如何在PostgreSQL的INSERT命令后使用调用函数更新表列?

时间:2018-11-13 18:28:02

标签: postgresql

我需要您的帮助,运行插入命令后更新特定列时遇到问题。 表格:

CREATE SEQUENCE public.llh_type_id_seq
INCREMENT 1
MINVALUE 1
MAXVALUE 9223372036854775807
START 261
CACHE 1;
ALTER TABLE public.llh_type_id_seq
OWNER TO postgres;

CREATE TABLE public.llh_type
(id bigint NOT NULL DEFAULT nextval('llh_type_id_seq'::regclass),
identifier text,
name text)
WITH (
OIDS=FALSE);
ALTER TABLE public.llh_type
OWNER TO postgres;

功能:

CREATE OR REPLACE FUNCTION public.generate_identifier(
id bigint,
prefix text)
RETURNS text AS
$BODY$
SELECT 
   CASE WHEN length($1::text) < 2 
      THEN UPPER($2 || '00000' || $1)
        WHEN length($1::text) >= 2 AND length($1::text) < 3  
      THEN UPPER($2 || '0000' || $1)
        WHEN length($1::text) >= 3 AND length($1::text) < 4  
      THEN UPPER($2 || '000' || $1)
        WHEN length($1::text) >= 4 AND length($1::text) < 5  
      THEN UPPER($2 || '00' || $1)
        WHEN length($1::text) >= 5 AND length($1::text) < 6  
      THEN UPPER($2 || '0' || $1)
    ELSE
    UPPER($2 || $1)
   END;
  $BODY$
  LANGUAGE sql IMMUTABLE STRICT
  COST 100;
  ALTER FUNCTION public.generate_identifier(bigint, text)
  OWNER TO postgres;

此后,我尝试在插入数据后调用函数:

WITH t AS(
INSERT INTO llh_type (name) values('one')
RETURNING id)

UPDATE llh_type SET identifier = generate_identifier((select id from   
t),'TC') WHERE id = (select id from t);

运行此代码后,我收到消息:“查询成功返回:受影响的行数为0,执行时间为12毫秒。”但是表格现在看起来像:http://joxi.ru/nAypMQjCYBggq2

在另一种情况下,我有解决方案,但是我不确定它是否正确:

INSERT INTO llh_type (name) values('one');
UPDATE llh_type SET identifier = generate_identifier((select id from   
llh_type order by id desc limit 1),'TC') 
WHERE id = (select id from llh_type order by id desc limit 1);

运行后,我收到一条消息:“查询成功返回:受影响的一行,执行时间为12毫秒”。结果看起来像预期的那样:http://joxi.ru/5md13oZCkM3el2

0 个答案:

没有答案