问题
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
I need result as
206209579 04/29/82,04/29/82
203276426 06/01/69,02/03/96
please suggest its need full
回答1:
In Oracle 11g and above, you can use listagg:
SELECT applicant_id,
LISTAGG(date_of_birth) WITHIN GROUP (ORDER BY date_of_birth)
FROM my_table
GROUP BY applicant_id
回答2:
As an alternative, for Oracle version prior to 11g where LISTAGG
is not supported, you could use ROW_NUMBER() and SYS_CONNECT_BY_PATH functions.
Test case:
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>
There are other ways demonstrated in Tim's article here.
来源:https://stackoverflow.com/questions/28690375/i-need-result-from-select-statement-as-multiple-data-of-birth-for-a-single-id