Question

我正在尝试创建一个聚合函数，它接受一组记录，进行一些计算并返回一个数字。

我已经得到了以下虚拟示例：

CREATE OR REPLACE FUNCTION dummy_func(text) RETURNS int    
AS
$$
  DECLARE
    rec record;
    num int := 0;
  BEGIN
     FOR rec IN EXECUTE $1
     LOOP
         IF POSITION('some text' IN rec.known_column_name) = 1
         THEN
           num := num + 1;
         END IF;
     END LOOP;
     RETURN num;
  END;
$$ LANGUAGE PLPGSQL;

如果我输入

SELECT dummy_func('select * from some_table');

，它按预期工作。

但是，我想这样做，以便我可以使用多种功能和条件，例如：

SELECT conditional_col_1, dummy_func_1(*), dummy_func_2(*)
FROM some_table
WHERE conditional_col_2 = 'some val'
GROUP BY 1

如何将我的虚拟示例重写为能够如此运行（或使用一堆UNIONS ALL等）？

回应Hambone

如果我有以下逻辑（计算'a'或计算'b'*列的东西3）

CREATE OR REPLACE FUNCTION count_a(text) RETURNS int
AS
$$
  DECLARE
    rec record;
    num int := 0;
  BEGIN
     FOR rec IN EXECUTE $1
     LOOP
        num := num + length(regexp_replace(rec.something1, '[^a]+', '', 'g'));
     END LOOP;
     RETURN num;
  END;
$$ LANGUAGE PLPGSQL;

CREATE OR REPLACE FUNCTION count_e_times_something3(text) RETURNS int
AS
$$
  DECLARE
    rec record;
    num int := 0;
  BEGIN
     FOR rec IN EXECUTE $1
     LOOP
        num := num + length(regexp_replace(rec.something1, '[^e]+', '', 'g')) * rec.something3::int;
     END LOOP;
     RETURN num;
  END;
$$ LANGUAGE PLPGSQL;

我目前必须单独执行它们，

$=# SELECT count_a('SELECT * FROM MY_TEST WHERE something2 = ''A'' and something1 = ''0'''); 
-[ RECORD 1 ]
count_a | 0

$=# SELECT count_a('SELECT * FROM MY_TEST WHERE something2 = ''A'' and something3 = ''0'''); 
-[ RECORD 1 ]
count_a | 2

$=# SELECT count_e_times_something3('SELECT * FROM MY_TEST WHERE something2 = ''A'' and something3 = ''0''');
-[ RECORD 1 ]------------+--
count_e_times_something3 | 0

$=# SELECT count_e_times_something3('SELECT * FROM MY_TEST WHERE something2 = ''A'' and something3 = ''1''');
-[ RECORD 1 ]------------+--
count_e_times_something3 | 7

如果我可以使用以下查询会更好（我知道我需要定义一个聚合函数，但我找不到任何返回标量的记录）：

SELECT something2, count_a(*), count_e_times_something3(*) FROM MY_TEST WHERE something3 = '1';

something: 'A'
count_a: 3
count_e_times_something3: 7

something: 'B'
count_a: 3
count_e_times_something3: 2

count_a导致3是巧合

将记录传递给Postgres功能

0 个答案: