Question

我试图在存储过程中运行insert语句后返回一个字符串id。

这是我的存储过程代码：

CREATE OR REPLACE FUNCTION create_quote(
    p_id character varying(40),
    p_name character varying(200),
    p_website character varying(200),
    p_phone character varying(100),
    p_email character varying(200),
    p_msg text,
    p_created_on timestamp without time zone,
    p_created_ip cidr,
    p_contacted boolean,
    p_contacted_on timestamp without time zone
)
RETURNS character varying(40) AS $body$
BEGIN
    INSERT INTO quote VALUES (p_id, p_name, p_website, p_phone, p_email, p_msg, p_created_on, p_created_ip, p_contacted, p_contacted_on)
    ON CONFLICT (id)
       DO UPDATE SET name = p_name, website = p_website, phone = p_phone, email = p_email, msg = p_msg, created_on = p_created_on, created_ip = p_created_ip, contacted 
= p_contacted, contacted_on = p_contacted_on RETURNING id;
END;
$body$ LANGUAGE plpgsql;

表架构：

                               Table "public.quote"
    Column    |            Type             |              Modifiers               
--------------+-----------------------------+--------------------------------------
 id           | character varying(40)       | not null
 name         | character varying(200)      | not null
 website      | character varying(200)      | not null
 phone        | character varying(100)      | not null
 email        | character varying(200)      | not null
 msg          | text                        | 
 created_on   | timestamp without time zone | default timezone('utc'::text, now())
 created_ip   | cidr                        | 
 contacted    | boolean                     | 
 contacted_on | timestamp without time zone | 
Indexes:
    "quote_pkey" PRIMARY KEY, btree (id)
    "quote_email_key" UNIQUE CONSTRAINT, btree (email)

这是我尝试从psql客户端调用该过程时遇到的错误。

select * from create_quote("qw2e3r4t5y6u7i8o9i8u7y6t5r4e3w2q1", "name", "website", "phone", "email", "msg", now()::timestamp, NULL, 0, now()::timestamp);
ERROR:  column "qw2e3r4t5y6u7i8o9i8u7y6t5r4e3w2q1" does not exist
LINE 1: select * from create_quote("qw2e3r4t5y6u7i8o9i8u7y6t5r4e3w2q...

我有点困惑，因为我没有传入一个列而只是第一个参数的字符串id。我做错了什么？

我做了下面提到的clemens的更改，但现在我收到了这个错误：

select create_quote('qw2e3r4t5y6u7i8o9i8u7y6t5r4e3w2q1', 'name', 'website', 'phone', 'email', 'msg', now()::timestamp, NULL, 0, now()::timestamp);
ERROR:  function create_quote(unknown, unknown, unknown, unknown, unknown, unknown, timestamp without time zone, unknown, integer, timestamp without time zone) does not exist
LINE 1: select create_quote('qw2e3r4t5y6u7i8o9i8u7y6t5r4e3w2q1', 'na...
               ^
HINT:  No function matches the given name and argument types. You might need to add explicit type casts.

由于

Answer 1

您的代码中有几处错误：

SQL中的字符串受限于单引号。双引号用于保护标识符（例如区分大小写的列名称）。例如，使用'qw2e3r4t5y6u7i8o9i8u7y6t5r4e3w2q1'代替"qw2e3r4t5y6u7i8o9i8u7y6t5r4e3w2q1"。

您的函数返回标量而不是集合。你不应该从FROM条款中调用它。

SELECT create_quote('qw2e3r4t5y6u7i8o9i8u7y6t5r4e3w2q1', 'name', 'website', 'phone', 'email', 'msg', now()::timestamp, NULL, 0, now()::timestamp);

您的功能没有返回值。您可以将 UPSERT 的结果分配给变量并返回该变量。或者更容易返回输入参数p_id，因为 UPSERT 将始终返回此值。

CREATE OR REPLACE FUNCTION create_quote(
    p_id character varying(40),
    p_name character varying(200),
    p_website character varying(200),
    p_phone character varying(100),
    p_email character varying(200),
    p_msg text,
    p_created_on timestamp without time zone,
    p_created_ip cidr,
    p_contacted boolean,
    p_contacted_on timestamp without time zone

)
RETURNS character varying(40) AS $body$
BEGIN
    INSERT INTO quote VALUES (p_id, p_name, p_website, p_phone, p_email, p_msg, p_created_on, p_created_ip, p_contacted, p_contacted_on)
    ON CONFLICT (id)
        DO UPDATE SET name = p_name, website = p_website, phone = p_phone, email = p_email, msg = p_msg, created_on = p_created_on, created_ip = p_created_ip, contacted 
= p_contacted, contacted_on = p_contacted_on;
    RETURN p_id;
END;
$body$ LANGUAGE plpgsql;

如何从postgresql

1 个答案: