postgres: COUNT, DISTINCT 没有为 window 函数实现

Question

当我使用COUNT + window function(OVER)时，我正在尝试使用COUNT(DISTINC column) OVER(PARTITION BY column) ) 。 我收到一个类似标题中的错误，无法正常工作。

我已经研究过如何处理这个错误，但我还没有找到如何处理下面这样一个复杂查询的示例。 我找不到如何处理如下所示的复杂查询的示例，我不知道如何处理它。

第 65 行存在COUNT部分的问题。如此复杂的查询如何在不减速的情况下解决？

WITH RECURSIVE "cte" AS((
        SELECT
            "videos_productvideocomment"."id",
            "videos_productvideocomment"."user_id",
            "videos_productvideocomment"."video_id",
            "videos_productvideocomment"."parent_id",
            "videos_productvideocomment"."text",
            "videos_productvideocomment"."commented_at",
            "videos_productvideocomment"."edited_at",
            "videos_productvideocomment"."created_at",
            "videos_productvideocomment"."updated_at",
            "videos_productvideocomment"."id" AS "root_id"
        FROM
            "videos_productvideocomment"
        WHERE
            (
                "videos_productvideocomment"."parent_id" IS NULL
            AND "videos_productvideocomment"."video_id" = 'f264433c-c0af-49cc-8b40-84453da71b2d'
            )
    ) UNION(
    SELECT
        "videos_productvideocomment"."id",
        "videos_productvideocomment"."user_id",
        "videos_productvideocomment"."video_id",
        "videos_productvideocomment"."parent_id",
        "videos_productvideocomment"."text",
        "videos_productvideocomment"."commented_at",
        "videos_productvideocomment"."edited_at",
        "videos_productvideocomment"."created_at",
        "videos_productvideocomment"."updated_at",
        "cte"."root_id" AS "root_id"
    FROM
        "videos_productvideocomment"
        INNER JOIN
            "cte"
        ON  "videos_productvideocomment"."parent_id" = "cte"."id"
))
SELECT
    *,
    EXISTS(
        SELECT
            (1) AS "a"
        FROM
            "videos_productvideolikecomment" U0
        WHERE
            (
                U0."comment_id" = t."id"
            AND U0."user_id" = '3bd3bc86-0335-481e-9fd2-eb2fb1168f48'
            )
        LIMIT 1
    ) AS "liked"
FROM
    (
        SELECT DISTINCT
            "cte"."id",
            "cte"."created_at",
            "cte"."updated_at",
            "cte"."user_id",
            "cte"."text",
            "cte"."commented_at",
            "cte"."edited_at",
            "cte"."parent_id",
            "cte"."video_id",
            "cte"."root_id" AS "root_id",
            COUNT(DISTINCT "cte"."root_id") OVER(PARTITION BY "cte"."root_id") AS "reply_count", <--- here
            COUNT("videos_productvideolikecomment"."id") OVER(PARTITION BY "cte"."id") AS "liked_count"
        FROM
            "cte"
            LEFT OUTER JOIN
                "videos_productvideolikecomment"
            ON  (
                    "cte"."id" = "videos_productvideolikecomment"."comment_id"
                )
    ) t
WHERE
    t."id" = t."root_id"
ORDER BY
    CASE
        WHEN t."user_id" = '3bd3bc86-0335-481e-9fd2-eb2fb1168f48' THEN 0
        ELSE 1
    END ASC,
    "liked_count" DESC

Answer 1

DISTINCT 会查找重复项并将其删除，但在大数据中处理此查询将花费大量时间，您应该在编程部分处理中间记录我认为它会比它快。 感谢