从每组预言中选择最高记录和计数？ [复制]答案

【问题标题】：select top record and count from each group oracle? [duplicate]从每组预言中选择最高记录和计数？ [复制]
【发布时间】：2020-01-06 04:30:43
【问题描述】：

获得最高记录并计数？

PKid    |    QId    |    QNumber    |    EmailId    |    FirstName    |    LastName    |
1    |    102    |    A1022    |    jsmith@test.com    |    John    |    Smith    |
2    |    103    |    A1021    |    jsmith@test.com    |    John    |    smith    |
3    |    104    |    A1031    |    jblack@test.com    |    Jack    |    Black    |
4    |    105    |    A1032    |    jblack@test.com    |    Jack    |    black    |
5    |    106    |    A1023    |    jsmith@test.com    |    John    |

我想按名称和按职业 desc 和计数的顺序获取记录组。像这样-

  S.no    |    QId    |    QNumber    |    EmailId    |    FirstName    |    LastName    |    Count
1    |    106    |    A1023    |    jsmith@test.com    |    John    |    |    3 
2    |    105    |    A1032    |    jblack@test.com    |    Jack    |    black    |    2

我尝试了类似的方法，但没有运气---

    SELECT
    ROW_NUMBER() OVER(
        ORDER BY
            COUNT(1) DESC
    ) AS S_NO,QId,
    MAX(QNUMBER) AS QNUMBER,
    EmailId,FirstName,LastName
    COUNT(1)
FROM
    TblEmp   
GROUP BY
    EmailId;

【问题讨论】：

标签： sql oracle oracle11g greatest-n-per-group

【解决方案1】：

我会做的

ROW_NUMBER() OVER(PARTITION BY EmailId ORDER BY qnumber DESC) AS rown

然后将其全部包装在具有WHERE rown = 1 的外部查询中。外部查询也会计算 S_NO，而不是内部查询

类似这样的：

SELECT 
  ROW_NUMBER() OVER(ORDER BY qnumber) AS S_NO,
  ee.*
FROM
(
    SELECT
      ROW_NUMBER() OVER(PARTITION BY EmailId ORDER BY qnumber DESC) AS rown,
      COUNT(*) OVER(PARTITION BY EmailId) AS count,
      e.*
    FROM
        TblEmp e   

) ee
WHERE ee.rown = 1

但我不太确定您的“按计数排序”在哪里；对我来说，看起来你正在获取最新的（根据 qnumber）记录，计算其他记录，并将一些任意递增的数字重新分配为 s_no

【讨论】：