從所有列中選擇不同的數據，但是1

Question

我正在使用select distinct來返回唯一不同的數據，但我希望它將所有列的區別基於不包含agg_source_tag_tag_name的區別。

這裏是我的查詢：

SELECT DISTINCT agg_article_title,agg_article_link,agg_article_media,agg_article_description,agg_article_source_name, agg_source_tag_tag_name, agg_source_url 
FROM agg_article join agg_source ON agg_article_source_name = agg_source_name join agg_source_tag ON agg_source_name = agg_source_tag_source_name 
WHERE agg_source_included = 1 

我得到以下結果：

agg_article_title | agg_article_link | agg_article_media | agg_article_description | agg_article_source_name | agg_source_tag_tag_name | agg_source_url | 
--------------------------------------------------------------------------------------------------------------------------------------------------------- 
some title 1  | some link 1  | some media 1  | some description 1  | some source name 1  | tag1     | someurl1  | 
--------------------------------------------------------------------------------------------------------------------------------------------------------- 
some title 1  | some link 1  | some media 1  | some description 1  | some source name 1  | tag2     | someurl1  | 

我碰到下面的返回，因爲所有的列是事實，因爲agg_source_tag_tag_name的不同，但是，我只想要返回一行，因爲我想省略agg_source_tag_tag_name。

Answer 1

嘗試對GROUP BY

SELECT agg_article_title,agg_article_link,agg_article_media,agg_article_description,agg_article_source_name, agg_source_tag_tag_name, agg_source_url 
    FROM agg_article 
    join agg_source 
    ON agg_article_source_name = agg_source_name 
    join agg_source_tag 
    ON agg_source_name = agg_source_tag_source_name 
    WHERE agg_source_included = 1 
    GROUP BY agg_article_title, agg_article_link 

，您可以按什麼都cloumn你想要的。