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 some characters to it  it always comes to the start of the line   test dat  which has carriage return at end    testVar value123   testScript sh  sources above file    source test dat echo  testVar got it   The output I get is  got it23   How can I remove the   r  from the variable

User · Answer

yet another solution uses tr   echo  testVar   tr -d   r  cat myscript   tr -d   r    the option -d stands for delete

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

use this command on your script file after copying it to Linux Unix  perl -pi -e  s  r    scriptfilename

User · Answer

Pipe to sed -e  s   r n   g  to remove both Carriage Returns   r  and Line Feeds   n  from each text line

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

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