1 有时候我们会需要将多条数据根据一些特别的字段做一些合并。比如下面这个查询，正常会查询出3条数据，但是我们会希望根据create_by 分成两列显示
2 这时候需要用到string_agg函数，先通过group by分组，在进行合并，当然查询结果需要满足group by的限制；sql语句：
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select create_by,string_agg(videoname,',') as videonames from w008_video_addr_info where id in (4248,538,546)
group by create_by
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查询结果：
3 复杂一些的应用场景（子查询）：
下面的语句是我用来查询一个学生在什么时间看了哪些视频：
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select
sa.id,
info.nickname,
(select string_agg(v.videoname,',')
from w008_school_assign_video sv
join w008_video_addr_info v on sv.videoaddrinfo =v.id
where sv.schoolassignment=sa.id and v.is_removed=0 and sv.is_removed=0
group by v.is_removed) as videos,
(select string_agg(to_char(sv.create_date, 'MM-DD HH24:MI'),',')
from w008_school_assign_video sv
join w008_video_addr_info v on sv.videoaddrinfo =v.id where sv.schoolassignment=sa.id and v.is_removed=0
and sv.is_removed=0 group by v.is_removed) as viewtime
from w008_school_assignment sa
join w008_user_business_info info on sa.userlongid=info.id where sa.shchoolworkid=2514505674916356
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结果：
当然，string_agg(field,'分隔符');分隔符可以填写其他任意的字符，方便后期处理即可；
补充：PostgreSql 聚合函数string_agg与array_agg，类似mysql中group_concat
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
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array_agg(expression)
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把表达式变成一个数组 一般配合 array_to_string() 函数使用
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string_agg(expression, delimiter)
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直接把一个表达式变成字符串
案例：
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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)
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查询同一个部门下的员工且合并起来
方法1：
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select deptno, string_agg(ename, ',') from jinbo.employee group by deptno;
deptno | string_agg
--------+--------------
20 | JONES
30 | ALLEN,MARTIN
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方法2：
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select deptno, array_to_string(array_agg(ename),',') from jinbo.employee group by deptno;
deptno | array_to_string
--------+-----------------
20 | JONES
30 | ALLEN,MARTIN
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在1条件的基础上，按ename 倒叙合并
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select deptno, string_agg(ename, ',' order by ename desc) from jinbo.employee group by deptno;
deptno | string_agg
--------+--------------
20 | JONES
30 | MARTIN,ALLEN
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按数组格式输出使用 array_agg
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select deptno, array_agg(ename) from jinbo.employee group by deptno;
deptno | array_agg
--------+----------------
20 | {JONES}
30 | {ALLEN,MARTIN}
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array_agg 去重元素，例如查询所有的部门
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select array_agg(distinct deptno) from jinbo.employee;
array_agg
-----------
{20,30}
(1 row)
#不仅可以去重，还可以排序
select array_agg(distinct deptno order by deptno desc) from jinbo.employee;
array_agg
-----------
{30,20}
(1 row)
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以上为个人经验，希望能给大家一个参考，也希望大家多多支持服务器之家。如有错误或未考虑完全的地方，望不吝赐教。
原文链接：https://blog.csdn.net/qxianx/article/details/81985615