How to remove carriage return and newline from a variable in shell script

Question

I am new to shell script. I am sourcing a file, which is created in Windows and has carriage returns, using the source command. After I source when I append som

Answer 1

Because the file you source ends lines with carriage returns, the contents of $testVar are likely to look like this:

$ printf '%q\n' "$testVar"
$'value123\r'

(The first line's $ is the shell prompt; the second line's $ is from the %q formatting string, indicating $'' quoting.)

To get rid of the carriage return, you can use shell parameter expansion and ANSI-C quoting (requires Bash):

testVar=${testVar//$'\r'}

Which should result in

$ printf '%q\n' "$testVar"
value123

Answer 2

for a pure shell solution without calling external program:

NL=$'\n'    # define a variable to reference 'newline'

testVar=${testVar%$NL}    # removes trailing 'NL' from string

Answer 3

You can use sed as follows:

MY_NEW_VAR=$(echo $testVar | sed -e 's/\r//g')
echo ${MY_NEW_VAR} got it

By the way, try to do a dos2unix on your data file.

Answer 4

use this command on your script file after copying it to Linux/Unix

perl -pi -e 's/\r//' scriptfilename

Answer 5

yet another solution uses tr:

echo $testVar | tr -d '\r'
cat myscript | tr -d '\r'

the option -d stands for delete.

Answer 6

Pipe to sed -e 's/[\r\n]//g' to remove both Carriage Returns (\r) and Line Feeds (\n) from each text line.