echo "testing:"
PROXY_URL="/proxyurl/"
proxyUrlSedEscaped=`echo "$PROXY_URL" | sed -e 's/[\/&]/\\&/g'`
echo "$PROXY_URL" | sed -e 's/[\/&]/\\&/g'
echo "$proxyUrlSedEscaped"
echo "???"
The above has the following output:
testing:
\/proxyurl\/
&proxyurl&
???
Why is the last echo output different from the previous echo output?
This command works correctly via the command line:
PROXY_URL="/proxyurl/"; echo "$PROXY_URL" | sed -e 's/[\/&]/\\&/g', output: \/proxyurl\/
Per the Bash manual (and POSIX too):
When the old-style backquote form of substitution is used, backslash retains its literal meaning except when followed by
$,`, or\.
That means that `echo "$PROXY_URL" | sed -e 's/[\/&]/\\&/g'` results in the sed command that's executed being sed -e 's/[\/&]/\&/g', because the double backslash has been escaped into a single one. \& then makes a literal & in sed.
You can use $(...)-style command substitution instead to avoid that:
proxyUrlSedEscaped=$(echo "$PROXY_URL" | sed -e 's/[\/&]/\\&/g')
The new-style substitution allows any command with its usual interpretation.
If you must use ` then the fairly-grotesque double-escape works too:
proxyUrlSedEscaped=`echo "$PROXY_URL" | sed -e 's/[\/&]/\\\\&/g'`
That turns \\\\ into \\, which sed then turns into \ for you.
... (backticks)?" : mywiki.wooledge.org/BashFAQ/082
proxyUrlSedEscaped=$(echo "$PROXY_URL" | sed -e 's/[\\/&]/\\&/g'). And according to tldp.org/LDP/abs/html/commandsub.html: "The $(...) form has superseded backticks for command substitution.".