Question

我正在运行Postgres 9.6。我有一个用户表：

 id                            | integer                     |           | not null | nextval('user_id_seq'::regclass)
 name                          | character varying           |           | not null |

我有一个action表：

 actioned_by      | integer                     |           |          | 
 category         | character varying           |           | not null | 
 t_actioned       | timestamp without time zone |           | not null |

我有一个查询，指示用户过去30天中有多少天处于活动状态：

 SELECT d.actioned_by, COUNT(*) AS cnt FROM 
  (SELECT date_trunc('day', t_actioned) AS day, actioned_by 
   FROM history 
   GROUP BY day, actioned_by) d 
 GROUP BY actioned_by
 ORDER BY cnt DESC;

但是有什么方法可以用此值注释每个用户，还是创建引用它的简便方法？

如果我可以执行“让我每月在活动中超过5天的所有法国用户”这样的事情，而不必每次都键入以上所有内容，那将非常有用。

也许我需要一个视图？

Answer 1

推荐的方法进行查看。

但是您也可以在复合类型（一种虚拟列）上创建函数：

-- Virtual column is a function that has a table type in its signature
CREATE FUNCTION last_activity("user") RETURNS text AS $$
    SELECT d.actioned_by, COUNT(*) AS cnt FROM 
      (SELECT date_trunc('day', t_actioned) AS day, actioned_by 
       FROM history
        --Filter
        WHERE actioned_by = $1.id
       GROUP BY day, actioned_by) d 
     GROUP BY actioned_by
     ORDER BY cnt DESC;    
$$ LANGUAGE SQL;

--In order to use virtual columns you must qualify it with table name/alias
SELECT u.last_activity, * FROM "user" u;

更多信息here。

Postgres：将常见查询另存为表属性？

1 个答案: