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dos assign carriage return to variable Code Example
dos assign carriage return to variable 
setlocal EnableDelayedExpansion
set LF=^


rem TWO empty lines are required
echo This text!LF!uses two lines
Source: stackoverflow.com
dos assign carriage return to variable 
(SET LF=^
%=this line is empty=%
)
Source: stackoverflow.com
dos assign carriage return to variable 
set LF=^


rem TWO empty lines are required
echo This text^%LF%%LF%uses two lines
echo This also^

uses two lines
Source: stackoverflow.com
dos assign carriage return to variable 
if "!LF!" NEQ "!LF:~0,1!" echo Error "Linefeed definition is defect, probably multiple invisble whitespaces at the line end in the definition of LF"

FOR /F "delims=" %%n in ("!LF!") do (
  echo Error "Linefeed definition is defect, probably invisble whitespaces at the line end in the definition of LF"
)
Source: stackoverflow.com
dos assign carriage return to variable 
set myLinefeed=^<LF1>
<LF2>
<LF3>
Source: stackoverflow.com



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Type:
 Code Example
Category:
 Coding
Sub Category:
 Code Example
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