error with echo command - bash

Question

I'm starting to learn how to script bash and I've run into a problem with the echo command and a variable.
#!/bin/bash
LOGINOUTPUT = "`wget --no-check-certificate --post-data 'login=redacted&password=redacted' https://nessusserver:8834/login -O -`"
echo $LOGINOUTPUT
Running this script returns the following:
--2013-08-15 15:07:32-- https://nessusserver:8834/login
Resolving nessussserver (nessusserver)... 172.23.80.88
Connecting to nessusserver (nessusserver)|172.23.80.88|:8834... connected.
WARNING: cannot verify nessusserver's certificate, issued by ‘/C=FR/ST=none/L=Paris/O=Nessus Users United/OU=Certification Authority for nessusserver.healthds.com/CN=nessusserver.healthds.com/emailAddress=ca#besecmisc1.healthds.com’:
Unable to locally verify the issuer's authority.
WARNING: certificate common name ‘nessusserver.healthds.com’ doesn't match requested host name ‘nessusserver’.
HTTP request sent, awaiting response... 200 OK
Length: 461 [text/xml]
Saving to: ‘STDOUT’
100%[=============================================================================================================================================================>] 461 --.-K/s in 0s
2013-08-15 15:07:33 (90.4 MB/s) - written to stdout [461/461]
./nessus-output.sh: line 2: LOGINOUTPUT: command not found
Why does it think that LOGINOUTPUT is a command? Thanks in advance for any help!
EDIT: Updated script
#!/bin/bash
LOGINOUTPUT=$(wget --no-check-certificate --post-data 'login=redacted&password=redacted' https://nessusserver:8834/login -O -)
echo $LOGINOUTPUT
Still yields the same error, same if I leave $(...) as backticks.

Answer 1

This happens because you have spaces before and after the = in the variable assignment. The correct assignment is:
LOGINOUTPUT="....
with no spaces.
If you add spaces, then the shell interprets LOGINOUTPUT as a command, and tries to pass it two arguments: the "=" and the the quoted string. This of course fails, with the error LOGINOUTPUT: command not found
As a sidenote, it is better to use this syntax $(command) than backticks when doing process substitution.