取下逗號分隔的重複清單，正則表達式

Question

我

contract, clause 1, Subsection 1.1, contract, clause 1, Subsection 1.2, 
paragraph (a), contract, clause 1, Subsection 1.2, paragraph (b), contract, 
clause 2 

，我想

contract, clause 1, Subsection 1.1, Subsection 1.2, paragraph (a), paragraph 
(b), clause 2 

我發現，正則表達式可以做到這一點，但我找不到要使用的字符串做

請幫助..

Answer 1

基於this link分裂逗號分隔值到行，我分裂串入行，保持中第一次出現的位置，取得了明顯的一個再聚集值

with test_string as ( 
select 1 as id, 
'contract, clause 1, Subsection 1.1, contract, clause 1, Subsection 1.2, paragraph (a), contract, clause 1, Subsection 1.2, paragraph (b), contract, clause 2' val 
from dual) 
select id, listagg(word,', ') WITHIN GROUP (order by position) FROM (
select distinct id, first_value(position) over (partition by word order by position) position, word from (
select 
    distinct t.id, 
    levels.column_value as position, 
    trim(regexp_substr(t.val, '[^,]+', 1, levels.column_value)) as word 
from 
    test_string t, 
    table(cast(multiset(select level from dual connect by level <= length (regexp_replace(t.val, '[^,]+')) + 1) as sys.OdciNumberList)) levels 
) 
) GROUP BY id 

如果你不感興趣，維持秩序

with test_string as ( 
select 1 as id, 
'contract, clause 1, Subsection 1.1, contract, clause 1, Subsection 1.2, paragraph (a), contract, clause 1, Subsection 1.2, paragraph (b), contract, clause 2' val 
from dual) 
select id, listagg(word,', ') WITHIN GROUP (order by 1) FROM (
select 
    distinct t.id, 
    trim(regexp_substr(t.val, '[^,]+', 1, levels.column_value)) as word 
from 
    test_string t, 
    table(cast(multiset(select level from dual connect by level <= length (regexp_replace(t.val, '[^,]+')) + 1) as sys.OdciNumberList)) levels 
) GROUP BY id