2017-10-15 5 views

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bien sûr cette analyse conditionnelle n'est pas fiable:

t=# with v(a) as(values('12 2 NOELA PLACE ST MARYS NSW 2760 AU') 
,('51 MALABAR ROAD SOUTH COOGEE NSW 2034 AU') 
,('12 LISTER STREET WINSTON HILLS NSW 2153 AU') 
) 
select reverse(split_part(reverse(a),' ',1)), reverse(split_part(reverse(a),' ',3)), case when split_part(a,' ',2) ~ '\d' then split_part(a,' ',2) end st, split_part(a,' ',1) un from v; 
reverse | reverse | st | un 
---------+---------+----+---- 
AU  | NSW  | 2 | 12 
AU  | NSW  | | 51 
AU  | NSW  | | 12 
(3 rows)