Variable expansion is not working as intended with curl - shell

Question

I am having some trouble figuring out the expansion of a variable in my shell script. If I replace the variable with the desired string it works.
#!/bin/zsh
KEY="$(curl -Ivs -X GET "http://admin:admin#192.168.1.1" &> >(awk '/^> Authorization/{ print $3 " " $4 }'))"
# The string returned by the curl and awk command is Basic "YWRtaW46YWRtaW4=" without double quotes.
curl -H "Authorization: $KEY" "http://192.168.1.1/userRpm/WlanMacFilterRpm.htm?Page=1&exclusive=1"
# This doesn't work
curl -H "Authorization: Basic YWRtaW46YWRtaW4=" "http://192.168.1.1/userRpm/WlanMacFilterRpm.htm?Page=1&exclusive=1"
# This works
The only thing thats different in the above two lines is.
-H "Authorization: $KEY"
-H "Authorization: Basic YWRtaW46YWRtaW4="

Answer 1

HTTP generally uses DOS style CR+LF line terminators. Whenever you parse data out of curl, you have to account for this.
To check if this is the problem, run your script with bash -x yourscript or zsh -x yourscript to see trace output that shows otherwise invisible carriage returns:
var=$'value\C-M' # zsh
var=$'value\r' # bash
(Dash and ash/busybox unfortunately doesn't highlight this problem, so try with one of the above shells)
To strip them, pipe your data through tr -d '\r'.