Question

给定一个返回查询的PostgreSQL函数：

CREATE OR REPLACE FUNCTION word_frequency(_max_tokens int)
  RETURNS TABLE (
    txt   text   -- visible as OUT parameter inside and outside function
  , cnt   bigint
  , ratio bigint) AS
$func$
BEGIN
   RETURN QUERY
   SELECT t.txt
        , count(*) AS cnt  -- column alias only visible inside
        , (count(*) * 100) / _max_tokens  -- I added brackets
   FROM  (
      SELECT t.txt
      FROM   token t
      WHERE  t.chartype = 'ALPHABETIC'
      LIMIT  _max_tokens
      ) t
   GROUP  BY t.txt
   ORDER  BY cnt DESC;  -- note the potential ambiguity 
END
$func$  LANGUAGE plpgsql;

如何检索此功能的结构？我的意思是，我知道此函数将返回txt，cnt和ratio列，但是如何创建返回这些列名称的查询？我试图在information_schema架构上找到这些列名，但我不能。

这个假设查询的预期结果将是这样的：

3 results found:
---------------------------------
?column_name?  |  ?function_name?
---------------------------------
txt                word_frequency
cnt                word_frequency
ratio              word_frequency

Answer 1

此信息存储在SELECT unnest(p.proargnames) as column_name, p.proname as function_name FROM pg_proc p JOIN pg_namespace n ON p.pronamespace = n.oid WHERE n.nspname = 'public' AND p.proname = 'word_frequency'

中

public function getGeneralSettings()
{

    $generalsetting = DB::table('generalsettings')
            ->where('id', '=', 1)
            ->first();
    return view('backend.generalsettings.add-general')->with('general', $generalsetting);
}

Answer 2

基于 a_horse_with_no_name 的答案，我带来了这个最终版本：

SELECT
  column_name,
  function_name
FROM
(
  SELECT 
    unnest(p.proargnames) as column_name,
    unnest(p.proargmodes) as column_type,
    p.proname as function_name
FROM pg_proc p 
   JOIN pg_namespace n ON p.pronamespace = n.oid  
WHERE n.nspname = 'public' 
  AND p.proname = 'my_function'
) as temp_table
WHERE column_type = 't';

我只是省略了参数，只返回函数返回的列

PostgreSQL结构返回查询的函数

2 个答案: