所以给定的数据是：

create table tbl1
(   
    date date,       
    family text, 
    class2 text,    
    sec2 text,  
    core1 text, 
    nlm text
);
insert into tbl1 values
('2020-07-22','f1','r1','p1','xyz','pqr'),
('2020-07-22','f2','r1','p1','xyz','pqrs'),
('2020-07-24','f5','r2','p4','sds','dsg');
create table tbl2
(   
    date date,       
    family text, 
    class2 text,    
    sec2 text,  
    core1 text, 
    nlm text
);
insert into tbl2 values
('2020-07-22','f4','r1','p1','xyz','pqr'),
('2020-07-24','f8','r6','p1','fds','sdg');
create table tbl3
(   
    date date,       
    family text, 
    class2 text,    
    sec2 text,  
    core1 text, 
    nlm text
);
insert into tbl3 values
('2020-07-28','f14','r1','p1','xyz','pqr'),
('2020-07-24','f18','r6','p1','fds','sdg');

查询：

create view vw_test as
select date,class,sec,core,string_agg(family,',') as group_of_family
from
(
    select date,class2 as class,
            sec2 as sec,
            core1 as core,
            family
    from tbl1
    union   
    select date,class2 as class,
            sec2 as sec,
            core1 as core,
            family
    from tbl2
    union
    select date,class2 as class,
            sec2 as sec,
            core1 as core,
            family
    from tbl3
)a 
group by date,class,sec,core;

检查预期结果：

select * from vw_test;

你必须使用 string_agg() 功能，

select date, class2 as class, sec2 as ssec, core1 as core, string_agg(col2,' , '  order by class, sec) as group_of_family from
(select date, family, class2, sec2, core1 from table1
union all
select date, family, class2, sec2, core1 from table2
union all
select date, family, class2, sec2, core1 from table3) qry
group by date, class2, sec2, core1;