我需要從select語句導致出身的多重數據單個ID

Question

APPLICANT_ID DATE_OF_BIRTH 
206209579  04/29/82 
206209579  04/29/82 
203276426  06/01/69 
203276426  02/03/96 
203276426  06/02/99 

我需要的結果是

206209579  04/29/82,04/29/82 
203276426  06/01/69,02/03/96 

請建議其需要充分

Answer 1

在Oracle 11g和上面，你可以使用listagg：

SELECT applicant_id, 
     LISTAGG(date_of_birth) WITHIN GROUP (ORDER BY date_of_birth) 
FROM  my_table 
GROUP BY applicant_id 

Answer 2

作爲替代方案，用於Oracle版本之前到11g其中LISTAGG不支持，您可以使用ROW_NUMBER（）和SYS_CONNECT_BY_PATH函數。

測試用例：

SQL> SELECT * FROM table1; 

APPLICANT_ID DATE_OF_B 
------------ --------- 
    206209579 29-APR-82 
    206209579 29-APR-82 
    203276426 01-JUN-69 
    203276426 03-FEB-96 
    203276426 02-JUN-99 

SQL> 
SQL> SELECT applicant_id, 
    2   LTRIM(MAX(SYS_CONNECT_BY_PATH(date_of_birth,',')) 
    3   KEEP (DENSE_RANK LAST ORDER BY curr),',') AS employees 
    4 FROM (SELECT applicant_id, 
    5     date_of_birth, 
    6     ROW_NUMBER() OVER (PARTITION BY applicant_id ORDER BY date_of_birth) AS curr, 
    7     ROW_NUMBER() OVER (PARTITION BY applicant_id ORDER BY date_of_birth) -1 AS prev 
    8   FROM table1) 
    9 GROUP BY applicant_id 
10 CONNECT BY prev = PRIOR curr AND applicant_id = PRIOR applicant_id 
11 START WITH curr = 1; 

APPLICANT_ID EMPLOYEES 
------------ -------------------------------------------------- 
    203276426 01-JUN-69,03-FEB-96,02-JUN-99 
    206209579 29-APR-82,29-APR-82 

SQL> 

有添的文章here證明的其他方式。