[译] Oracle Database 19c 中的 LISTAGG DISTINCT

原创 Lucifer三思而后行 2022-01-06

1955

原文地址：https://oracle-base.com/articles/19c/listagg-distinct-19c
原文作者：Tim Hall

LISTAGG 函数是在 Oracle 11gR2 中引入的，以使字符串聚合更简单。在 Oracle 12cR2 中，它被扩展为包括溢出错误处理。Oracle 19c 更新了 LISTAGG 功能，通过 DISTINCT 关键字从结果中删除重复项。

配置

本文中的示例使用下表：

-- 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;

问题

该LISTAGG函数的默认操作如下所示：

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>

让我们将一些名为 “MILLER” 的额外人员添加到第 10 部门，以便在汇总列表中为我们提供重复项：

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 中看到名称“MILLER”的多个条目：

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>

如果这就是我们所期待的，那就太好了。如果我们想删除重复项，我们该怎么做？

解决方案：19c之前

19C 之前，我们可以通过多种方式解决这个问题。在下面的示例中，我们使用 ROW_NUMBER 分析函数删除任何重复项，然后使用常规 LISTAGG 函数聚合数据。

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 DISTINCT 在内联视图中使用来删除重复的行，然后使用传统的 LISTAGG 函数调用来聚合数据。

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>

解决方案：19c 以后

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

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>

默认功能是包含所有结果，我们可以使用 ALL 关键字明确表达。

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>