1

My Table table1 looks like this:

 no    name

 1      C_GT_2013-2014,C_GT_AcMaster,C_GT_Master
 1      C_GT_2014-2015,C_GT_AcMaster,C_GT_Master
 2      C_TGYY_2013-2014,G_New_AcMaster,G_New_Master
 2      C_TGYY_2014-2015,G_New_AcMaster,G_New_Master
 3      C_TGYN_2013-2014,G_ACYMAN_AcMaster,C_TGYN_Master
 3      C_TGYN_2014-2015,G_ACYMAN_AcMaster,C_TGYN_Master
 4      C_TGNY_2013-2014,C_TGNY_AcMaster,G_ACNMAY_Master
 4      C_TGNY_2014-2015,C_TGNY_AcMaster,G_ACNMAY_Master
 5      C_GYY_2013-2014,G_New_AcMaster,G_New_Master
 6      C_DD_2013-2014,C_DD_AcMaster,G_ACNMAY_Master
 7      C_YN_2013-2014,G_ACYMAN_AcMaster,C_YN_Master

I want to retrieve the value without duplicate no and split the name using "," so I tried like this:

select (string_to_array(name,','))[3] as master
      ,(string_to_array(name,','))[2] as acmaster
      ,string_agg((string_to_array(name,','))[1],',') as trans
      ,no
 from table t
 where no in (2,3,4,6,7)
 group by (string_to_array(schemaname,','))[3]
        , (string_to_array(schemaname,','))[2]
        , no
 order by no

But it return no duplicate. I want to get the value

  1. No duplicate in no column
  2. Name have same no. It merage using ',' like this (no 1 name with split(',') C_GT_2013-2014,C_GT_2014-2015)

I am using Postgresql 9.3.

My result would be:

I want to split Name into three columns which word have master its comes to Master column, Which word have its Comes to AcMaster, Others Comes to Trans With out No duplicate

 Master            AcMaster                  Trans                                   No

 G_New_Master     G_New_AcMaster             C_TGYY_2013-2014,C_TGYY_2014-2015        1
 C_TGYN_Master,   G_ACYMAN_AcMaster          C_TGYN_2014-2015,C_TGYN_2013-2014,       2
 C_YN_Master                                             C_YN_2013-2014    

 G_ACNMAY_Master  C_DD_AcMaster,              C_DD_2013-2014                          3
                  C_TGNY_AcMaster             C_TGNY_2013-2014,C_TGNY_2014-2015
4

1 回答 1

1

这是您的查询：

select no,
string_agg(case when name like '%\_Master' escape '\' then name end, ','),
string_agg(case when name like '%\_AcMaster' escape '\' then name end, ','),
string_agg(case when (name not like '%\_Master' and name not like '%\_AcMaster' escape '\') then name end, ',')
from
(select no, (string_to_array(name,','))[1] as name
from   table1
union
select no, (string_to_array(name,','))[2] as name
from   table1
union
select no, (string_to_array(name,','))[3] as name
from   table1) t
where no in (2,3,4,6,7)
group by no
order by no

这个查询结果是：

no  Master              AcMaster        Trans

2   G_New_Master        G_New_AcMaster      C_TGYY_2014-2015,C_TGYY_2013-2014
3   C_TGYN_Master       G_ACYMAN_AcMaster   C_TGYN_2014-2015,C_TGYN_2013-2014
4   G_ACNMAY_Master     C_TGNY_AcMaster     C_TGNY_2013-2014,C_TGNY_2014-2015
6   G_ACNMAY_Master     C_DD_AcMaster       C_DD_2013-2014
7   C_YN_Master         G_ACYMAN_AcMaster   C_YN_2013-2014

现在有一些解释：

1-我把你的桌子弄平了：

(select no, (string_to_array(name,','))[1] as name
from   table1
union
select no, (string_to_array(name,','))[2] as name
from   table1
union
select no, (string_to_array(name,','))[3] as name
from   table1)

所以我可以得到这样的结果：

4   G_ACNMAY_Master
1   C_GT_Master
3   G_ACYMAN_AcMaster
1   C_GT_2013-2014
2   G_New_Master
2   G_New_AcMaster
1   C_GT_2014-2015
...

2- 我选择了 3 个必需的列：

case when name like '%\_Master' escape '\' then name end -- Master
case when name like '%\_AcMaster' escape '\' then name end -- AcMaster
case when (name not like '%\_Master' and name not like '%\_AcMaster' escape '\') then name end -- Trans

得到以下结果：

2   G_New_Master
2                                       C_TGYY_2014-2015
2                   G_New_AcMaster
2                                       C_TGYY_2013-2014
3                                       C_TGYN_2014-2015
...

3- 之后，我使用 string_agg 函数分组聚合结果no：

string_agg(case when name like '%\_Master' escape '\' then name end, ',')
于 2014-02-05T00:24:13.547 回答