Question

表数据集：


SSN        ID  F_NAME  L_NAME
000000000  1   JANE    DOE
123456789  2   PETE    PETER
123456789  3   JOHN    SHAW
234567890  4   ALAN    DAVID
234567890  5   MAY     APRI
345677891  6   JUNE    ALYSSA
456789123  7   LISA    LINH
567891234  8   HAL     LEY
567891234  9   DEREK   LI
567891234  10  JIM     JAMES

从上面的数据集中，我的目的是识别具有相同SSN的ID /人口统计数据，并水平列出所有名称和结果，如下所示。我们还对与每个SSN关联的唯一ID计数感兴趣。


SSN       COUNT ID  F_NAME  L_NAME ID_2 F_NAME_2 L_NAME_2 ID_3 F_NAME_3 L_NAME_3
123456789 2     2   PETE    PETER  3    JOHN     SHAW
234567889 2     4   ALAN    DAVID  5    MAY      APRI
567891234 3     8   HAL     LEY    9    DEREK    LI       10   JIM      JAMES

帮助？

Answer 1

我会以下列方式使用listagg，您也可以添加唯一ID：

create table demographics(
ssn number(9),
id number(12),
f_name varchar2(50),
l_name varchar2(50));


insert into demographics values(000000000,  1,   'JANE',    'DOE');
insert into demographics values(123456789,  2,   'PETE',    'PETER');
insert into demographics values(123456789,  3,   'JOHN',    'SHAW');
insert into demographics values(234567890,  4,   'ALAN',    'DAVID');
insert into demographics values(234567890,  5,   'MAY',     'APRI');
insert into demographics values(345677891,  6,   'JUNE',    'ALYSSA');
insert into demographics values(456789123,  7,   'LISA',    'LINH');
insert into demographics values(567891234,  8,   'HAL',     'LEY');
insert into demographics values(567891234,  9,   'DEREK',   'LI');
insert into demographics values(567891234,  10,  'JIM',     'JAMES');

SELECT ssn, count(ssn) as count, LISTAGG(f_name || ' ' || l_name, ', ') WITHIN GROUP (ORDER BY f_name || ' ' || l_name) "Names"
FROM demographics
group by ssn;



SSN COUNT   Names
0   1   JANE DOE
123456789   2   JOHN SHAW, PETE PETER
234567890   2   ALAN DAVID, MAY APRI
345677891   1   JUNE ALYSSA
456789123   1   LISA LINH
567891234   3   DEREK LI, HAL LEY, JIM JAMES

Answer 2

根据@ lad2025的建议

查看有关pivot here的更多信息

根据您可以使用的格式，我将使用的快速解决方案是SQL (Oracle)中的listagg函数，类似于MySQL group_concat，例如< / p>

SELECT SSN,
       count(*),
       LISTAGG(ID, ', ') WITHIN GROUP (ORDER BY ID),
       LISTAGG(F_NAME, ', ') WITHIN GROUP (ORDER BY F_NAME)
FROM table GROUP BY SSN HAVING count(*) > 1;

会导致

123456789 2     2,3      PETE, JOHN    
234567889 2     4,5      ALAN, MAY   
567891234 3     8,9,10   HAL, DEREK, JIM

或者，您可以识别具有重复SSN的所有行 - 这将避免使用pivot并将结果保持为相同的关系格式

SELECT *
FROM table
WHERE SSN in (SELECT SSN from table group by SSN having count(*) > 1);

将返回包含重复SSN的所有行：

SSN        ID  F_NAME  L_NAME
123456789  2   PETE    PETER
123456789  3   JOHN    SHAW
234567890  4   ALAN    DAVID
234567890  5   MAY     APRI
567891234  8   HAL     LEY
567891234  9   DEREK   LI
567891234  10  JIM     JAMES

以水平格式显示具有公共元素的行

2 个答案: