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,deptno) values (7654,'MARTIN','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,' 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}