regex - Oracle SQL: Return first line of string using regexp_substr -

i trying return first line of text text box in sql query (oracle 11). content of text box looks this:

   x withdrawn     explanation. 

i want return top line, i.e. x withdrawn. i'm not sure if can specify @ first line only, or return text before carriage return - either work.

i think need use regexp_substr i'm not quite sure on syntax. have tried:

   regexp_substr(table.text,'^.*$') 

but didn't work, assistance appreciated!

edit: solution used:

   select regexp_substr(table.text, '[^,]+['||chr(10)||']') tab 

edit: noticed getting mixture of line feed , carriage returns returned in answer, i've use following solution return text , no additional characters.

    select       replace(replace(regexp_substr(table.text, '[^,]+['||chr(10)||']'),chr(10),''),chr(13),'')       tab  

edit: following @ben's answer, i've amended solution following:

select initcap(replace(regexp_substr(table.text, '.*$', 1, 1, 'm'),chr(13),'')) tab 

parado's regular expression matches that's not comma multiple times followed carriage return. means won't work line-feed or if there's comma in text.

oracle supports multi-line expressions using m match parameter. when using mode, $ matches end of each line end of string. can use expression massively to:

regexp_substr(str, '.*$', 1, 1, 'm') 

that match first occurrence (the first line) of string matches anything, followed end of string, counting first character.

as example:

with strings (   select 'hi          hi again' str    dual   union  select 'bye          , again'    dual         )  select regexp_substr(str, '.*$', 1, 1, 'm')    strings