表3
OWNERS: pk name
POSTS: pk, post_date, owner_pk
REPLIES: pk,reply_date, post_pk,owner_pk,comment_count
两个日期:日期1和日期2 我希望能取得以下成果:
OWNER:名称 COUNT (posts) SUM(comment_count)
我的问题是,我无法理解如何编写一份单点文件,将定位和定位程序表分开。
页: 1 第2天,没有发表任何意见。 答复表列出了某一天/天的评论总数。
PostA Day1
CommentA1 Day1
CommentA2 Day1
PostB Day2
CommentA3 Day2
PostC Day3
第1天:
Owner:Name count(post) sum(comment_count)
OwnerName 1 2
• 为2天提出询问:
OwnerName 1 1
• 主持3日工作:
OwnerName 1 0
你也许需要更详细地回答你的问题,但我怀疑你在这样做之后:
SELECT name, COUNT(posts.pk), COUNT(replies.pk)
FROM owners, OUTER (posts, OUTER replies)
WHERE owners.pk = posts.owner_pk
AND posts.pk = replies.post_pk
AND (posts.post_date BETWEEN ? AND ? OR replies.reply_date BETWEEN ? AND ?)
GROUP BY name
而不是<代码>SUM(),什么是<编码>COUNT(DISTINCT)?
SELECT
o.name,
COUNT(DISTINCT p.pk) AS posts,
COUNT(DISTINCT r.pk) AS replies
FROM
owners o
LEFT JOIN posts p ON p.owner_pk = o.pk AND p.post_date BETWEEN @date1 AND @date2
LEFT JOIN replies r ON r.post_pk = p.pk AND r.post_date BETWEEN @date1 AND @date2
GROUP BY
o.pk,
o.name
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