Bash see's whitespace as new line

Question

Take this small example:

#!/bin/bash -x
myvar="\"Hello Stackoverflow\""
ping $myvar

The result of this is that any commands (ping is just for example here) will automatically see the whitespace between 'Hello' and 'Stackoverflow' as a newline or carridge return.

#  ./test.sh
+ myvar='"Hello Stakeoverflow"'
+ ping '"Hello' 'Stakeoverflow"'
ping: unknown host "Hello

Is there anyway I can overide this?

Ignore the use of ping, I understand you can't ping words, its just helpful to draw the example on how variables are processed when supplied as an argument to a shell command and so I need an error to show that.

Answer 1

The internal quotes do you no good -- don't try to use them. In the case of myvar="\"foo\"", the outer quotes are syntax, but the inner quotes are data. Quotes that are data aren't used by the shell for string-splitting purposes, and they don't turn back into syntax later (unless you use eval, which introduces a host of new problems, many of them security-impacting); all they do is add data that your actual command doesn't want or need.

You need syntactic quotes around your expansion to prevent string-splitting and glob expansion. Compare these four examples (only the first of which is correct):

do_something_with() {
   printf '<%s> ' "$@"
   echo
}

myvar="Hello Stackoverflow"
do_something_with "$myvar"  # output: <Hello Stackoverflow>
do_something_with $myvar    # output: <Hello> <Stackoverflow>

myvar2="\"Hello Stackoverflow\""
do_something_with $myvar2   # output: <"Hello> <Stackoverflow">
do_something_with "$myvar2" # output: <"Hello Stackoverflow">

If you want to store something which should be interpreted as multiple arguments in a variable, the proper thing to use is an array:

my_array=( Hello "Stack Overflow" )
do_something_with "${my_array[@]}" # output: <Hello> <Stack Overflow>

See also http://mywiki.wooledge.org/BashFAQ/050

Answer 2

you need to double quote the variable

ping "$myvar"