80

我检查了 Oracle 提供的文档,并找到了一种无需删除表即可修改约束的方法。问题是,它在修改时出错,因为它无法识别关键字。

为 PostgreSQL 使用 EMS SQL 管理器。

Alter table public.public_insurer_credit MODIFY CONSTRAINT public_insurer_credit_fk1
    deferrable, initially deferred;

我可以通过使用以下方式删除约束来解决它:

ALTER TABLE "public"."public_insurer_credit"
  DROP CONSTRAINT "public_insurer_credit_fk1" RESTRICT;

ALTER TABLE "public"."public_insurer_credit"
  ADD CONSTRAINT "public_insurer_credit_fk1" FOREIGN KEY ("branch_id", "order_id", "public_insurer_id")
    REFERENCES "public"."order_public_insurer"("branch_id", "order_id", "public_insurer_id")
    ON UPDATE CASCADE
    ON DELETE NO ACTION
    DEFERRABLE 
    INITIALLY DEFERRED;
4

4 回答 4

145

Postgres中没有ALTER约束命令。完成此操作的最简单方法是删除约束并使用所需参数重新添加它。当然,约束的任何更改都将针对当前表数据运行。

BEGIN;
ALTER TABLE t1 DROP CONSTRAINT ...
ALTER TABLE t1 ADD CONSTRAINT ...
COMMIT;
于 2011-11-30T17:50:34.543 回答
45

根据正确的手册(由 PostgreSQL 提供,而不是由 Oracle 提供),在 ALTER TABLE 语句中没有可用的修改约束:

这是正确手册的链接:

http://www.postgresql.org/docs/current/static/sql-altertable.html

于 2011-11-30T17:49:07.343 回答
42

从 9.4 版开始,PostgreSQL 支持ALTER TABLE ... ALTER CONSTRAINT外键。

此功能将"Allow constraint attributes to be altered, so the default setting of NOT DEFERRABLE can be altered to DEFERRABLE and back."查看您的问题,我认为这(某种)是您一直在寻找的。

可以在此处找到更多详细信息和示例:http:
//www.depesz.com/2013/06/30/waiting-for-9-4-alter-table-alter-constraint-for-fks/

于 2013-08-27T20:00:29.403 回答
6

ALTER CONSTRAINT 需要知道外键名称,这并不总是很方便。

这是函数,您只需要知道表名和列名。用法:

select replace_foreign_key('user_rates_posts', 'post_id', 'ON DELETE CASCADE');

功能:

CREATE OR REPLACE FUNCTION 
    replace_foreign_key(f_table VARCHAR, f_column VARCHAR, new_options VARCHAR) 
RETURNS VARCHAR
AS $$
DECLARE constraint_name varchar;
DECLARE reftable varchar;
DECLARE refcolumn varchar;
BEGIN

SELECT tc.constraint_name, ccu.table_name AS foreign_table_name, ccu.column_name AS foreign_column_name 
FROM 
    information_schema.table_constraints AS tc 
    JOIN information_schema.key_column_usage AS kcu
      ON tc.constraint_name = kcu.constraint_name
    JOIN information_schema.constraint_column_usage AS ccu
      ON ccu.constraint_name = tc.constraint_name
WHERE constraint_type = 'FOREIGN KEY' 
   AND tc.table_name= f_table AND kcu.column_name= f_column
INTO constraint_name, reftable, refcolumn;

EXECUTE 'alter table ' || f_table || ' drop constraint ' || constraint_name || 
', ADD CONSTRAINT ' || constraint_name || ' FOREIGN KEY (' || f_column || ') ' ||
' REFERENCES ' || reftable || '(' || refcolumn || ') ' || new_options || ';';

RETURN 'Constraint replaced: ' || constraint_name || ' (' || f_table || '.' || f_column ||
 ' -> ' || reftable || '.' || refcolumn || '); New options: ' || new_options;

END;
$$ LANGUAGE plpgsql;

注意:这个函数不会复制初始外键的属性。它只需要外部表名/列名,删除当前键并替换为新键。

于 2015-10-03T09:25:13.270 回答