Bashrc function errors

Shell Programming and Scripting
1

So I am messing around customizing my bashrc and I created a function that will move the apt proxy file depending on what parameter I pass.

It checks for two conditions either home or work (user provided) and moves the file according to where the file is currently located and what parameter it is given.

Function:

function aptproxy()
{
  echo $1
  if [ "$1" == 'work' -a -f /etc/apt/apt.conf.d/80proxy ]; then
      echo 'You are at work'
  elif [ $1 == 'work' -a !-f /etc/apt/apt.conf.d/80proxy ]; then
      echo 'You are set to home, switching to work'
      mv "/etc/apt/conf.d_tmp/80proxy" "/etc/apt/apt.conf.d/80proxy"
  elif [ $1 == 'home' -a -f /etc/apt/apt.conf.d/80proxy ]; then
      echo "You are setup for work, moving proxy"
      mv "/etc/apt/apt.conf.d/80proxy" "/etc/apt/conf.d_tmp/80proxy"
  elif [ $1 == 'home' a- !-f /etc/apt/apt.conf.d/80proxy ]; then
    echo "You are already configured for home"
  fi
}

output when calling it with one of the parameters (home or work) with set -xv

root@xubuntu:/home/user# aptproxy work
aptproxy work
+ aptproxy work
+ echo work
work
+ '[' work == work -a -f /etc/apt/apt.conf.d/80proxy ']'
+ '[' work == work -a '!-f' /etc/apt/apt.conf.d/80proxy ']'
bash: [: too many arguments
+ '[' work == home -a -f /etc/apt/apt.conf.d/80proxy ']'
+ '[' work == home -a '!-f' /etc/apt/apt.conf.d/80proxy ']'
bash: [: too many arguments

From this it looks like it is the mv commands causing the issue. I have tried them with the directories in quotes and without quotes and it yields the same results. From everything I've been reading that too many arguments error usually means a string variable is being split up into multiple pieces but with that output I don't see that happening?

Any help would be appreciated.

TIA

2

Without digging deeper: I think the ! needs spaces around it.

3

annnnd I feel dumb. Ty for the quick response that worked.