如何从 Bash 中的字符串中删除换行符

Question

I have the following variable.

echo "|$COMMAND|"

which returns

|
REBOOT|

How can I remove that first newline?

Answer 1

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.

Answer 2

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.

Answer 3

通过删除所有回车符来清理变量：

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

Answer 4

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|

Answer 5

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

Where testVar contained my variable/script-output

Answer 6

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

Answer 7

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.

Answer 8

You can simply use echo -n "|$COMMAND|" .

$ man echo

-n do not output the trailing newline

Answer 9

To address one possible root of the actual issue, there is a chance you are sourcing a crlf file.

CRLF Example:

.env (crlf)

VARIABLE_A="abc"
VARIABLE_B="def"

run.sh

#!/bin/bash
source .env
echo "$VARIABLE_A"
echo "$VARIABLE_B"
echo "$VARIABLE_A $VARIABLE_B"

Returns:

abc
def
 def

If however you convert to LF:

.env (lf)

VARIABLE_A="abc"
VARIABLE_B="def"

run.sh

#!/bin/bash
source .env
echo "$VARIABLE_A"
echo "$VARIABLE_B"
echo "$VARIABLE_A $VARIABLE_B"

Returns:

abc
def
abc def

Answer 10

Short answers are the best (please upwote) Use this bashism if you don't want spawn processes like (tr, sed or awk) for such a simple task, bash can do that alone:

COMMAND=${COMMAND//$'\n'/}

From the Documentation:

${FOO//from/to} Replace all
${FOO/from/to}  Replace first match

Thanks for upwoting. This is the best answer.

Answer 11

Grep can be used to filter out any blank lines. This is useful if you have several lines of data and want to remove any empty.

echo "$COMMAND" | grep .