postgresql:枚举和字符变化,更新

Question

我在postgresql中有ENUM类型

CREATE TYPE user_state AS ENUM ('unconfirmed', 'locked', 'active');

我有一个字段state表users与类型user_state.

我尝试执行下一个查询:

UPDATE "users" SET "state" = 'active'::character varying WHERE "id" = 1 (*1)

但是查询失败了.错误是:Column "state" is of type user_state but expression is of type character varying. Hint: You will need to rewrite or cast the expression.

好吧,有一些阅读勒芒,写了一些代码,我有一个隐式的类型转换为字符变化到user_state:

CREATE OR REPLACE FUNCTION
    character_varying_to_user_status(user_state)
    RETURNS user_state AS $$
    SELECT $1::user_state
    $$ LANGUAGE SQL;

CREATE CAST (character varying AS user_state)
    WITH FUNCTION character_varying_to_user_status (character varying)
    AS IMPLICIT

但这不起作用.它仅适用于id不存在的情况(因此,查询不执行任何操作,因此,我认为语法和类型检查是正确的).但是如果id存在,postgresql会告诉我堆栈溢出(这就是为什么我现在在这个网站上,ha-ha).错误是:

ERROR:  stack depth limit exceeded
HINT:  Increase the configuration parameter "max_stack_depth" (currently 2048kB), after ensuring the platform's stack depth limit is adequate.
CONTEXT:  SQL function "character_varying_to_user_status" during inlining
SQL function "character_varying_to_user_status" during startup
SQL function "character_varying_to_user_status" statement 1
SQL function "character_varying_to_user_status" statement 1
SQL function "character_varying_to_user_status" statement 1
...
...
SQL function "character_varying_to_user_status" statement 1

In operator:
UPDATE "users" SET "state" = 'unconfirmed'::character varying WHERE "id" = 8

我不知道如何解决它.有任何想法吗？

Postgresql 9.1和lift-squeryl-record(版本2.5).

(*1)查询UPDATE "users" SET "state" = 'active'::character varying WHERE "id" = 1不是真实的.我使用squeryl(scala语言orm),并且squeryl生成查询,所以,我不能删除::character varying它让它工作.真正的查询看起来像这样(在错误答案中):update "users" set "state" = ? where "id" = ? jdbcParams:[active,10]和我完全按照我上面写的那样回答(关于重写或转换表达式).所以,给出了查询as is,我无法改变它:我不能只删除:: character vary或add :: user_state或CAST(.. as ..).

UPD.还有人可以尝试在较新版本的postgresql(9.2,9.3,9.4)上运行代码吗？如果它有效,那也是答案.

Answer 1

一个简单的尝试的问题是你有一个演员调用演员,调用演员,调用演员......

你需要以某种方式远离你的演员阵容中的varchar-> enum.最简单(最易理解)的方法是手动转换.请注意,在case语句中强制转换的字符串文字不是它们引用的文本 - unknown-type,它会回避无限递归.

BEGIN;

CREATE TYPE t_tl AS ENUM ('red', 'amber', 'green');

CREATE FUNCTION dummy_cast(varchar) RETURNS t_tl AS $$
    SELECT CASE $1
        WHEN 'red' THEN 'red'::t_tl
        WHEN 'amber' THEN 'amber'::t_tl
        WHEN 'green' THEN 'green'::t_tl
    END;
$$ LANGUAGE SQL;

CREATE CAST (varchar AS t_tl) WITH FUNCTION dummy_cast(varchar) AS ASSIGNMENT;

CREATE TABLE t (x t_tl);

INSERT INTO t VALUES ('red');
INSERT INTO t VALUES ('green'::varchar);

SELECT * FROM t;

ROLLBACK;