Question

我有一个约会数据库，每个人都有一个人格，序列号，约会日期和两个问卷评分。

我需要做的是每个人带回一行，显示他们的第一次约会，以及他们最后的约会，以及两者的得分。

我已经花了很长时间在一个非常简单的查询上，我已经尝试使用以下内容给我第一次约会分数：

SELECT 
PERSONID,
MIN(CONTACTDATE) AS 'FIRST_CONTACT',
QUEST_1,
QUEST_2

FROM TBL_APPOINTMENTS

GROUP BY
PERSONID,
CONTACTDATE,
QUEST_1,
QUEST_2

ORDER BY 
PERSONID,
FIRST_CONTACT

正如您可能猜到的那样，每次QUEST_1或QUEST_2中的值发生变化时，这都会给我重复的行。

有人可以帮忙吗？我确信这一切都非常简单，但却让我不得不在墙上！

提前致谢！

Answer 1

使用TypeError: theForm.submit is not a function theForm.submit();您可以按约ROW_NUMBER()从1开始为约会分配一个连续值。

在下面的例子中，我创建了两个这样的序数，一个是时间前进，另一个是时间倒退。然后我可以选择PERSONID以及first going forward。

first going backwards

编辑 将第一个和最后一个汇总到一行......

WITH
  sorted AS
(
  SELECT
    PERSONID,
    CONTACTDATE,
    QUEST_1,
    QUEST_2,
    ROW_NUMBER() OVER (PARTITION BY PERSONID ORDER BY CONTACTDATE  ASC)   AS ORD_FWD,
    ROW_NUMBER() OVER (PARTITION BY PERSONID ORDER BY CONTACTDATE DESC)   AS ORD_REV
  FROM
    TBL_APPOINTMENTS
)
SELECT
  PERSONID,
  CONTACTDATE,
  QUEST_1,
  QUEST_2
FROM
  sorted
WHERE
  ORD_FWD = 1 OR ORD_REV = 1
ORDER BY
  PERSONID,
  CONTACTDATE

注意：第一次和最后一次约会可能是同一个约会（如果该人只有一次约会）。

Answer 2

如果您要查找的结果是这样的：

PERSONID    FIRST_CONTACT FIRST_QUEST_1 FIRST_QUEST_2 LAST_CONTACT LAST_QUEST_1 LAST_QUEST_2
----------- ------------- ------------- ------------- ------------ ------------ ------------
1           2015-01-01    10            11            2015-01-21   21           211
2           2015-01-01    11            24            2015-01-31   12           25
3           2015-02-01    13            21            2015-03-01   14           28
4           2015-03-01    15            29            2015-04-01   16           21

然后下面的查询会给你这个。请注意，如果一个人只有一条记录，则最大和最小日期和分数将相同。

SELECT 
    PERSONID, 
    FIRST_CONTACT, FIRST_QUEST_1, FIRST_QUEST_2, 
    LAST_CONTACT, LAST_QUEST_1, LAST_QUEST_2
FROM (
    SELECT PERSONID, MIN(CONTACTDATE) AS 'FIRST_CONTACT', MAX(CONTACTDATE) AS 'LAST_CONTACT' 
    FROM TBL_APPOINTMENTS a 
    GROUP BY PERSONID
) a
CROSS APPLY (
    SELECT QUEST_1 AS 'FIRST_QUEST_1', QUEST_2  AS 'FIRST_QUEST_2'
    FROM TBL_APPOINTMENTS 
    WHERE PERSONID = a.PERSONID AND CONTACTDATE = A.FIRST_CONTACT
) o_first
CROSS APPLY (
    SELECT QUEST_1 AS 'LAST_QUEST_1', QUEST_2  AS 'LAST_QUEST_2'
    FROM TBL_APPOINTMENTS 
    WHERE PERSONID = a.PERSONID AND CONTACTDATE = A.LAST_CONTACT
) o_last
ORDER BY PERSONID, FIRST_CONTACT;

可以使用连接编写相同的查询：

SELECT 
    A.PERSONID, 
    FIRST_CONTACT, 
    F.QUEST_1 AS FIRST_QUEST_1, 
    F.QUEST_2 AS FIRST_QUEST_2, 
    LAST_CONTACT, 
    L.QUEST_1 AS LAST_QUEST_1, 
    L.QUEST_2 AS LAST_QUEST_2
FROM (
    SELECT PERSONID, MIN(CONTACTDATE) AS 'FIRST_CONTACT', MAX(CONTACTDATE) AS 'LAST_CONTACT' 
    FROM TBL_APPOINTMENTS a 
    GROUP BY PERSONID
) a
JOIN TBL_APPOINTMENTS F ON F.PERSONID = a.PERSONID AND A.FIRST_CONTACT = F.CONTACTDATE
JOIN TBL_APPOINTMENTS L ON L.PERSONID = a.PERSONID AND A.LAST_CONTACT = L.CONTACTDATE
ORDER BY A.PERSONID, FIRST_CONTACT;

Answer 3

    SELECT PERSONID,
    MIN(CONTACTDATE) AS 'FIRST_CONTACT', MAX(CONTACTDATE) AS 'LAST_CONTACT'
    QUEST_1,
    QUEST_2
    FROM TBL_APPOINTMENTS
    GROUP BY PERSONID,
CONTACTDATE,
QUEST_1,
QUEST_2
 ORDER BY 
    PERSONID,
    FIRST_CONTACT

从约会列表中选择第一个和最后一个分数

3 个答案: