How to remove a newline from a string in Bash

Question

Under bash, there are some bashisms:

The tr command could be replaced by // bashism:

COMMAND=$'\nREBOOT\r   \n'
echo "|${COMMAND}|"
|
   OOT
|

echo "|${COMMAND//[$'\t\r\n']}|"
|REBOOT   |

echo "|${COMMAND//[$'\t\r\n ']}|"
|REBOOT|

See Parameter Expansion and QUOTING in bash's man page:

man -Pless\ +/\/pattern bash
man -Pless\ +/\\\'string\\\' bash

man -Pless\ +/^\\\ *Parameter\\\ Exp bash
man -Pless\ +/^\\\ *QUOTING bash

Further...

As asked by @AlexJordan, this will suppress all specified characters. So what if $COMMAND do contain spaces...

COMMAND=$'         \n        RE BOOT      \r           \n'
echo "|$COMMAND|"
|
           BOOT      
|

CLEANED=${COMMAND//[$'\t\r\n']}
echo "|$CLEANED|"
|                 RE BOOT                 |

shopt -q extglob || { echo "Set shell option 'extglob' on.";shopt -s extglob;}

CLEANED=${CLEANED%%*( )}
echo "|$CLEANED|"
|                 RE BOOT|

CLEANED=${CLEANED##*( )}
echo "|$CLEANED|"
|RE BOOT|

Shortly:

COMMAND=$'         \n        RE BOOT      \r           \n'
CLEANED=${COMMAND//[$'\t\r\n']} && CLEANED=${CLEANED%%*( )}
echo "|${CLEANED##*( )}|"
|RE BOOT|

Note: bash have extglob option to be enabled (shopt -s extglob) in order to use *(...) syntax.

echo "|$COMMAND|"|tr '\n' ' '

will replace the newline (in POSIX/Unix it's not a carriage return) with a space.

To be honest I would think about switching away from bash to something more sane though. Or avoiding generating this malformed data in the first place.

Hmmm, this seems like it could be a horrible security hole as well, depending on where the data is coming from.

Clean your variable by removing all the carriage returns:

COMMAND=$(echo $COMMAND|tr -d '\n')

Using bash:

echo "|${COMMAND/$'\n'}|"

^{(Note that the control character in this question is a 'newline' (\n), not a carriage return (\r); the latter would have output REBOOT| on a single line.)}

Explanation

Uses the Bash Shell Parameter Expansion ${parameter/pattern/string}:

The pattern is expanded to produce a pattern just as in filename expansion. Parameter is expanded and the longest match of pattern against its value is replaced with string. [...] If string is null, matches of pattern are deleted and the / following pattern may be omitted.

Also uses the $'' ANSI-C quoting construct to specify a newline as $'\n'. Using a newline directly would work as well, though less pretty:

echo "|${COMMAND/
}|"

Full example

#!/bin/bash
COMMAND="$'\n'REBOOT"
echo "|${COMMAND/$'\n'}|"
# Outputs |REBOOT|

Or, using newlines:

#!/bin/bash
COMMAND="
REBOOT"
echo "|${COMMAND/
}|"
# Outputs |REBOOT|

What worked for me was echo $testVar | tr "\n" " "

Where testVar contained my variable/script-output

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

If you are using bash with the extglob option enabled, you can remove just the trailing whitespace via:

shopt -s extglob
COMMAND=$'\nRE BOOT\r   \n'
echo "|${COMMAND%%*([$'\t\r\n '])}|"

This outputs:

|
RE BOOT|

Or replace %% with ## to replace just the leading whitespace.