postgreSQL 应用case when的例子

select
name,md5(indvl_id_nbr) as indvl_id_nbr,case when char_length(indvl_id_nbr)=18 or char_length(indvl_id_nbr)=15 then substring(indvl_id_nbr from 1 for 6)
end as id_prefix,case when char_length(indvl_id_nbr)=18 then substring(indvl_id_nbr from 7 for 8) 
when char_length(indvl_id_nbr)=15 then  '19' || substring(indvl_id_nbr from 7 for 6) 
else null end as id_brth_dt
from myschema.table_test;

（编辑：李大同）

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