regex - How to convert LineFeed delimited file to one line using repl.bat -

for example:

i have set of strings:

aaaa~ bbbb~ 

expected:

aaaa~bbbb~ 

im using repl.bat find , replace tool impelemented in batch file follows:

for /f "tokens=2,3" %%a in ('type c:\user\delim.txt') (      copy "c:\user\linefeed\%%a" "c:\user\linefeed\%%a.in.tmp"         /f "tokens=1,*" %%s in (findnreplace.txt) (             type "c:\user\linefeed\%%a.in.tmp"  | c:\user\bat\repl.bat "%%s" "%%t" x > "c:\user\linefeed\%%a.out.tmp"             move /y "c:\user\linefeed\%%a.out.tmp" "c:\user\linefeed\%%a.in.tmp"         )   move "c:\user\linefeed\%%a.in.tmp" "c:\user\clientout\%%a" ) 

example contents of delim.txt

^(aaaaa.{5}).{4}(.*)$       $17777$2 

^(aaaaa.{5}).{4}(.*)$ - find pattern starts aaaaa followed 5 chars , 4 succeeding characters

$17777$2 - match value replaced 7777

what equivalent value in regex in order convert line feed delimited single line?

i think m42 right:

how replacing linefeed nothing? ie. : replace \r ""; \r stands line break, \n or \r or \r\n.

if want linefeed character, \r only.