postgres获取数据,其中两列引用相同的父ID
postgres 数据库中有 3 个表,
CREATE TABLE tab_name
(
name_id integer NOT NULL,
cust_name character varying NOT NULL, -- contains names like david,jones,athur
CONSTRAINT tab_name_pkey PRIMARY KEY (name_id)
)
CREATE TABLE tab_rel
(
rel_id integer NOT NULL,
rel_desc character varying NOT NULL,-- contains relation description father son, sister brother
CONSTRAINT tab_rel_pkey PRIMARY KEY (rel_id)
)
CREATE TABLE tab_rel_map
(
rel_id integer NOT NULL,
name_id1 integer NOT NULL,
name_id2 integer NOT NULL,
CONSTRAINT tab_rel_map_name_id1_fkey FOREIGN KEY (name_id1)
REFERENCES tab_name (name_id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT tab_rel_map_name_id2_fkey FOREIGN KEY (name_id2)
REFERENCES tab_name (name_id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT tab_rel_map_rel_id_fkey FOREIGN KEY (rel_id)
REFERENCES tab_rel (rel_id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION
)
我正在尝试编写以 rel_id 作为输入的函数,并应输出相对于 name_id1 和 name_id2 的 cust_name 。因为 name_id1 和 name_id2 都引用相同的父 ID,所以我无法获取各自的名称。
rel_id | relation | cust_name1 | cust_name2
------------------------------------------------
1 | Father son | David | Jones
There are 3 table in postgres database
CREATE TABLE tab_name
(
name_id integer NOT NULL,
cust_name character varying NOT NULL, -- contains names like david,jones,athur
CONSTRAINT tab_name_pkey PRIMARY KEY (name_id)
)
CREATE TABLE tab_rel
(
rel_id integer NOT NULL,
rel_desc character varying NOT NULL,-- contains relation description father son, sister brother
CONSTRAINT tab_rel_pkey PRIMARY KEY (rel_id)
)
CREATE TABLE tab_rel_map
(
rel_id integer NOT NULL,
name_id1 integer NOT NULL,
name_id2 integer NOT NULL,
CONSTRAINT tab_rel_map_name_id1_fkey FOREIGN KEY (name_id1)
REFERENCES tab_name (name_id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT tab_rel_map_name_id2_fkey FOREIGN KEY (name_id2)
REFERENCES tab_name (name_id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT tab_rel_map_rel_id_fkey FOREIGN KEY (rel_id)
REFERENCES tab_rel (rel_id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION
)
I am trying to write function which take rel_id as input and should output cust_name respect to name_id1 and name_id2. Because both name_id1 and name_id2 references to same parent id I am not able to get respective name.
rel_id | relation | cust_name1 | cust_name2
------------------------------------------------
1 | Father son | David | Jones
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您需要两次加入同一个表 - 这是使用 别名 执行此操作的方法:
You need to join to the same table twice - here's how you do that using aliases: