斯卡拉油滑3.1.0 - 從2個表計數中列左連接查詢

Question

我試圖構建留在油滑3.1本類似的聯接查詢重複計算：

SELECT 
    COUNT(p.id) AS replies, 
    COUNT(f.id) AS images 
    FROM posts AS p 
    LEFT JOIN files AS f 
    ON p.id = f.post_id 
    WHERE thread = :thread_id 

加入查詢的一部分是相當簡單，看起來像這樣：

val joinQ = (threadId: Rep[Long]) => 
    (postDAO.posts joinLeft fileRecordDAO.fileRecords on (_.id === _.postId)) 
    .filter(_._1.thread === threadId) 

使用joinQ(1L).map(_._1.id).length產生COUNT(1)，其對所有行 - 而不是結果，我想獲得的。使用joinQ(1L).map(_._1.id).countDistinct產生COUNT(DISTINCT p.id)這是有點什麼我要找的，但試圖做其中的兩個產生這個畸形：

select x2.x3, x4.x5 
    from (select count(distinct x6.`id`) as x3 
    from `posts` x6 
    left outer join `files` x7 on x6.`id` = x7.`post_id` 
    where x6.`thread` = 1) x2, 
    (select 
    count(distinct (case when (x8.`id` is null) then null else 1 end)) as x5 
    from `posts` x9 
    left outer join `files` x8 on x9.`id` = x8.`post_id` 
    where x9.`thread` = 1) x4 

這裏的雙countDistinct代碼：

val q = (threadId: Rep[Long]) => { 
    val join = joinQ(threadId) 
    val q1 = join.map(_._1.id).countDistinct // map arg type is 
    val q2 = join.map(_._2).countDistinct // (Post, Rep[Option[FileRecord]]) 
    (q1, q2) 
} 

任何想法？ :)

Answer 1

計數是聚合函數，所以您還需要通過某個字段（e.x. id）進行分組。試試下面的查詢：

def joinQ(threadId: Long) = (postDAO.posts joinLeft fileRecordDAO.fileRecords on (_.id === _.postId)) 
.filter(_._1.thread === threadId) 

val q = (threadId: Long) => { 
    joinQ(threadId).groupBy(_._1.id).map { 
    case (id, qry) => (id, qry.map(_._1.id).countDistinct, qry.map(_._2.map(_.id)).countDistinct) 
    } 
} 

它產生下一個SQL：

SELECT x2.`id`, 
     Count(DISTINCT x2.`id`), 
     Count(DISTINCT x3.`id`) 
FROM `posts` x2 
     LEFT OUTER JOIN `files` x3 
        ON x2.`id` = x3.`post_id` 
WHERE x2.`thread` = 10 
GROUP BY x2.`id`