使用存储过程在LEFTOUTERJOIN中返回SETOF记录-UsingstoredprocedurereturningSETOFrecordinLEFTOUTERJOIN

5  

Generally, you can expand well known row types (a.k.a. record type, complex type, composite type) with the simple syntax @Daniel supplied:

通常,您可以使用提供的简单语法@Daniel扩展众所周知的行类型(a.k.a。记录类型,复杂类型,复合类型):

SELECT i.name, (compute_prices(i.id, current_date)).*
FROM   items i
WHERE  i.type = 404;

However, if your description is accurate ...

但是,如果您的描述准确无误......

The compute_prices sp returns a setof record.

compute_prices sp返回一组记录。

... we are dealing with anonymous records. Postgres does not know how to expand anonymous records and throws an EXCEPTION in despair:

......我们正在处理匿名记录。 Postgres不知道如何扩展匿名记录并在绝望中抛出一个EXCEPTION:

ERROR:  a column definition list is required for functions returning "record"

PostgreSQL 9.3

There is a solution for that in Postgres 9.3. LATERAL, as mentioned by @a_horse in the comments:

Postgres 9.3中有一个解决方案。正如@a_horse在评论中提到的那样:LATERAL:

SELECT i.name, sp.*
FROM   items i
LEFT   JOIN LATERAL compute_prices(i.id,current_date) AS sp (
                       price    numeric(15,2)
                      ,discount numeric(5,2)
                      ,taxes    numeric(5,2)
                      ) ON TRUE
WHERE i.type = 404;

Details in the manual.

手册中的详细信息。

PostgreSQL 9.2 and earlier

Things get hairy. Here's a workaround: write a wrapper function that converts your anonymous records into a well known type:

事情变得多毛了。这是一个解决方法:编写一个包装函数,将您的匿名记录转换为众所周知的类型:

CREATE OR REPLACE FUNCTION compute_prices_wrapper(int, date)
  RETURNS TABLE (
            price    numeric(15,2)
           ,discount numeric(5,2)
           ,taxes    numeric(5,2)
          ) AS
$func$
    SELECT * FROM compute_prices($1, $2)
    AS t(price    numeric(15,2)
        ,discount numeric(5,2)
        ,taxes    numeric(5,2));
$func$ LANGUAGE sql;

Then you can use the simple solution by @Daniel and just drop in the wrapper function:

然后你可以使用@Daniel的简单解决方案,只需输入包装函数:

SELECT i.name, (compute_prices_wrapper(i.id, current_date)).*
FROM   items i
WHERE  i.type = 404;

PostgreSQL 8.3 and earlier

PostgreSQL 8.3 has just reached EOL and is unsupported as of now (Feb. 2013).
So you'd better upgrade if at all possible. But if you can't:

PostgreSQL 8.3刚刚达到EOL,截至目前(2013年2月)尚未得到支持。因此,如果可能的话,你最好升级。但如果你不能:

CREATE OR REPLACE FUNCTION compute_prices_wrapper(int, date
           ,OUT price    numeric(15,2)
           ,OUT discount numeric(5,2)
           ,OUT taxes    numeric(5,2))
  RETURNS SETOF record AS
$func$
    SELECT * FROM compute_prices($1, $2)
    AS t(price    numeric(15,2)
        ,discount numeric(5,2)
        ,taxes    numeric(5,2));
$func$ LANGUAGE sql;

Works in later versions, too.

也适用于更高版本。

The proper solution would be to fix your function compute_prices() to return a well know type to begin with. Functions returning SETOF record are generally a PITA. I only poke those with a five-meter-pole.

正确的解决方案是修复函数​​compute_prices()以返回一个众所周知的类型。返回SETOF记录的函数通常是PITA。我只戳那些五米高的杆子。

3  

Assuming the compute_prices function always return a record with 3 prices, you could make its return type to TABLE (price numeric(15,2), discount numeric(5,2),taxes numeric(5,2)), and then I believe what you want could be expressed as:

假设compute_prices函数总是返回一个有3个价格的记录,你可以将它的返回类型设为TABLE(价格数字(15,2),折扣数字(5,2),税数字(5,2)),然后我相信你想要什么可以表达为:

SELECT i.name, (compute_prices(i.id,current_date)).*
  FROM items i
WHERE i.type=404;

Note that its seems to me that LEFT JOIN ON 1=1 does not differ from an unconstrained normal JOIN (or CROSS JOIN), and I interpreted the question as actually unrelated to the left join.

请注意,在我看来,LEFT JOIN ON 1 = 1与无约束的正常JOIN(或CROSS JOIN)没有区别,我将该问题解释为与左连接实际上无关。

1  

I believe Daniel's answer will work also but haven't tried it yet. I do know that I have an SP called list_failed_jobs2 in a schema called logging, and a dummy table called Dual (like in Oracle) and the following statement works for me:

我相信丹尼尔的回答也会奏效但尚未尝试过。我知道我在名为logging的模式中有一个名为list_failed_jobs2的SP,以及一个名为Dual的虚拟表(如在Oracle中),以下语句适用于我:

select * from Dual left join 
              (select * from logging.list_failed_jobs2()) q on 1=1;

Note, the SP call will not work without the parens, the correlation (q), or the ON clause. My SP returns a SETOF also.

注意,没有parens,相关(q)或ON子句,SP调用将不起作用。我的SP也返回SETOF。

Thus, I suspect something like this will work for you:

因此,我怀疑这样的事情对你有用:

select i.name,sp.*
from items i
left join (select * from compute_prices(i.id,current_date)) as sp on 1=1
where i.type = 404;

Hope that helps.

希望有所帮助。