sql查询oracle以适应树结构格式

Question

我已经创建了这些示例表来创建 json 树结构，以便我可以使用 jqtree 创建树布局。

我希望我的 json 格式为

[
  {"id":1, "parentid": 0, "name": "Carnivores"},
  {"id":2, "parentid": 0, "name": "Herbivores"},
  {"id":3, "parentid": 1, "name": "Dogs"},
  {"id":4, "parentid": 3, "name": "Labradors"},  
  {"id":5, "parentid": 3, "name": "Pugs"},
  {"id":6, "parentid": 3, "name": "Terriers"}

]

表格如下。

| catg_id       |   catg_name       | 
| —————-        |————————-          |
| 1             |   Carnivores      |
| 2             |   Herbivores      |



| animal_catg_id    | animal_catg_name      |   catg_id |
| —————-        |————————-                  |————————-  |
| 1             |   Dogs                    |   1       |
| 2             |   Cats                    |   1       |
| 3             |   Cows                    |   2       |
| 4             |   Buffalo                 |   2       |



| animal_id     | animal_name   | animal_catg_id    |
| —————-        |————————-      |   ————————-       |
| 1             |   labs        |   1               |
| 2             |   pugs        |   1               |
| 3             |   terriers    |   1               |
| 4             |   german      |   1               |
| 5             |   lion        |   2               |
| 6             |   tiger       |   2               |

我假设它是分层查询，我以前从未写过，我需要一些帮助。 我不知道从哪里开始以及如何开始。

编辑

答案中的一个 cmets 是架构设计不清楚。 我应该做些什么改变来获取 json 格式的数据，以便它保持层次结构

EDIT2

我当前的查询返回这个表

Carnivores     |  Dogs    | labs
Carnivores     |  Dogs    | pugs
Carnivores     |  Dogs    | terriers
.......

Answer 1

您提议的 JSON 似乎在您分配的 IDs 和表中的 IDs 之间没有关联，这将使得很难将任何东西从客户端连接回数据库。

您最好重新组织您的表格，以便您可以将所有内容放入一个单一的层次结构中。类似于林奈分类法：

SQL Fiddle

Oracle 11g R2 架构设置：

CREATE TABLE Taxonomies ( ID, PARENT_ID, Category, Taxonomy, Common_Name ) AS
          SELECT  1, CAST(NULL AS NUMBER),  'Kingdom',     'Animalia',    'Animal'       FROM DUAL
UNION ALL SELECT  2,  1,    'Phylum',      'Chordata',    'Chordate'     FROM DUAL
UNION ALL SELECT  3,  2,    'Class',       'Mammalia',    'Mammal'       FROM DUAL
UNION ALL SELECT  4,  3,    'Order',       'Carnivora',   'Carnivore'    FROM DUAL
UNION ALL SELECT  5,  4,    'Family',      'Felidae',     'Feline'       FROM DUAL
UNION ALL SELECT  6,  5,    'Genus',       'Panthera',    'Tiger'         FROM DUAL
UNION ALL SELECT  7,  5,    'Genus',       'Felis',       'Cat'           FROM DUAL
UNION ALL SELECT  8,  5,    'Genus',       'Lynx',        'Lynx'          FROM DUAL
UNION ALL SELECT  9,  4,    'Family',      'Canidae',     'Canid'        FROM DUAL
UNION ALL SELECT 10,  9,    'Genus',       'Canis',       'Canine'       FROM DUAL
UNION ALL SELECT 11, 10,    'Species',     'Canis Lupus', 'Gray Wolf'     FROM DUAL
UNION ALL SELECT 12, 11,    'Sub-Species', 'Canis Lupus Familiaris', 'Domestic Dog' FROM DUAL
UNION ALL SELECT 13, 12,    'Breed',       NULL,          'Pug'           FROM DUAL
UNION ALL SELECT 14, 12,    'Breed',       NULL,          'German Shepherd' FROM DUAL
UNION ALL SELECT 15, 12,    'Breed',       NULL,          'Labradors'     FROM DUAL
UNION ALL SELECT 16,  7,    'Species',     'Felis Catus', 'Domestic Cat'  FROM DUAL
UNION ALL SELECT 17,  8,    'Species',     'Lynx Lynx',   'Eurasian Lynx' FROM DUAL
UNION ALL SELECT 18,  8,    'Species',     'Lynx Rufus',  'Bobcat'        FROM DUAL;

那么就可以比较简单的提取数据了：

查询 1 - 获取与“猫”分类相关的所有内容：

SELECT *
FROM (
  SELECT *
  FROM   Taxonomies
  START WITH Common_Name = 'Cat'
  CONNECT BY PRIOR PARENT_ID = ID
  ORDER BY LEVEL DESC
)
UNION
SELECT *
FROM (
  SELECT *
  FROM   Taxonomies
  START WITH Common_Name = 'Cat'
  CONNECT BY PRIOR ID = PARENT_ID
  ORDER SIBLINGS BY Common_Name
)

Results：

| ID | PARENT_ID | CATEGORY |    TAXONOMY |  COMMON_NAME |
|----|-----------|----------|-------------|--------------|
|  1 |    (null) |  Kingdom |    Animalia |       Animal |
|  2 |         1 |   Phylum |    Chordata |     Chordate |
|  3 |         2 |    Class |    Mammalia |       Mammal |
|  4 |         3 |    Order |   Carnivora |    Carnivore |
|  5 |         4 |   Family |     Felidae |       Feline |
|  7 |         5 |    Genus |       Felis |          Cat |
| 16 |         7 |  Species | Felis Catus | Domestic Cat |

查询 2 - 获取与“犬”相关的所有分类：

SELECT *
FROM (
  SELECT *
  FROM   Taxonomies
  START WITH Common_Name = 'Canine'
  CONNECT BY PRIOR PARENT_ID = ID
  ORDER BY LEVEL DESC
)
UNION
SELECT *
FROM (
  SELECT *
  FROM   Taxonomies
  START WITH Common_Name = 'Canine'
  CONNECT BY PRIOR ID = PARENT_ID
  ORDER SIBLINGS BY Common_Name
)

Results：

| ID | PARENT_ID |    CATEGORY |               TAXONOMY |     COMMON_NAME |
|----|-----------|-------------|------------------------|-----------------|
|  1 |    (null) |     Kingdom |               Animalia |          Animal |
|  2 |         1 |      Phylum |               Chordata |        Chordate |
|  3 |         2 |       Class |               Mammalia |          Mammal |
|  4 |         3 |       Order |              Carnivora |       Carnivore |
|  9 |         4 |      Family |                Canidae |           Canid |
| 10 |         9 |       Genus |                  Canis |          Canine |
| 11 |        10 |     Species |            Canis Lupus |       Gray Wolf |
| 12 |        11 | Sub-Species | Canis Lupus Familiaris |    Domestic Dog |
| 13 |        12 |       Breed |                 (null) |             Pug |
| 14 |        12 |       Breed |                 (null) | German Shepherd |
| 15 |        12 |       Breed |                 (null) |       Labradors |

Answer 2

以下是 Oracle 文档中的分层查询示例：

SELECT last_name, employee_id, manager_id, LEVEL
  FROM employees
  START WITH employee_id = 100
  CONNECT BY PRIOR employee_id = manager_id
  ORDER SIBLINGS BY last_name;

http://docs.oracle.com/cd/B19306_01/server.102/b14200/queries003.htm

在你的情况下是这样的，但你的架构设计不清楚

 SELECT animal_name, level
  FROM animals
  START WITH parentid is null
  CONNECT BY PRIOR id = parentid;