Why does xargs strip quotes from input?

Question

Why does xargs strip quotes from input text?

Here is a simplified example:

echo "/Place/='http://www.google.com'" | xargs echo

outputs

/Place/=http://www.google.com

Is there any way to work-around this? (xargs -0 doesn't help me)

Answer 1

if you want xargs to ignore quotes one of the good soultion can be the use of xargs flag xargs -0

Directly from Man page OPTIONS

OPTIONS -0, --null

Input items are terminated by a null character instead of by whitespace, and the quotes and backslash are not special (every character is taken literally). Disables the end of file string, which is treated like any other argument. Useful when input items might contain white space, quote marks, or backslashes. The GNU find -print0 option produces input suitable for this mode.

I've checked on a GNU system that setting the delimiter to a specific value (like a newline) with -d option (and not just -0) would also cause xargs not to treat the quotes etc specially:

-bash-4.3$ { echo "a'b'c"; echo d; } | xargs -d$'\n' echo
a'b'c d
-bash-4.3$ rpm -qf "$(which xargs)"
findutils-4.6.0.0.99.11a05-alt1.x86_64
-bash-4.3$ { echo "a'b'c"; echo d; } | xargs echo
abc d
-bash-4.3$ 

Answer 2

For GNU xargs, I found another solution in the manpage: explicitly specify delimiter to be '\n'. This turns off special handling for quotes:

--delimiter=delim, -d delim

Input items are terminated by the specified character. The specified delimiter may be a single character, a C-style character escape such as \n, or an octal or hexadecimal escape code. Octal and hexadecimal escape codes are understood as for the printf command. Multibyte characters are not supported. When processing the input, quotes and backslash are not special; every character in the input is taken literally.

So,

echo "/Place/='http://www.google.com'" | xargs -d'\n' echo

outputs

/Place/='http://www.google.com'

Answer 3

From the xargs manual:

If you want an input argument to contain blanks or horizontal tabs, enclose it in double quotes or apostrophes. If the argument contains a double quote character ("), you must enclose the argument in apostrophes. Conversely, if the argument contains an apostrophe ('), you must enclose the argument in double quotes. You can also put a backslash (\) in front of a character to tell xargs to ignore any special meaning the character may have (for example, white space characters, or quotes).

This means you can escape quotes if the quotes are quoted themselves:

$ echo "/Place/=\'http://www.google.com\'" | xargs echo
/Place/='http://www.google.com'

will work but echo /Place/=\'http://www.google.com\' | xargs echo will not.

Answer 4

You could use GNU Parallel instead:

$ echo "/Place/='http://www.google.com'" | parallel echo
/Place/='http://www.google.com'

Then you do not have to do the quoting yourself.

Learn more: https://www.youtube.com/playlist?list=PL284C9FF2488BC6D1

Answer 5

I found another solution here https://stackoverflow.com/a/17468560/1795821 that suggests using sed to escape quotes.

For example:

sh-3.2$ echo "/Place/='http://www.google.com'" | sed "s/\'/\\\'/g" | xargs echo
/Place/='http://www.google.com'

Answer 6

From others answer one importation aspect is missing. "strip quotes" is not accurate description what xargs do. Better would be say its "evaluate" quotes similar how normal bash handling do:

echo "'a "'"'"a' "'"'"a b"'"'" '\\' \\\"a b\\\" " | xargs -n1 echo

output:

a "a
a b
\
"a
b"

And for "raw" output:

echo "'a "'"'"a' "'"'"a b"'"'" '\\' \\\"a b\\\" " | xargs -d$'\n' -n1 echo

output:

'a "a' "a b" '\' \"a b\"

As we see some " get processed where other " will not be changed. This mean adding random \ to escape input will butcher nested quotes. This mean is better to disable quote processing in xargs than trying to work around it.