C语言数据类型中 struct 的基本内容 ,存在于所有流行语言中,并且期望这些语言的函数也可以返回 struct ...。并且,根据正交性原理,期望您可以访问返回的 struct但是,PostgreSQL不提供对 FUNCTION的 struct itens 的访问。 ..返回记录。
但是程序员使用PostgreSQL时不会抱怨... 有一个简单而直观的解决方法吗?
类似的问题: PostgreSQL v9.X具有真实的记录数组?
以典型案例为例 创建函数foo(int)返回记录为$$ 选择$ 1作为a,'Hello#'|| $ 1作为b; $语言SQL; SELECT foo(6); -有效,但是我只需要一项在 SQL上下文中访问记录项:
SELECT(foo(6))。a; -不起作用(但没有歧义!) -对于语法讨论: WITH f AS(SELECT foo(6)as r)SELECT r.a FROM f; -不起作用-语法不明确;将r与表格混淆,在 fra中,将f与架构-也许r ['a']是一个不错的语法解决方案在 PLpgSQL上下文中访问记录项目:
怎么说 x:=(foo(6))。a 或 y:= foo(6); x:= y.a ?现在在PLpgSQL中有一些预期的行为,至少允许命名记录:
CREATE FUNCTION bar()以$ F $ DECLARE tmp记录的形式返回文本; 的文字; 开始-s:=(foo(7))。b; NOT WORKS,就像匿名记录(不允许) tmp:= foo(6); s:= tmp.b; - 有用!!就像命名记录(允许) RETURN s ||’! ’|| tmp.a; -...正在访问其他任何个人,并使其延误 END; $ F $语言plpgsql不可用;解决方案
如果使用更灵活的返回表,而不是(有些过时的)返回记录,那么事情就变得很容易了:
CREATE FUNCTION foo(int)RETURNS table(a int,b text) AS $ SELECT $ 1 as,'Hello #'|| $ 1 as b; $语言SQL;现在您可以使用:
从foo(6)中选择b ;如果您担心表与记录,还可以定义一个 type 来克服其他结果集定义:
创建类型foo_return为(int ,b文字); CREATE FUNCTION foo(int)返回foo_return AS $ SELECT $ 1, Hello# || $ 1; $语言SQL;您仍然可以在上面进行选择:
从foo(6)中选择b ;第三个可能更像 C的方法是使用参数(如手册中所示)
创建函数foo(p1 int,out a int,out b text) AS $$ SELECT $ 1,'Hello#'| | $ 1; $$ 语言SQL;那么您就不需要来自 :
select(foo(1))。b;
Basic things as struct in C data types, exist in all popular languages, and is expected that functions, of these languages, also can return a struct... And, by an orthogonality principle, is expected you can access the returned struct itens.
PostgreSQL, nevertheless, did not offer access to the struct itens of a FUNCTION ... RETURNS RECORD. It is correct?
But programmers use PostgreSQL without complaining... There are a simple and intuitive workaround?
Similar question: PostgreSQL v9.X have real "array of record"?
Illustrating by typical cases CREATE FUNCTION foo(int) RETURNS RECORD AS $$ SELECT $1 as a, 'Hello #'||$1 as b; $$ LANGUAGE SQL; SELECT foo(6); -- works, but I need only one itemAccess of record itens in a SQL context:
SELECT (foo(6)).a; -- NOT works (but no ambiguity!) -- For syntax discussion: WITH f AS (SELECT foo(6) as r) SELECT r.a FROM f; -- NOT works -- ambiguous syntax; confused r with table, in "f.r.a", f with schema -- perhaps r['a'] would be a good syntax solutionAccess of record itens in a PLpgSQL context:
How to say x:=(foo(6)).a or y:=foo(6); x:=y.a? Now there are some expected behaviuor, in PLpgSQL, at least "named record" is permitted:
CREATE FUNCTION bar() RETURNS text AS $F$ DECLARE tmp record; s text; BEGIN -- s:=(foo(7)).b; NOT WORKS, is like an "anonymous record" (not permitted) tmp := foo(6); s:=tmp.b; -- IT WORKS!! is like a "named record" (permitted) RETURN s||'! '||tmp.a; -- ...works accessing any other individual itens END; $F$ LANGUAGE plpgsql IMMUTABLE;解决方案
If you use the more flexible returns table instead of (the somewhat outdated) returns record, then things get really easy:
CREATE FUNCTION foo(int) RETURNS table (a int, b text) AS $$ SELECT $1 as a, 'Hello #'||$1 as b; $ LANGUAGE SQL;now you can use:
select b from foo(6);If you are concerned about "tables" vs. "records" you can also define a type to overcome the additional result set definition:
create type foo_return as (a int, b text); CREATE FUNCTION foo(int) RETURNS foo_return AS $$ SELECT $1, 'Hello #'||$1; $$ LANGUAGE SQL;You can still the above select then:
select b from foo(6);A third maybe more "C" like approach would be to use out parameters (as shown in the manual)
CREATE FUNCTION foo(p1 int, out a int, out b text) AS $$ SELECT $1, 'Hello #'||$1; $$ LANGUAGE SQL;Then you don't need a from :
select (foo(1)).b;
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函数不能返回记录的单个项目,是吗?任何解决方法?
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