Trying to create a regular expression [ORACLE]

Question

Good,

I need help to create a regular expression to just take the name and extension of file of the following directories.

/home/user/work/file1.dbf
/opt/user/file2.dfb

I am trying to create an expression in Oracle12C to only output " file1.dbf " and " file2.dbf ".

I am currently trying to do the regular expression on the next page and reading the following documentation .

Thanks in advance and I hope I have explained correctly.

Answer 1

You don't need a regular expression to do this. A combination of substr and instr would be sufficient.

instr(colname,'/',-1) gets the last occurrence of / in the string. And the substring after that position would be the filename as per the data shown.

The filter instr(colname,'/') > 0 restricts the rows which don't have a / in them.

select substr(colname,instr(colname,'/',-1)+1) as filename
from tablename
where instr(colname,'/') > 0

A regular expression for the same would be

select regexp_substr(colname,'(.*/|^)(.+)$',1,1,null,2) as filename
from tablename

• (.*/|^) - All the characters upto the last / occurence in the string or the start of the string if there are no / characters.

• (.+)$ - All the characters after the last / if it exists in the string or the full string if / doesn't exist.

They are extracted as 2 groups and we are interested in the 2nd group. Hence the argument 2 at the end of regexp_substr .

Read about the arguments to REGEXP_SUBSTR here .

Answer 2

An alternative regex approach would be to strip everything up to the last / :

with demo as
     ( select '/home/user/work/file1.dbf' as path from dual union all
       select '/opt/user/file2.dfb' from dual )
select path
     , regexp_replace(path,'^..*/') as filename
from   demo;