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社区首页 >专栏 >Oracle 19c 中的 LISTAGG 函数结果去重

Oracle 19c 中的 LISTAGG 函数结果去重

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Yunjie Ge
发布2022-04-24 10:05:41
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发布2022-04-24 10:05:41
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文章被收录于专栏:数据库与编程

Oracle 11gR2 中引入了 LISTAGG 函数,以简化字符串聚合。在Oracle 12cR2中,它已扩展为包括溢出错误处理。Oracle 19c 中通过包含 DISTINCT 关键字,可以从 LISTAGG 结果中删除重复项。

1、初始化环境

代码语言:javascript
复制
-- DROP TABLE EMP PURGE;

CREATE TABLE EMP (
  EMPNO NUMBER(4) CONSTRAINT PK_EMP PRIMARY KEY,
  ENAME VARCHAR2(10),
  JOB VARCHAR2(9),
  MGR NUMBER(4),
  HIREDATE DATE,
  SAL NUMBER(7,2),
  COMM NUMBER(7,2),
  DEPTNO NUMBER(2)
);

INSERT INTO EMP VALUES (7369,'SMITH','CLERK',7902,to_date('17-12-1980','dd-mm-yyyy'),800,NULL,20);
INSERT INTO EMP VALUES (7499,'ALLEN','SALESMAN',7698,to_date('20-2-1981','dd-mm-yyyy'),1600,300,30);
INSERT INTO EMP VALUES (7521,'WARD','SALESMAN',7698,to_date('22-2-1981','dd-mm-yyyy'),1250,500,30);
INSERT INTO EMP VALUES (7566,'JONES','MANAGER',7839,to_date('2-4-1981','dd-mm-yyyy'),2975,NULL,20);
INSERT INTO EMP VALUES (7654,'MARTIN','SALESMAN',7698,to_date('28-9-1981','dd-mm-yyyy'),1250,1400,30);
INSERT INTO EMP VALUES (7698,'BLAKE','MANAGER',7839,to_date('1-5-1981','dd-mm-yyyy'),2850,NULL,30);
INSERT INTO EMP VALUES (7782,'CLARK','MANAGER',7839,to_date('9-6-1981','dd-mm-yyyy'),2450,NULL,10);
INSERT INTO EMP VALUES (7788,'SCOTT','ANALYST',7566,to_date('13-JUL-87','dd-mm-rr')-85,3000,NULL,20);
INSERT INTO EMP VALUES (7839,'KING','PRESIDENT',NULL,to_date('17-11-1981','dd-mm-yyyy'),5000,NULL,10);
INSERT INTO EMP VALUES (7844,'TURNER','SALESMAN',7698,to_date('8-9-1981','dd-mm-yyyy'),1500,0,30);
INSERT INTO EMP VALUES (7876,'ADAMS','CLERK',7788,to_date('13-JUL-87', 'dd-mm-rr')-51,1100,NULL,20);
INSERT INTO EMP VALUES (7900,'JAMES','CLERK',7698,to_date('3-12-1981','dd-mm-yyyy'),950,NULL,30);
INSERT INTO EMP VALUES (7902,'FORD','ANALYST',7566,to_date('3-12-1981','dd-mm-yyyy'),3000,NULL,20);
INSERT INTO EMP VALUES (7934,'MILLER','CLERK',7782,to_date('23-1-1982','dd-mm-yyyy'),1300,NULL,10);
COMMIT;

2、问题

LISTAGG 函数的默认使用如下所示。

代码语言:javascript
复制
COLUMN employees FORMAT A40

SELECT deptno, LISTAGG(ename, ',') WITHIN GROUP (ORDER BY ename) AS employees
FROM   emp
GROUP BY deptno
ORDER BY deptno;

    DEPTNO EMPLOYEES
---------- ----------------------------------------
        10 CLARK,KING,MILLER
        20 ADAMS,FORD,JONES,SCOTT,SMITH
        30 ALLEN,BLAKE,JAMES,MARTIN,TURNER,WARD

3 rows selected.

SQL>

让我们在部门10中添加一些额外的,名称为“ MILLER”的人,在汇总列表中提供重复项。

代码语言:javascript
复制
INSERT INTO emp VALUES (9998,'MILLER','ANALYST',7782,to_date('23-1-1982','dd-mm-yyyy'),1600,NULL,10);
INSERT INTO emp VALUES (9999,'MILLER','MANADER',7782,to_date('23-1-1982','dd-mm-yyyy'),1500,NULL,10);
COMMIT;

正如预期的那样,我们现在在部门10中看到多个名称为“ MILLER”的条目。

代码语言:javascript
复制
COLUMN employees FORMAT A40

SELECT deptno, LISTAGG(ename, ',') WITHIN GROUP (ORDER BY ename) AS employees
FROM   emp
GROUP BY deptno
ORDER BY deptno;

    DEPTNO EMPLOYEES
---------- ----------------------------------------
        10 CLARK,KING,MILLER,MILLER,MILLER
        20 ADAMS,FORD,JONES,SCOTT,SMITH
        30 ALLEN,BLAKE,JAMES,MARTIN,TURNER,WARD

3 rows selected.

SQL>

如果那是我们的期望,那就太好了。如果我们要删除重复项,我们该怎么办?

3、19c 之前的解决方案

我们可以通过多种方式解决此问题。在以下示例中,我们使用 ROW_NUMBER 分析函数删除所有重复项,然后使用常规的 LISTAGG 函数聚合数据。

代码语言:javascript
复制
COLUMN employees FORMAT A40

SELECT e2.deptno, LISTAGG(e2.ename, ',') WITHIN GROUP (ORDER BY e2.ename) AS employees
FROM   (SELECT e.*,
               ROW_NUMBER() OVER (PARTITION BY e.deptno, e.ename ORDER BY e.empno) AS myrank
        FROM   emp e) e2
WHERE  e2.myrank = 1
GROUP BY e2.deptno
ORDER BY e2.deptno;

    DEPTNO EMPLOYEES
---------- ----------------------------------------
        10 CLARK,KING,MILLER
        20 ADAMS,FORD,JONES,SCOTT,SMITH
        30 ALLEN,BLAKE,JAMES,MARTIN,TURNER,WARD

3 rows selected.

SQL>

或者,我们可以在内联视图中使用 DISTINCT 删除重复的行,然后使用常规的 LISTAGG 函数调用来聚合数据。

代码语言:javascript
复制
COLUMN employees FORMAT A40

SELECT e2.deptno, LISTAGG(e2.ename, ',') WITHIN GROUP (ORDER BY e2.ename) AS employees
FROM   (SELECT DISTINCT e.deptno, e.ename
        FROM   emp e) e2
GROUP BY e2.deptno
ORDER BY e2.deptno;

    DEPTNO EMPLOYEES
---------- ----------------------------------------
        10 CLARK,KING,MILLER
        20 ADAMS,FORD,JONES,SCOTT,SMITH
        30 ALLEN,BLAKE,JAMES,MARTIN,TURNER,WARD

3 rows selected.

SQL>

4、从19c开始的解决方案

Oracle 19c 引入了一个更简单的解决方案。现在,我们可以直接在 LISTAGG 函数调用中包含 DISTINCT 关键字。

代码语言:javascript
复制
COLUMN employees FORMAT A40

SELECT deptno, LISTAGG(DISTINCT ename, ',') WITHIN GROUP (ORDER BY ename) AS employees
FROM   emp
GROUP BY deptno
ORDER BY deptno;

    DEPTNO EMPLOYEES
---------- ----------------------------------------
        10 CLARK,KING,MILLER
        20 ADAMS,FORD,JONES,SCOTT,SMITH
        30 ALLEN,BLAKE,JAMES,MARTIN,TURNER,WARD

3 rows selected.

SQL>

默认功能是包括所有结果,我们可以使用 ALL 关键字明确表示这些结果。

代码语言:javascript
复制
SELECT deptno, LISTAGG(ALL ename, ',') WITHIN GROUP (ORDER BY ename) AS employees
FROM   emp
GROUP BY deptno
ORDER BY deptno;

    DEPTNO EMPLOYEES
---------- ----------------------------------------
        10 CLARK,KING,MILLER,MILLER,MILLER
        20 ADAMS,FORD,JONES,SCOTT,SMITH
        30 ALLEN,BLAKE,JAMES,MARTIN,TURNER,WARD

3 rows selected.

SQL>
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