Find and replace value in Delimited String

Question

Hi All,
I have a requirement, where i need to find and replace values in delimited string.
For example, the string is "GL~1001~157747~FEB-13~CREDIT~A~N~USD~NULL~". The 4th column gives month and year. I need to replace it with previous month name. For example: "GL~1001~157747~JAN-13~CREDIT~A~N~USD~NULL~". I need to do same for last 12 months.
I thought of first devide the values and store it in variable and then after replacing it with required value, join it back.
I just wanted to know if there is any better way to do it?

Answer 1

for example (Assumption: the abbreviated month is the first occurance of 3 consecutive alphabetic charachters)
with testdata as (
select 'GL~1001~157747~FEB-13~CREDIT~A~N~USD~NULL~' str from dual
select
str
,regexp_substr(str, '[[:alpha:]]{3}') part
,to_date('01'||regexp_substr(str, '[[:alpha:]]{3}')||'2013', 'DDMONYYYY') part_date
,replace (str
         ,regexp_substr(str, '[[:alpha:]]{3}')
         ,to_char(add_months(to_date('01'||regexp_substr(str, '[[:alpha:]]{3}')||'2013', 'DDMONYYYY'),-1),'MON')
) res
from testdata
STR
PART
PART_DATE
RES
GL~1001~157747~FEB-13~CREDIT~A~N~USD~NULL~
FEB
02/01/2013
GL~1001~157747~JAN-13~CREDIT~A~N~USD~NULL~
with year included
with testdata as (
select 'GL~1001~157747~JAN-13~CREDIT~A~N~USD~NULL~' str from dual
select
str
,regexp_substr(str, '[[:alpha:]]{3}-\d{2}') part
,to_date(regexp_substr(str, '[[:alpha:]]{3}-\d{2}'), 'MON-YY') part_date
,replace (str
         ,regexp_substr(str, '[[:alpha:]]{3}-\d{2}')
         ,to_char(add_months(to_date(regexp_substr(str, '[[:alpha:]]{3}-\d{2}'), 'MON-YY'),-1),'MON-YY')
) res
from testdata
STR
PART
PART_DATE
RES
GL~1001~157747~JAN-13~CREDIT~A~N~USD~NULL~
JAN-13
01/01/2013
GL~1001~157747~DEC-12~CREDIT~A~N~USD~NULL~
Message was edited by: chris227 year included