加入对 generate_series() 的计数查询并将 Null 值检索为“0" [英] Join a count query on generate_series() and retrieve Null values as '0'

问题描述

我想使用 generate_series() 计算每个月的 ID.此查询适用于 PostgreSQL 9.1:

I want to count ID's per month using generate_series(). This query works in PostgreSQL 9.1:

SELECT (to_char(serie,'yyyy-mm')) AS year, sum(amount)::int AS eintraege FROM (
    SELECT  
       COUNT(mytable.id) as amount,   
       generate_series::date as serie   
       FROM mytable  
    
    RIGHT JOIN generate_series(     
       (SELECT min(date_from) FROM mytable)::date,   
       (SELECT max(date_from) FROM mytable)::date,  
       interval '1 day') ON generate_series = date(date_from)  
       WHERE version = 1   
       GROUP BY generate_series       
       ) AS foo
GROUP BY Year   
ORDER BY Year ASC;  

这是我的输出:

"2006-12" | 4  
"2007-02" | 1  
"2007-03" | 1  

但我想得到的是这个输出(一月份的'0'值):

But what I want to get is this output ('0' value in January):

"2006-12" | 4  
"2007-01" | 0  
"2007-02" | 1  
"2007-03" | 1  

没有id的月份应该被列出.
任何想法如何解决这个问题?

Months without id should be listed nevertheless.
Any ideas how to solve this?

示例数据:

drop table if exists mytable;
create table mytable(id bigint, version smallint, date_from timestamp);
insert into mytable(id, version, date_from) values
(4084036, 1, '2006-12-22 22:46:35'),
(4084938, 1, '2006-12-23 16:19:13'),
(4084938, 2, '2006-12-23 16:20:23'),
(4084939, 1, '2006-12-23 16:29:14'),
(4084954, 1, '2006-12-23 16:28:28'),
(4250653, 1, '2007-02-12 21:58:53'),
(4250657, 1, '2007-03-12 21:58:53')
;

推荐答案

解开、简化和固定，它可能看起来像这样:

Untangled, simplified and fixed, it might look like this:

SELECT to_char(s.tag,'yyyy-mm') AS monat
     , count(t.id) AS eintraege
FROM  (
   SELECT generate_series(min(date_from)::date
                        , max(date_from)::date
                        , interval '1 day'
          )::date AS tag
   FROM   mytable t
   ) s
LEFT   JOIN mytable t ON t.date_from::date = s.tag AND t.version = 1   
GROUP  BY 1
ORDER  BY 1;

db<>fiddle 这里

在所有的噪音、误导性标识符和非常规格式中，真正的问题隐藏在这里:

Among all the noise, misleading identifiers and unconventional format the actual problem was hidden here:

WHERE version = 1

您正确使用了 右[外部]加入.但是添加需要 mytable 中的现有行的 WHERE 子句会将 RIGHT [OUTER] JOIN 转换为 [INNER] JOIN 有效.

You made correct use of RIGHT [OUTER] JOIN. But adding a WHERE clause that requires an existing row from mytable converts the RIGHT [OUTER] JOIN to an [INNER] JOIN effectively.

将该过滤器移动到 JOIN 条件中以使其工作.

Move that filter into the JOIN condition to make it work.

我在做的时候简化了其他一些事情.

I simplified some other things while being at it.

SELECT to_char(mon, 'yyyy-mm') AS monat
     , COALESCE(t.ct, 0) AS eintraege
FROM  (
   SELECT date_trunc('month', date_from)::date AS mon
        , count(*) AS ct
   FROM   mytable
   WHERE  version = 1     
   GROUP  BY 1
   ) t
RIGHT JOIN (
   SELECT generate_series(date_trunc('month', min(date_from))
                        , max(date_from)
                        , interval '1 mon')::date
   FROM   mytable
   ) m(mon) USING (mon)
ORDER  BY mon;

db<>fiddle 这里

先聚合后加入要便宜得多 - 每月加入一行而不是每天加入一行.

It's much cheaper to aggregate first and join later - joining one row per month instead of one row per day.

将 GROUP BY 和 ORDER BY 基于 date 值而不是呈现的 text 更便宜.

It's cheaper to base GROUP BY and ORDER BY on the date value instead of the rendered text.

count(*) 比 count(id) 快一点，而在 this 查询中等效.

count(*) is a bit faster than count(id), while equivalent in this query.

generate_series() 基于 timestamp 而不是 date 会更快更安全.见:

generate_series() is a bit faster and safer when based on timestamp instead of date. See:

• 在 PostgreSQL 中生成两个日期之间的时间序列

这篇关于加入对 generate_series() 的计数查询并将 Null 值检索为“0"的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！