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.comSolution 1:
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'
Solution 2:
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 -doption (and not just -0) would also cause xargs not to treat the quotes etc specially.
A Passionate Techie 