Postgresql LEFT JOIN json_agg()忽略/删除NULL

Question

SELECT C.id, C.name, json_agg(E) AS emails FROM contacts C
LEFT JOIN emails E ON C.id = E.user_id
GROUP BY C.id;

例如,Postgres 9.3创建输出

  id  |  name  |  emails
-----------------------------------------------------------
   1  |  Ryan  |  [{"id":3,"user_id":1,"email":"hello@world.com"},{"id":4,"user_id":1,"email":"again@awesome.com"}]
   2  |  Nick  |  [null]

由于我使用LEFT JOIN,因此会出现没有右表匹配的情况,因此将空(null)值替换为右表列.因此,我将[null]作为JSON聚合之一获得.

我如何忽略/删除null所以[]当右表列为空时我有一个空的JSON数组？

干杯!

Answer 1

在9.4中,您可以使用coalesce和聚合过滤器表达式.

SELECT C.id, C.name, 
  COALESCE(json_agg(E) FILTER (WHERE E.user_id IS NOT NULL), '[]') AS emails 
FROM contacts C
LEFT JOIN emails E ON C.id = E.user_id
GROUP BY C.id, C.name
ORDER BY C.id;

过滤器表达式可防止聚合处理空行,因为不满足左连接条件,因此最终使用数据库null而不是json [null].一旦数据库为null,就可以像往常一样使用coalesce.

http://www.postgresql.org/docs/9.4/static/sql-expressions.html#SYNTAX-AGGREGATES

Answer 2

这样的事情可能会是什么？

select
    c.id, c.name,
    case when count(e) = 0 then '[]' else json_agg(e) end as emails
from contacts as c
    left outer join emails as e on c.id = e.user_id
group by c.id

sql fiddle demo

你也可以在加入之前进行分组(我更喜欢这个版本,它更清楚一点):

select
    c.id, c.name,
    coalesce(e.emails, '[]') as emails
from contacts as c
    left outer join (
        select e.user_id, json_agg(e) as emails from emails as e group by e.user_id
    ) as e on e.user_id = c.id

sql fiddle demo

Answer 3

如果这实际上是一个 PostgreSQL 错误，我希望它已在 9.4 中修复。很烦人。

SELECT C.id, C.name, 
  COALESCE(NULLIF(json_agg(E)::TEXT, '[null]'), '[]')::JSON AS emails 
FROM contacts C
LEFT JOIN emails E ON C.id = E.user_id
GROUP BY C.id;

我个人不做 COALESCE 位，只返回 NULL。您的来电。

Answer 4

这种方法可行，但必须有更好的方法:(

SELECT C.id, C.name, 
  case when exists (select true from emails where user_id=C.id) then json_agg(E) else '[]' end
FROM contacts C
LEFT JOIN emails E ON C.id = E.user_id
GROUP BY C.id, C.name;

演示： http: //sqlfiddle.com/#!15/ddefb /16