Question

SELECT *, SUM(CASE WHEN "token"."name" = ''  THEN 5 ELSE 9 END) AS "n"
FROM "word"
LEFT OUTER JOIN "token" ON ("word"."id" = "token"."word_id")
where ("word"."id" = 1)
GROUP BY "word"."id";

这给出了14的预期结果，因为只有一个空标记和一个非空标记。

SELECT *, SUM(CASE WHEN "token"."name" = ''  THEN 5 ELSE 9 END) AS "n"
FROM "word"
LEFT OUTER JOIN "token" ON ("word"."id" = "token"."word_id")
INNER JOIN "token" T3 ON ("word"."id" = T3."word_id")
where ("word"."id" = 1)
GROUP BY "word"."id";

此查询输出n=28，这是14加倍。

为什么？

Answer 1

这是您修改过的查询：

SELECT *, SUM(CASE WHEN t."name" = ''  THEN 5 ELSE 9 END) AS "n"
FROM "word" w LEFT OUTER JOIN
     "token" t
     ON w."id" = t."word_id" INNER JOIN
     "token" T3
     ON w."id" = T3."word_id"
WHERE w."id" = 1
GROUP BY w."id";

没有意义。您将word中的同一列加入token中的同一列 - 两次。您获得重复的原因是因为您在token中有多个行用于给定的单词。您可以轻松查看：

select word_id, count(*)
from token
group by word_id
having count(*) > 1;

但是，您的查询毫无意义。我可能会建议您使用示例数据，所需结果以及您希望查询执行操作的说明来询问另一个问题。

Postgres Additional Join增加总和

1 个答案: