2

我有以下脚本可以动态地在 PostgreSQL 数据库中创建视图。

CREATE OR REPLACE FUNCTION cs_refresh_mviews() RETURNS integer AS $$
DECLARE
    mviews RECORD;
    query text;
    park_name text;
    ppstatements int;
BEGIN
    RAISE NOTICE 'Creating views...';

    FOR mviews IN SELECT name FROM "Canadian_Parks" LOOP
        park_name := mviews.name;
        RAISE NOTICE 'Creating or replace view %s...', mviews.name; 
        query := 'CREATE OR REPLACE VIEW %_view AS
          SELECT * from "Canadian_Parks" where name=''%'';
          ALTER TABLE %_view OWNER TO postgres', park_name, park_name, park_name;
      --  RAISE NOTICE query;
        EXECUTE query;
    END LOOP;

    RAISE NOTICE 'Done refreshing materialized views.';
    RETURN 1;
END;
$$ LANGUAGE plpgsql;

我已确认字符串的完整性，例如

CREATE OR REPLACE VIEW Saguenay_St__Lawrence_view AS
SELECT * from "Canadian_Parks" where name='Saguenay_St__Lawrence';
ALTER TABLE Saguenay_St__Lawrence_view OWNER TO postgres

通过手动将其提交到数据库并获得成功的响应来分配给查询变量。

但是，如果我尝试使用

SELECT cs_refresh_mviews();

显示以下错误：

ERROR:  query "SELECT 'CREATE OR REPLACE VIEW %_view AS SELECT * from "Canadian_Parks" where name=''%''; ALTER TABLE %_view OWNER TO postgres', park_name, park_name, park_name" returned 4 columns
CONTEXT:  PL/pgSQL function "cs_refresh_mviews" line 32 at assignment

********** Error **********

ERROR: query "SELECT 'CREATE OR REPLACE VIEW %_view AS SELECT * from "Canadian_Parks" where name=''%''; ALTER TABLE %_view OWNER TO postgres', park_name, park_name, park_name" returned 4 columns
SQL state: 42601
Context: PL/pgSQL function "cs_refresh_mviews" line 32 at assignment

为什么这被转换为 SELECT 语句，而不是纯粹的 CREATE？

4

2 回答 2

4

你的设置很扭曲。为什么要将视图名称的一部分保存在表的复合类型中，而不是将其保存在纯文本列中？

无论如何，它可以像这样工作：

设置匹配问题：

CREATE SCHEMA x;  -- demo in test schema
SET search_path = x;
CREATE TYPE mviews AS (id int, name text); -- composite type used in table

CREATE TABLE "Canadian_Parks" (name mviews);
INSERT INTO "Canadian_Parks"(name) VALUES
 ('(1,"canadian")')
,('(2,"islandic")');  -- composite types, seriously?

SELECT name, (name).* from "Canadian_Parks";

CREATE OR REPLACE FUNCTION cs_refresh_mviews()
  RETURNS int LANGUAGE plpgsql SET search_path = x AS  -- search_path for test
$func$
DECLARE
    _parkname text;
BEGIN

FOR _parkname IN SELECT (name).name FROM "Canadian_Parks" LOOP
   EXECUTE format('
      CREATE OR REPLACE VIEW %1$I AS
      SELECT * FROM "Canadian_Parks" WHERE (name).name = %2$L;
      ALTER TABLE %1$I OWNER TO postgres'
      , _parkname || '_view', _parkname);
END LOOP;

RETURN 1;

END
$func$;

SELECT cs_refresh_mviews();

DROP SCHEMA x CASCADE; -- clean up

要点

  • 当您使用 execute 执行文本时，您需要防范SQL 注入。我将format()函数用于标识符和文字

  • 我使用语法SELECT (name).name来应对您奇怪的设置并name立即提取我们需要的内容。

  • 同样，需要读取 VIEW 才能WHERE (name).name = ..在此设置中工作。

  • 我删除了很多与问题无关的噪音。

  • 拥有该功能也可能毫无意义RETURN 1。只需用 定义函数RETURNS void。不过，我保留了它以匹配问题。

解开的设置

它应该是怎样的：

CREATE SCHEMA x;
SET search_path = x;

CREATE TABLE canadian_parks (id serial primary key, name text);
INSERT INTO canadian_parks(name) VALUES ('canadian'), ('islandic');

SELECT * from canadian_parks;

CREATE OR REPLACE FUNCTION cs_refresh_mviews()
  RETURNS void LANGUAGE plpgsql SET search_path = x AS
$func$
DECLARE
    parkname text;
BEGIN

FOR parkname IN SELECT name FROM canadian_parks LOOP
   EXECUTE format('
      CREATE OR REPLACE VIEW %1$I AS
      SELECT * FROM canadian_parks WHERE name = %2$L;
      ALTER TABLE %1$I OWNER TO postgres'
      , parkname || '_view', parkname);
END LOOP;

END
$func$;

SELECT cs_refresh_mviews();

DROP SCHEMA x CASCADE;
于 2012-10-10T01:16:34.287 回答
2

您误解了赋值表达式中逗号的用法。它变成query数组 ( RECORD) 而不是标量。使用串联：

park_name := quote_ident(mviews.name||'_view');
query := 'CREATE OR REPLACE VIEW '||park_name||' AS SELECT * from "Canadian_Parks" where name='||quote_literal(mviews.name)||'; ALTER TABLE '||park_name||' OWNER TO postgres';
于 2012-10-09T20:29:43.413 回答