我有两张桌子:
医务人员
CREATE TABLE "medic" (
"id" BIGINT NOT NULL,
"name" CHARACTER VARYING(255) NOT NULL,
PRIMARY KEY ("id")
);
评论
CREATE TABLE IF NOT EXISTS "comment" (
"id" BIGINT NOT NULL,
"medic_id" BIGINT NOT NULL,
"comment" CHARACTER VARYING(1024) NOT NULL,
"created_at" TIMESTAMP WITHOUT TIME ZONE NOT NULL DEFAULT now(),
CONSTRAINT pk_comment PRIMARY KEY (id),
CONSTRAINT fk_comment_medic FOREIGN KEY (medic_id)
REFERENCES medic(id) ON UPDATE NO ACTION ON DELETE NO ACTION
);
现在我想获得medic_id, name, comments_count
和所有ordered by created_at
这是我到目前为止所尝试的内容:
SELECT m.id, m.name, COUNT(c.id)
FROM COMMENT AS c
JOIN medic AS m ON m.id = c.medic_id
GROUP BY m.id, m.name, c.created_at
ORDER BY c.created_at DESC
但显然这不起作用,因为按日期分组是没有意义的,但是当我想按日期订购时我必须这样做。
另一个appraoch是使用窗口函数。特别是rank() over (partition by m.id order by c.created_at desc)
。但在这种情况下,我失去了对所有记录的排序。
这里有一些SQLFiddle。
我正在使用Postgres 9.3
答案 0 :(得分:0)
我猜你想通过最近的评论日期订购:
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