在redshift中,如何在创建表后向表中添加多个约束？

Question

在Redshift中,有没有办法为已创建的表添加多个约束？

第一步,创建表(说一个月前)

CREATE TABLE test_user 
(
account_id          BIGINT DEFAULT NULL encode mostly32,
ingest_timestamp    BIGINT DEFAULT NULL ENCODE mostly32,
ingest_date         TIMESTAMP DEFAULT NULL ENCODE delta32k,
user_id             BIGINT NOT NULL encode mostly32,
group_id            BIGINT NOT NULL encode mostly32,
department_id       BIGINT NOT NULL encode mostly32,
name                VARCHAR(255) DEFAULT NULL ENCODE LZO,
mobile              VARCHAR(15) DEFAULT NULL ENCODE LZO,
phone               VARCHAR(15) DEFAULT NULL ENCODE LZO,
user_language       VARCHAR(15) DEFAULT NULL ENCODE runlength,
deleted             INT DEFAULT NULL ENCODE mostly8,
created_at          TIMESTAMP DEFAULT NULL ENCODE delta32k,
updated_at          TIMESTAMP DEFAULT NULL ENCODE delta32k,
PRIMARY KEY (user_id)
)
sortkey (account_id)

第二步,创建了更多的表,如'test_group','test_department'

第三步,一个月后,我打算像这样在'test_user'表中添加多个约束(但在单独的查询中)

ALTER TABLE test_user ADD CONSTRAINT FK_1 FOREIGN KEY (group_id) REFERENCES test_group (group_id); 
ALTER TABLE test_user ADD CONSTRAINT FK_1 FOREIGN KEY (department_id) REFERENCES test_department (department_id); 

以上成功.

但是,有没有办法在单个查询中添加多个约束？试过这样的,

选项1 - > ALTER TABLE test_user ADD CONSTRAINT FK_1 FOREIGN KEY (group_id) REFERENCES test_group (group_id),ADD CONSTRAINT FK_1 FOREIGN KEY (department_id) REFERENCES test_department (department_id);

选项2 - > ALTER TABLE test_user ADD CONSTRAINT FK_1 FOREIGN KEY (group_id) REFERENCES test_group (group_id), FK_1 FOREIGN KEY (department_id) REFERENCES test_department (department_id);

但是这两种情况都有这个错误,

An error occurred when executing the SQL command:
ALTER TABLE test_user ADD CONSTRAINT FK_1 FOREIGN KEY (group_id) REFERENCES test_group (group_id), FK_2 FOREIGN KEY (department_id) REFERENCES test_department (department_id); REFERENCES helpdesk_user (user_id)

ERROR: syntax error at or near "FK_2"

有没有办法为已创建的表添加多个约束？请帮忙

Answer 1

在定义它们时,或者即使您完全定义它们也没关系.

Redshift忽略了约束.定义它们没有任何效果,除了优化器可以在创建访问计划时将它们用作附加信息.

但是,我已经在redshift中创建并使用了大量(数十亿行)数据库,并且我没有定义任何类型的单个约束,并且它运行得非常快.

答:不要打扰.