postgres STRING_AGG（）返回重复项？

Question

我看过一些类似的帖子，要求提供有关从查询中获得不同结果的建议。可以用子查询解决，但是我正在聚合的列image_name是唯一的image_name VARCHAR(40) NOT NULL UNIQUE。我不认为那应该是必要的。

这是spot_images表中的数据

spotdk=# select * from spot_images;
 id | user_id | spot_id |              image_name              
----+---------+---------+--------------------------------------
  1 |       1 |       1 | 81198013-e8f8-4baa-aece-6fbda15a0498
  2 |       1 |       1 | 21b78e4e-f2e4-4d66-961f-83e5c28d69c5
  3 |       1 |       1 | 59834585-8c49-4cdf-95e4-38c437acb3c1
  4 |       1 |       1 | 0a42c962-2445-4b3b-97a6-325d344fda4a
(4 rows)

SELECT Round(Avg(ratings.rating), 2) AS rating, 
       spots.*, 
       String_agg(spot_images.image_name, ',') AS imageNames
FROM   spots 
       FULL OUTER JOIN ratings 
                    ON ratings.spot_id = spots.id 
       INNER JOIN spot_images 
               ON spot_images.spot_id = spots.id 
WHERE  spots.id = 1 
GROUP  BY spots.id; 

这是图像行的结果：

81198013-e8f8-4baa-aece-6fbda15a0498,
21b78e4e-f2e4-4d66-961f-83e5c28d69c5,
59834585-8c49-4cdf-95e4-38c437acb3c1,
0a42c962-2445-4b3b-97a6-325d344fda4a,
81198013-e8f8-4baa-aece-6fbda15a0498,
21b78e4e-f2e4-4d66-961f-83e5c28d69c5,
59834585-8c49-4cdf-95e4-38c437acb3c1,
0a42c962-2445-4b3b-97a6-325d344fda4a,
81198013-e8f8-4baa-aece-6fbda15a0498,
21b78e4e-f2e4-4d66-961f-83e5c28d69c5,
59834585-8c49-4cdf-95e4-38c437acb3c1,
0a42c962-2445-4b3b-97a6-325d344fda4a

不带换行符，我添加它们是为了提高可见性。

我应该怎么做来一次检索图像名称？

Answer 1

如果您不想重复，请使用DISTINCT：

   String_agg(distinct spot_images.image_name, ',') AS imageNames

Answer 2

类似，ratings中有几行与给定的spot相匹配，和 spot_images中有几行也与给定的sport相匹配。结果，行变得重复了。

避免这种情况的一种选择是在子查询中进行汇总：

SELECT r.avg_raging
       s.*, 
       si.image_names
FROM   spots s
       FULL OUTER JOIN (
           SELECT spot_id, Round(Avg(ratings.rating), 2) avg_rating
           FROM ratings
           GROUP BY spot_id
       ) r ON r.spot_id = s.id
       INNER JOIN (
           SELECT spot_id, string_agg(spot_images.image_name, ',') image_names
           FROM spot_images
           GROUP BY spot_id
       ) si ON si.spot_id = s.id 
WHERE  s.id = 1 

实际上可能比外部聚合更有效率。

注意：如果不查看数据就很难分辨，但是我不确定这里是否确实需要FULL JOIN。 LEFT JOIN实际上可能就是您想要的。