REGEXP_REPLACE - replace '\n''s inside a xml string

Question

I have this xml string and i need to replace the '\n' for one # but only the \n's inside the tag's not the first \n how do that with REGEXP_REPLACE?
<?xml version="1.0" encoding="UTF-8"?>\n
<ipb>
<address>Rua são Joao\n
Bragança
</address>
<calendar>
1º Semestre:\n
Início: 21/09/2009\n
Fim: 19/02/2010\n
Pausa Natal: de 21/12/2009 a 03/01/2010\n
\n
2º Semestre:\n
Início: 22/02/2010\n
Fim: 16/07/2010\n
</calendar>

Answer 1

Well, i'm no regular expression master, but no one's responded so i'll venture an answer (not using regular expression though).
ME_XE?with data as
select
'<?xml version="1.0" encoding="UTF-8"?>\n
<ipb>
<address>Rua são Joao\n
Bragança
</address>
<calendar>
1º Semestre:\n
Início: 21/09/2009\n
Fim: 19/02/2010\n
Pausa Natal: de 21/12/2009 a 03/01/2010\n
\n
2º Semestre:\n
Início: 22/02/2010\n
Fim: 16/07/2010\n
</calendar>' as col1 from dual
select
  replace(replace(replace(col1, '>\n', '~*~'), '\n', NULL), '~*~', '>\n') as new_col1
from data;
NEW_COL1
<?xml version="1.0" encoding="UTF-8"?>\n
<ipb>
<address>Rua s??o Joao
Bragan??a
</address>
<calendar>
1?? Semestre:
In??cio: 21/09/2009
Fim: 19/02/2010
Pausa Natal: de 21/12/2009 a 03/01/2010
2?? Semestre:
In??cio: 22/02/2010
Fim: 16/07/2010
</calendar>
1 row selected.
Elapsed: 00:00:00.17
ME_XE?Just used the regular replace to decode the instances of \n you want to keep (using a string you're not likely to find in your source data) and then replaced all the \n's with null, then put back the /n replacing the string we subbed in before.
If Michaels is around, maybe you'll get a flashy XML answer to your question ... sorry, but i'm not that much flash :)