Question

说我们有下表

Name Date                 reg_no
aa   12/12/2013 10:13:15  rty003
aa   13/12/2013 11:14:16  jyu887
bb   13/14/2013 09:45:10  rty003
bb   12/12/2013 10:13:27  rty003

我想只显示具有匹配的reg_no和日期值（没有秒数）的名称。结果应如下所示。

Name1 Name2  Date              reg_no
aa    bb     12/12/2013 10:13  rty003

我如何在oracle中实现这一目标？

Answer 1

尝试使用listagg功能：

SELECT listagg( NAME,',') WITHIN GROUP ( ORDER BY NAME), 
       to_char(date_time,'mm/dd/yyyy hh24:mi'), 
       reg_no 
FROM t
GROUP BY to_char(date_time,'mm/dd/yyyy hh24:mi'), reg_no
HAVING count(*) > 1;

Answer 2

可能更好但值得，

WITH TAB(NAME,DDATE, REG_NO) AS (
SELECT 'aa', TO_DATE('12/12/2013 10:13:15', 'MM/DD/YYYY HH24:MI:SS'), 'rty003' FROM DUAL UNION ALL
SELECT 'aa', TO_DATE('03/12/2013 11:14:16', 'MM/DD/YYYY HH24:MI:SS'), 'jyu887' FROM DUAL UNION ALL
SELECT 'bb', TO_DATE('03/14/2013 09:45:10', 'MM/DD/YYYY HH24:MI:SS'), 'rty003' FROM DUAL UNION ALL
SELECT 'bb', TO_DATE('12/12/2013 10:13:27', 'MM/DD/YYYY HH24:MI:SS'), 'rty003' FROM DUAL),
----------------
-- End of Data Preparation
----------------
TAB2 AS 
(SELECT NAME, 
        TRUNC(DDATE, 'MI') DDATE, 
        REG_NO, 
        ROW_NUMBER() OVER(PARTITION BY TRUNC(DDATE, 'MI'), REG_NO ORDER BY NAME) RN,
        COUNT(1) OVER(PARTITION BY TRUNC(DDATE, 'MI'), REG_NO) CNT
    FROM TAB) 
SELECT MIN(DECODE(RN, 1, NAME, NULL)) NAME1,
       MIN(DECODE(RN, 2, NAME, NULL)) name2,
       DDATE,
       REG_NO
  FROM TAB2
 WHERE CNT = 2
 group by ddate, reg_no;

输出：

| NAME1 | NAME2 |                           DDATE | REG_NO |
|-------|-------|---------------------------------|--------|
|    aa |    bb | December, 12 2013 10:13:00+0000 | rty003 |

您的最终查询将是

WITH TAB2 AS 
(SELECT NAME, 
        TRUNC(DDATE, 'MI') DDATE, 
        REG_NO, 
        ROW_NUMBER() OVER(PARTITION BY TRUNC(DDATE, 'MI'), REG_NO ORDER BY NAME) RN,
        COUNT(1) OVER(PARTITION BY TRUNC(DDATE, 'MI'), REG_NO) CNT
    FROM <your_table_name>) 
SELECT MIN(DECODE(RN, 1, NAME, NULL)) NAME1,
       MIN(DECODE(RN, 2, NAME, NULL)) name2,
       DDATE,
       REG_NO
  FROM TAB2
 WHERE CNT = 2
 group by ddate, reg_no;

Answer 3

SELECT MIN(NAME),MAX(NAME), 
    TRUNC(DDATE, 'MI'), 
    REG_NO
FROM tab
GROUP BY  
    TRUNC(DDATE, 'MI'), 
    REG_NO
HAVING COUNT(*) >= 2

如何在oracle中查找具有相同值的所有字段

3 个答案: