编辑1：09-05-2013

Question

我已阅读过文档和几篇SO帖子，但仍然感觉不清楚如何在Postgres中声明和分配变量......这就是我最初尝试做的事情：

SET SEARCH_PATH = PsychoProductions;
declare var_person_id   INT;
select var_person_id = (
        select id 
        from person 
        where 
                first_name = prm_first_name AND
                last_name = prm_last_name AND
                organization = prm_organization
        );
/* 
Result:
ERROR:  syntax error at or near "INT"
LINE 2:     DECLARE var_person_id   INT;
                                    ^

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

ERROR: syntax error at or near "INT"
SQL state: 42601
Character: 70
*/

这很简单，但它不起作用......我是Postgres语法的新手，我希望得到一些帮助来解决这个问题，因为我想明确声明数据类型如果可能，在将数据设置为变量之前。

你们是否知道在设置变量时明确声明数据类型我缺少什么？

编辑1：09-05-2013

有些人指出必须在函数中使用变量。我不相信Postgres的情况完全如此（我知道它与MySQL有关，而不是T-SQL）但到了这一点，这里是上面的代码，在函数的上下文中。我将在短期内测试一个答案中指出的内容，并据此接受。所以这是目前为止的完整代码：

set search_path = PsychoProductions;
create or replace function fcn_Insert_person (
        -- "person" table
        prm_role_id             text, -- table default 'Customer'
        prm_first_name          text,
        prm_last_name           text,
        prm_organization        text,
        prm_website             text,
        prm_default_Billing_Method_ID   text, -- table default 'Net 30'
        prm_active              BOOLEAN,
        -- "address" table
        prm_address_type_id     text, -- table default 'Unique'
        prm_address             text,
        prm_city                text,
        prm_state               text,
        prm_zip_code            text,
        -- "email" table
        prm_email_address       text,
        prm_email_type_id       text, -- table default 'Business'
        -- "phone" table
        prm_phone_number        text,
        prm_phone_type_id       text  -- table default 'Main'
        ) 
        returns setof Person 
as

$delimiter$

begin
set search_patch = PsychoProductions;
start transaction;
insert into person (
        role_id,
        first_name,
        last_name,
        organization,
        website,
        default_billing_method_id,
        active
        )
values (
        prm_role_id,
        prm_first_name,
        prm_last_name,
        prm_organization,
        prm_website,
        prm_default_Billing_Method_ID,
        prm_active
        );
commit; 

start transaction;
declare var_person_id   int;
select var_person_id = (
        select id 
        from person 
        where 
                first_name = prm_first_name AND
                last_name = prm_last_name AND
                organization = prm_organization
        );
-- and this is where I got stuck

编辑2：09-06-2013

我使用了select into myVar ...语法，但现在我遇到了一个完全不同的问题......我应该问一个新问题吗？

错误：

ERROR:  column "var_person_id" does not exist
LINE 81:         var_person_id,
                 ^
********** Error **********

ERROR: column "var_person_id" does not exist
SQL state: 42703
Character: 2220

代码：

start transaction;
-- declare var_person_id   int;
select id into var_person_id 
        from person 
        where 
                first_name = prm_first_name AND
                last_name = prm_last_name AND
                organization = prm_organization;

insert into address (
        person_id,
        address_type_id,
        address,
        city,
        state,
        zip_code
        )
values (
        var_person_id, -- error here
        prm_address_type_id,
        prm_address,
        prm_city,
        prm_state,
        per_zip_code
        );

Answer 1

您只能在函数（存储过程）中声明变量。尝试这样的事情：

CREATE FUNCTION my_func(prm_first_name text,prm_last_name text) 
RETURNS SET OF int $$
DECLARE
    var_person_id int;
BEGIN
    SELECT id INTO var_person_id
        FROM person,var_parameter
        WHERE first_name = prm_first_name 
        AND last_name = prm_last_name
        AND organization = prm_organization;

    -- OR

    var_person_id := (SELECT id FROM ...);

    -- use it later like this:

    RETURN QUERY SELECT ... FROM ... WHERE id=var_person_id;
END;
$$ LANGUAGE PLPGSQL;

PostgreSQL使用显式数据类型声明变量

编辑1：09-05-2013

编辑2：09-06-2013

1 个答案: