SQL REGEXP_SUBSTR returnerer null-streng ved opdeling

Brug bare REPLACE og din standardkode med ,

SqlFiddleDemo

 with temp as
(
    select 108 Name, 'test' Project, 'Err1:::Err2:::Err3' Error  from dual
    union all
    select 109, 'test2', 'Err1' from dual
)
select distinct
  t.name, t.project,
  trim(regexp_substr(REPLACE(t.error, ':::', ', '), '[^,]+', 1, levels.column_value))  as error
from 
  temp t,
  table(cast(multiset(select level from dual connect by  level <= length (regexp_replace(REPLACE(t.error, ':::', ', '), '[^,]+'))  + 1) as sys.OdciNumberList)) levels
order by name

Eller du skal dividere med længden af ​​delimeter:

SqlFiddle

with temp as
(
    select 108 Name, 'test' Project, 'Err1:::Err2:::Err3' Error  from dual
    union all
    select 109, 'test2', 'Err1:::Err2' from dual
)
select distinct
  t.name, t.project,
  trim(regexp_substr(t.error, '[^:::]+', 1, levels.column_value))  as error
from 
  temp t,
  table(cast(multiset(select level from dual connect by  level <= length (
       regexp_replace(t.error, '[^:::]+'))/3  + 1) as sys.OdciNumberList)) levels
order by name

Du kan se, hvorfor du udfører:

SELECT length (regexp_replace('Err1:::Err2:::Err3', '[^:::]+')) + 1 AS l
FROM dual

Dette vil returnere 7 og din:

SELECT DISTINCT  t.name, t.project,
trim(regexp_substr(t.error, '[^:::]+', 1, levels.column_value))  as error

vil prøve at hente regexp_substr for 7 forekomster, hvor 4 af dem vil være NULL og til sidst 4 NULL vil blive klemt til én NULL ved DISTINCT .