postgresql合并string_agg函数的实例

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：

select create_by,string_agg(videoname,',') as videonames from w008_video_addr_info where id in (4248,538,546)
group by create_by

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：

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

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

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)

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

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

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}

array_agg 去重元素，例如查询所有的部门

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)

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。如有错误或未考虑完全的地方，望不吝赐教。