                              WHEN 6 THEN NULL
                              WHEN 8 THEN NULL
                              ELSE '/'
                          END
                       || r.ref_id,
                       ','))    AS ref_label
      FROM refs r
START WITH parent_ref_id is null
CONNECT BY PRIOR ref_id = parent_ref_id;