PostgreSql 聚合函数string_agg与array_agg

string_agg,array_agg 这两个函数的功能大同小异,只不过合并数据的类型不同。
https://www.postgresql.org/docs/9.6/static/functions-aggregate.html

array_agg(expression) 把表达式变成一个数组 一般配合 array_to_string() 函数使用
string_agg(expression, delimiter) 直接把一个表达式变成字符串

案例:

create table(empno smallint, ename varchar(20), job varchar(20), mgr smallint, hiredate date, sal bigint, comm bigint, deptno smallint);

insert into jinbo.employee(empno,ename,job, mgr, hiredate, sal, comm, deptno) values (7499, 'ALLEN', 'SALEMAN', 7698, '2014-11-12', 16000, 300, 30);

insert into jinbo.employee(empno,ename,job, mgr, hiredate, sal, comm, deptno) values (7499, 'ALLEN', 'SALEMAN', 7698, '2014-11-12', 16000, 300, 30);

insert into jinbo.employee(empno,ename,job, mgr, hiredate, sal, comm, deptno) values (7654, 'MARTIN', 'SALEMAN', 7698, '2016-09-12', 12000, 1400, 30);

select * from jinbo.employee;
 empno | ename  |   job   | mgr  |  hiredate  |  sal  | comm | deptno 
-------+--------+---------+------+------------+-------+------+--------
  7499 | ALLEN  | SALEMAN | 7698 | 2014-11-12 | 16000 |  300 |     30
  7566 | JONES  | MANAGER | 7839 | 2015-12-12 | 32000 |    0 |     20
  7654 | MARTIN | SALEMAN | 7698 | 2016-09-12 | 12000 | 1400 |     30
(3 rows)

1.查询同一个部门下的员工且合并起来

方法1:
select deptno, string_agg(ename, ',') from jinbo.employee group by deptno;

 deptno |  string_agg  
--------+--------------
     20 | JONES
     30 | ALLEN,MARTIN

方法2:
select deptno, array_to_string(array_agg(ename),',') from jinbo.employee group by deptno;
 deptno | array_to_string 
--------+-----------------
     20 | JONES
     30 | ALLEN,MARTIN

2、在1条件的基础上,按ename 倒叙合并

select deptno, string_agg(ename, ',' order by ename desc) from jinbo.employee group by deptno;
 deptno |  string_agg  
--------+--------------
     20 | JONES
     30 | MARTIN,ALLEN

3、按数组格式输出使用 array_agg

select deptno, array_agg(ename) from jinbo.employee group by deptno;
 deptno |   array_agg    
--------+----------------
     20 | {JONES}
     30 | {ALLEN,MARTIN}
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