一道比较难的sql题，筛选出重复字段的行数

select * from 导入数据表;    


id    city_column
1    北京,上海,广州
2    上海,上海,深圳
3    北京,杭州,北京
4    上海,广州,深圳

select substring_index(khmc,',',1), *  from 导入数据表    

truncate table 导入数据表    


select count(distinct khmc) from 导入数据表;    

select count(*)     
from 导入数据表    
where length(khmc)-length(replace(khmc),',','') +1    

select  trim(substring_index(substring_index(khmc,',',n),',',-1)), * from 导入数据表    
select  length(khmc), length(khmc)-length(REGEXP_REPLACE(khmc,',','')) +1 from 导入数据表    

-----    

select xh, city, count(*)    
from(    
select /*+ MAPJOIN ( t1 ) */    
xh,    
trim(substring_index(substring_index(khmc,',',n),',',-1)) as city    
from 导入数据表 t1     
join     
(select 1 as n union select 2 union select 3) numbers    
on numbers.n <= length(t1.khmc)-length(REGEXP_REPLACE(t1.khmc,',','')) +1    
) group by xh, city     
having count(*) > 1    

select sum(cnt) from    
(    
select count(distinct xh) as cnt     
from(    
select /*+ MAPJOIN ( t1 ) */    
xh,    
trim(substring_index(substring_index(khmc,',',n),',',-1)) as city    
from 导入数据表 t1     
join     
(select 1 as n union select 2 union select 3) numbers    
on numbers.n <= length(t1.khmc)-length(REGEXP_REPLACE(t1.khmc,',','')) +1    
) group by xh, city     
having count(*) > 1