t=# with s(v) as (values('6A CAMBOON ROAD MORLEY WA 6062 AU'),('8 115 MACKIE STREET VICTORIA PARK WA 6100 A'))
, split as (select *,count(1) over (partition by v) from s, regexp_matches(v,'([A-Z]+)','g') with ordinality t(m,o))
select distinct v,string_agg(m[1],'') over (partition by v) from split where o <= count-(3-1);
v | string_agg
---------------------------------------------+------------------------------
8 115 MACKIE STREET VICTORIA PARK WA 6100 A | MACKIE STREET VICTORIA PARK
6A CAMBOON ROAD MORLEY WA 6062 AU | CAMBOON ROAD MORLEY
(2 rows)
我排除索引(或任何不嵌合掩模[A-Z]+
),从而切割从所述端不三个位置,而是两个(3-1)
其中1是提前公知的指数。
另外,我不从第二个空间开始,因为它会违背你想要的结果...