用sql实现18位身份证校验代码分享身份证校验位计算

with t as(select '34052419800101001X'id from dual)
select id
from t
where exists(select 1
from dual connect by level<=length(id)-1 --17
having mod(sum(substr(id,level,1)*power(2,18-level)),11)=
case substr(id,-1,1)
when '1' then 0
when '0' then 1
when 'X' then 2
else
12-substr(id,-1,1)
end);