Question

我正在尝试在Postgres中创建一个函数，如下所示：

Create function Samplefunc() 
returns resultdata( Tested int, Score int,
       Growth int) as
$BODY$
Select 
       Count(distinct student_id) Tested, 
       Cast(Avg(R.raw_score) as Int)  Score,
       Avg(R.growth) as Growth
from results R
where R.id=1 and test_id='ME04';
$BODY$
LANGUAGE sql;

但是我收到以下错误：

ERROR:  syntax error at or near "int"
LINE 2: returns resultdata( NTested int, RawScore int,
                                    ^

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

ERROR: syntax error at or near "int"
SQL state: 42601
Character: 59

我哪里错了？

Answer 1

CREATE FUNCTION samplefunc() 
  RETURNS TABLE(ntested int, rawscore int, growth int) AS
$func$
SELECT count(DISTINCT r.student_id) -- AS NTested
      ,avg(r.raw_score)::int        -- AS RawScore
      ,avg(r.growth)::int           -- AS Growth
FROM   reports_results r
WHERE  r.test_type_id = 1
AND    r.test_id = '201403MAME04'
$func$ LANGUAGE sql;

返回表格的子句是RETURNS TABLE。
小心避免OUT参数和列名之间的冲突。（我的初稿中有这样的冲突）。表限定列以消除歧义。 RETURNS TABLE中的所有字段名称都是有效的OUT参数，并且在函数内部（几乎）无处可见。

此外：

avg(growth)会导致类型与声明的返回类型int不匹配。你也需要施展它。使用短Postgres-specific syntax ::type，顺便说一句更好的是：返回numeric or a floating point number以保留平均数中的小数位数。
列别名仅在函数内可见。如果你不打算在里面引用它们这个函数，它们只是文档。
大写是什么？ Unquoted identifiers are cast to lower case in Postgres automatically.

如果保证查询返回单个行，您可能希望将OUT参数与RETURNS record合并：

CREATE FUNCTION samplefunc(OUT ntested int, OUT rawscore int, OUT growth int) 
  RETURNS record  AS ...

微妙的区别：这样，如果找不到任何内容，您将获得一个NULL个值的行，其中第一个表单不会返回任何行/没有行。

添加IN参数（评论中的请求）

CREATE FUNCTION samplefunc(_test_type_id int, _test_id text) 
  RETURNS TABLE(ntested int, rawscore int, growth int) AS
$func$
SELECT count(DISTINCT r.student_id)
      ,avg(r.raw_score)::int
      ,avg(r.growth)::int
FROM   reports_results r
WHERE  r.test_type_id = $1   -- or: = _test_type_id in Postgres 9.2+
AND    r.test_id = $2        -- or: = _test_id
$func$ LANGUAGE sql;

这里的许多相关答案都有更多的代码示例。像：

PostgreSQL return a function with a Custom Data Type

Try a search.

Answer 2

尽量不指定复合返回类型的详细信息。但是，我认为返回结构必须首先存在。

Create table resultdata (NTested int, RawScore int, Growth int);

Create function Samplefunc() returns resultdata as
$BODY$
Select 
   Count(distinct student_id) as NTested, 
   Cast(Avg(R.raw_score) as Int) as RawScore,
   Avg(R.growth) as Growth
from reports_results R
where R.test_type_id=1 and test_id='201403MAME04';
$BODY$
LANGUAGE sql;

或尝试显式返回表格：

Create function Samplefunc() 
    returns Table (NTested int, RawScore int, Growth int) as
$BODY$
Select 
   Count(distinct student_id) as NTested, 
   Cast(Avg(R.raw_score) as Int) as RawScore,
   Avg(R.growth) as Growth
from reports_results R
where R.test_type_id=1 and test_id='201403MAME04';
$BODY$
LANGUAGE sql;

我认为您还可以使用输出参数返回一组记录：

Create function Samplefunc(OUT NTested int, OUT RawScore int, OUT Growth int) 
    returns SetOf Record as
$BODY$
Select 
   Count(distinct student_id) as NTested, 
   Cast(Avg(R.raw_score) as Int) as RawScore,
   Avg(R.growth) as Growth
from reports_results R
where R.test_type_id=1 and test_id='201403MAME04';
$BODY$
LANGUAGE sql;

Answer 3

您正在尝试返回包含多个输出参数的记录。你应该这样做：

Create function Samplefunc(out NTested int, out RawScore int, out Growth int) as
$BODY$
Select 
       Count(distinct student_id) NTested, 
       Cast(Avg(R.raw_score) as Int)  RawScore,
       Avg(R.growth) as Growth
from reports_results R
where R.test_type_id=1 and test_id='201403MAME04';
$BODY$
LANGUAGE sql;

您可以使用明确命名的复合类型更详细地执行相同的操作;像

CREATE TYPE resultdata AS (NTested int, RawScore int, Growth int);

CREATE FUNCTION Samplefunc() RETURNS resultdata
    AS .......

或使用像{/ p>这样的TABLE函数

CREATE FUNCTION Samplefunc() RETURNS TABLE(NTested int, RawScore int, Growth int)
AS
......

有关相同

的更多信息，请参阅PostgreSQL Documentation

Answer 4

如果你没有像Luke那样定义一个“结果集”类型，那么请尝试返回一个表。

Create function Samplefunc() 
returns table( NTested int, RawScore int,
       Growth int) as
$BODY$
Select 
       Count(distinct student_id) NTested, 
       Cast(Avg(R.raw_score) as Int)  RawScore,
       Avg(R.growth) as Growth
from reports_results R
where R.test_type_id=1 and test_id='201403MAME04';
$BODY$
LANGUAGE sql;

Postgres创建函数时出错

4 个答案:

添加IN参数（评论中的请求）