列出一个人及其第 n 个电话号码答案

【问题标题】：Listing a person and their nth phone number列出一个人及其第 n 个电话号码
【发布时间】：2019-11-04 14:38:34
【问题描述】：

我正在使用 PowerSchool (Oracle DB)，为了导出我们的紧急机器人呼叫服务，我需要提取学生报告以及与他们关联的前 7 个紧急电话号码。我有一个查询，它只提取紧急号码，但无法将第 n 个电话记录拉到正确的上下文中。当我运行当前代码时，我得到“ORA-00923: FROM keyword not found where expected”。

我可以从 WITH 语句中的查询中选择所有值，并按预期运行。但是，我不知道如何在主查询中正确抓取它们。

WITH emerglist as (
    select emergnumbers.stu_dcid, emergnumbers.contact_priority_order, 1 as phone_order, emergnumbers.contact_phone_1 as phone_number
    from emergnumbers where contact_phone_1 is not null
    UNION
    select emergnumbers.stu_dcid, emergnumbers.contact_priority_order, 2 as phone_order, emergnumbers.contact_phone_2 as phone_number
    from emergnumbers where contact_phone_2 is not null
    UNION
    select emergnumbers.stu_dcid, emergnumbers.contact_priority_order, 3 as phone_order, emergnumbers.contact_phone_3 as phone_number
    from emergnumbers where contact_phone_3 is not null
)
--select * from emerglist;  -- this works

select distinct 
    s.student_number as referencecode,
    s.first_name as firstname,
    s.last_name as lastname,
    --emergency numbers
    (select enum.phone_number from enum where rn=1) as emerg_1
    (select enum.phone_number from enum where rn=2) as emerg_2,
    (select enum.phone_number from enum where rn=3) as emerg_3,
    (select enum.phone_number from enum where rn=4) as emerg_4,
    (select enum.phone_number from enum where rn=5) as emerg_5,
    (select enum.phone_number from enum where rn=6) as emerg_6,
    (select enum.phone_number from enum where rn=7) as emerg_7
from students s
    left outer join (  select emerglist.*, rownum rn from emerglist order by contact_priority_order, phone_order  ) enum on s.dcid = enum.stu_dcid             
where s.enroll_status=0
;

编辑添加：我的目标是每个学生有一行，填写尽可能多的紧急号码列，没有间隙（例如，不能有 emerg1 = #、emerg_2 = null、emerg_3 = #） .每个学生可以有 0-10 个联系人，每个联系人可以有 0-5 个号码。我们限制为每位学生 7 个号码。这是一个更大查询的子集，它还为普通消息（例如下雪天、电子邮件地址和其他人口统计数据）提取主要数字。除了紧急号码外，所有这些都有效。这些只有在真正的紧急情况下才会使用，并且我们希望联系尽可能多的号码。

来自 EmergList 的样本数据：

STU_DCID   CONTACT_PRIORITY_ORDER   PHONE_ORDER   PHONE_NUMBER
52  1   1   (222) 222-6476
52  2   1   (222) 555-9649
52  2   2   (222) 888-2212
58  1   1   (222) 222-1734
58  1   2   (222) 555-7222

期望的结果：

REFERENCECODE   FIRSTNAME   LASTNAME   EMERG_1   EMERG_2   EMERG_3   EMERG_4   EMERG_5   EMERG_6   EMERG_7
52   John   Smith   (222) 222-6476   (222) 555-9649   (222) 888-2212
58   Mary   Jones   (222) 222-1734   (222) 555-7222

【问题讨论】：

您能提供示例数据和预期输出吗？认为使用 PIVOT 进行正常连接会更好，而不是许多内部选择...
我以前从未使用过 Pivot...我会研究一下 - 谢谢！
我仍在试图弄清楚如何在此站点中嵌入示例数据。
您不能嵌入它，您必须手动将其编写为文本，将其格式化为表格并放置空格，以便网站将其视为代码......看看这个例子stackoverflow.com/questions/42955002/…

标签： sql oracle subquery

【解决方案1】：

接下来是你如何做到这一点的一种方法......

select STU_DCID, 
   MAX("1") "1", 
   MAX("2") "2",
   MAX("3") "3",
   MAX("4") "4", 
   MAX("5") "5",
   MAX("6") "6",
   MAX("7") "7", 
   MAX("8") "8",
   MAX("9") "9",
   MAX("10") "10"
from (
   select STU_DCID, 
          CONTACT_PRIORITY_ORDER, 
          PHONE_ORDER, 
          PHONE_NUMBER,
          ROW_NUMBER() OVER (PARTITION BY STU_DCID 
                             ORDER BY CONTACT_PRIORITY_ORDER, PHONE_ORDER) RN
   from EmergList) src
PIVOT (
MAX(PHONE_NUMBER)
FOR RN IN (1 "1", 2 "2", 3 "3",
           4 "4", 5 "5", 6 "6",
           7 "7", 8 "8", 9 "9",
           10 "10")
) GROUP BY STU_DCID;

然后您只需更改列的别名并在src 上添加额外的连接以获得学生姓名...您将得到的当前结果是这样的

STU_DCID    |       1       |       2        |      3         |4|5|6|7|8|9|10|
    52      |(222) 222-6476 | (222) 555-9649 | (222) 888-2212 | | | | | | | |
    58      |(222) 222-1734 | (222) 555-7222 |                | | | | | | | |

【讨论】：