同一个表，用sql查询今天和昨天的差值，然后排序

版权声明： https://blog.csdn.net/Dongguabai/article/details/82776674

SELECT tagId2, yesterday, cha, absSum
FROM (
	SELECT tt.tagId2, tt.yesterday, tt.absSum, tt.cha, rownum AS rn
	FROM (
		SELECT tagId2, yesterday, before_yesterday - yesterday AS cha
			, abs(before_yesterday - yesterday) AS absSum
		FROM (
			SELECT x.tagId1 AS tagId2, SUM(CASE 
					WHEN x.VN = 20180529 THEN x.totalMarkNum1
					ELSE 0
				END) AS yesterday, SUM(CASE 
					WHEN x.VN = 20180528 THEN x.totalMarkNum1
					ELSE 0
				END) AS before_yesterday
			FROM (
				SELECT t.tag_id AS tagId1, t.vn, SUM(t.key_count) AS totalMarkNum1
				FROM bdatag_hist_tag_result_sum t
				WHERE t.org_id LIKE '3%'
					AND t.vn BETWEEN '20180528' AND '20180529'
				GROUP BY t.tag_id, t.vn
			) x
			GROUP BY x.tagId1
		)
		ORDER BY absSum DESC
	) tt
)
WHERE rn <= 10