[bash] How to remove a newline from a string in Bash

Adding answer to show example of stripping multiple characters including \r using tr and using sed. And illustrating using hexdump.

In my case I had found that a command ending with awk print of the last item |awk '{print $2}' in the line included a carriage-return \r as well as quotes.

I used sed 's/["\n\r]//g' to strip both the carriage-return and quotes.

I could also have used tr -d '"\r\n'.

Interesting to note sed -z is needed if one wishes to remove \n line-feed chars.

$ COMMAND=$'\n"REBOOT"\r   \n'

$ echo "$COMMAND" |hexdump -C
00000000  0a 22 52 45 42 4f 4f 54  22 0d 20 20 20 0a 0a     |."REBOOT".   ..|

$ echo "$COMMAND" |tr -d '"\r\n' |hexdump -C
00000000  52 45 42 4f 4f 54 20 20  20                       |REBOOT   |

$ echo "$COMMAND" |sed 's/["\n\r]//g' |hexdump -C
00000000  0a 52 45 42 4f 4f 54 20  20 20 0a 0a              |.REBOOT   ..|

$ echo "$COMMAND" |sed -z 's/["\n\r]//g' |hexdump -C
00000000  52 45 42 4f 4f 54 20 20  20                       |REBOOT   |

And this is relevant: What are carriage return, linefeed, and form feed?

  • CR == \r == 0x0d
  • LF == \n == 0x0a