正如标题所说,我需要使用此查询(无字符串聚合)将结果查询与字符串agg连接
select pdet.dept_id, pdet.grade_id
from psa_aso_target ptar
inner join psa_aso_targetdetails pdet on pdet.target_id=ptar.target_id and ptar.branch_id='18'
结果就是这个
然后我添加字符串agg
select pdet.dept_id, string_agg(distinct pdet.grade_id::char,'|') as grade
from psa_aso_target ptar
inner join psa_aso_targetdetails pdet on pdet.target_id=ptar.target_id and ptar.branch_id='18'
group by pdet.dept_id
我希望结果结果是
dept_id | grade_id
2 | 1|2|3
3 | 4|13|14|15|18
5 | 6|10|17
63 | 2|4|7
但我得到的结果是
dept_id | grade_id
2 | 1|2|3
3 | 1|4
5 | 1|6
63 | 2|4|7
任何想法?
答案 0 :(得分:2)
这是因为您将数值转换为char(1)。通过将它转换为char(1),您只将值截断为1个字符,因此13,14,15,18全部变为1
您需要将其强制转换为varchar或text:
select pdet.dept_id, string_agg(distinct pdet.grade_id::varchar,'|') as grade
from psa_aso_target ptar
inner join psa_aso_targetdetails pdet on pdet.target_id=ptar.target_id and ptar.branch_id='18'
group by pdet.dept_id
select pdet.dept_id, string_agg(distinct pdet.grade_id::text,'|') as grade
from psa_aso_target ptar
inner join psa_aso_targetdetails pdet on pdet.target_id=ptar.target_id and ptar.branch_id='18'
group by pdet.dept_id
感谢下面的评论者为答案做出贡献