相关疑难解决方法(0)

我想在Postgres函数中传递一个表名作为参数.我试过这段代码:

CREATE OR REPLACE FUNCTION some_f(param character varying) RETURNS integer 
AS $$
    BEGIN
    IF EXISTS (select * from quote_ident($1) where quote_ident($1).id=1) THEN
     return 1;
    END IF;
    return 0;
    END;
$$ LANGUAGE plpgsql;

select some_f('table_name');

我得到了这个:

ERROR:  syntax error at or near "."
LINE 4: ...elect * from quote_ident($1) where quote_ident($1).id=1)...
                                                             ^

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

ERROR: syntax error at or near "."

以下是更改为此时出现的错误select * from quote_ident($1) tab where tab.id=1:

ERROR:  column tab.id does not exist
LINE 1: ...T EXISTS …

我写了一个函数,输出一个SELECT以文本形式组成的PostgreSQL 查询.现在我不想再输出文本,但实际上SELECT对数据库运行生成的语句并返回结果 - 就像查询本身一样.

到目前为止我所拥有的:

CREATE OR REPLACE FUNCTION data_of(integer)
  RETURNS text AS
$BODY$
DECLARE
   sensors varchar(100);   -- holds list of column names
   type    varchar(100);   -- holds name of table
   result  text;           -- holds SQL query
       -- declare more variables

BEGIN
      -- do some crazy stuff

      result := 'SELECT\r\nDatahora,' || sensors ||
      '\r\n\r\nFROM\r\n' || type ||
      '\r\n\r\nWHERE\r\id=' || $1 ||'\r\n\r\nORDER BY Datahora;';

      RETURN result;
END;
$BODY$
LANGUAGE 'plpgsql' VOLATILE;
ALTER FUNCTION data_of(integer) OWNER TO postgres;

sensors …

它必须简单,但我正在迈出Postgres函数的第一步,我找不到任何有效的东西......

我想创建一个修改表和/或列的函数,我找不到在我的函数中指定我的表和列作为参数的正确方法.

就像是:

CREATE OR REPLACE FUNCTION foo(t table)
RETURNS void AS $$
BEGIN
   alter table t add column c1 varchar(20);
   alter table t add column c2 varchar(20);
   alter table t add column c3 varchar(20);
   alter table t add column c4 varchar(20);
END;
$$ LANGUAGE PLPGSQL;

select foo(some_table)

在另一种情况下,我想要一个改变某个表中某个列的函数:

CREATE OR REPLACE FUNCTION foo(t table, c column)
RETURNS void AS $$
BEGIN
   UPDATE t SET c = "This is a test";
END;
$$ LANGUAGE PLPGSQL;

有可能吗？

我正在PostgreSQL DB中创建一个存储过程(函数),它根据输入更新表.为了创建一个可变数量的参数函数,我正在创建一个名为mode的额外输入参数,我用它来控制我在更新查询中使用的参数.

CREATE OR REPLACE FUNCTION update_site(
    mode integer,
    name character varying,
    city character varying,
    telephone integer,
)
RETURNS integer AS
$$
BEGIN
IF mode = 0 THEN
BEGIN
    UPDATE "Sites" SET 
    ("City","Telephone") = (city,telephone)
    WHERE "SiteName" = name;
    RETURN 1;
    EXCEPTION WHEN others THEN
    RAISE NOTICE 'Error on site update: %, %',SQLERRM,SQLSTATE;
    RETURN 0;
END;
ELSIF mode = 1 THEN
BEGIN
    UPDATE "Sites" SET "City" = city
    WHERE "SiteName" = name;
    RETURN 1;
    EXCEPTION WHEN others THEN
    RAISE NOTICE …

我有一个表address_all,它由几个地址表继承.address_history从父表继承history_all并保留当前地址信息.我正在创建一个新表继承address_all表并将信息复制address_history到新表.

我的存储过程如下所示.我打电话时遇到一些错误.为了更好地解释错误,我正在使用行号.

1  CREATE OR REPLACE FUNCTION somefunc()
2  RETURNS void AS
3  $BODY$
4  DECLARE
5   year_id INTEGER;
6   month_id INTEGER;
7   week_id INTEGER;
8   addresstablename text; 
9   backupdays text;
10 BEGIN
11  week_id := EXTRACT(DAY FROM TIMESTAMP 'now()');
12  month_id := EXTRACT(MONTH FROM TIMESTAMP 'now()');
13  year_id := EXTRACT(YEAR FROM TIMESTAMP 'now()');
14  addresstablename := 'address_history_' || week_id || '_' || month_id || '_' || year_id;
15  backupdays:= …