T-SQL GROUP BY：消除多次出现

Question

由于日期值不同，如何防止以下查询中出现多个相同的价格行？我只需要显示价格与先前记录不同的记录的第一次出现，但是跳过直到价格字段发生另一次更改。

SELECT STOCKID, PRICE, TRANSTYPENAME, TRANSDATE
FROM VE_STOCKTRANS 
WHERE STOCKID = 6000 AND TANSTYPE IN (3500,3553)
GROUP BY PRICE, STOCKID, TRANSTYPENAME, TRANSDATE

即我需要省略多个35.83333333行，而仅采用它的首次出现（24/4/2018）

对戈登答案的图片答复：

Answer 1

使用LAG()查看价格是否发生变化：

SELECT STOCKID, PRICE, TRANSTYPENAME, TRANSDATE
FROM (SELECT st.*,
             LAG(st.price) OVER (PARTITION BY stockId ORDER BY transdate) as prev_price
      FROM VE_STOCKTRANS st
      WHERE STOCKID = 6000 AND TANSTYPE IN (3500, 3553)
     ) st
WHERE prev_price IS NULL OR prev_price <> price;

编辑：

如果您只希望第一次看到每个价格的列表（与您在问题中提到并描述的价格变化大不相同），请使用row_number()：

SELECT STOCKID, PRICE, TRANSTYPENAME, TRANSDATE
FROM (SELECT st.*,
             ROW_NUMBER() OVER (PARTITION BY st.stockId, st.price ORDER BY st.transdate) as seqnum
      FROM VE_STOCKTRANS st
      WHERE STOCKID = 6000 AND TANSTYPE IN (3500, 3553)
     ) st
WHERE seqnum = 1;

Answer 2

您可以这样获得第一个TRANSDATE：

SELECT STOCKID, PRICE, TRANSTYPENAME, min(TRANSDATE)
FROM VE_STOCKTRANS 
WHERE STOCKID = 6000 AND TRANSTYPE IN (3500,3553)
GROUP BY PRICE, STOCKID, TRANSTYPENAME

Answer 3

glm()

这可能会有帮助。