syntax error near unexpected token `then'

Question

I tried changing some things in my code and still it doesn't work :S

#!/bin/bash 



function auto_net() {
    welcome
}

function welcome() {    
    echo "Will your net include a server?"
    read choice 
    if [ $choice -eq "Yes" ]; then
        interface Server;   
    elif [ $choice -eq "No" ];
    then
        interface Default;
    fi
}

function user_input() {
    declare -a devices=("${!1}")
    for((i=1; i <= $2; ++i)); do
         device=$3
         device="$device$i"
         devices+=("$device")   
    done
    for i in ${devices[@]}; do
        echo $i
    done
}

function interface() {
    if[ "$1" -eq "Server" ];
    then
        set_up Server;
        set_up Router;
        set_up Switch;
        set_up Host;
    elif[ "$1" -eq "Default" ]; then
        set_up Router;
        set_up Switch;
        set_up Host;
    fi
}

function set_up (){
    local routers=()
    local hosts=()      
    local switches=()
    if [ "$1" -eq  "Router" ];
    then    
        echo "How many routers are you going to configure?"
        read router_number
        user_input routers[@] router_number R 
    elif [ "$1" -eq "Switch" ];
    then
        echo "How many switches are there?"
        read switch_number
        user_input switches[@] switch_number S 

       elif [ $1 -eq "Host" ];
       then
            echo "How many hosts per switch? Type it in order"
        read host_number
        user_input hosts[@] host_number H
       fi 
}

auto_net
echo $?

Try https://www.shellcheck.net/ – Stéphane Chazelas Jun 02 '17 at 21:22 — Stéphane Chazelas, Jun 02 '17 at 21:22

Kusalananda · Answer 1 · 2017-06-03T06:20:15.130

You are missing some spaces.

if[ "$1" -eq "Server" ];
elif[ "$1" -eq "Default" ]; then

These should be a space after the keywords if and elif.

for((i=1; i <= $2; ++i)); do

This (missing space after for) is not strictly an error, but should be fixed for consistency with the rest of the code.

You are also using -eq throughout for string comparisons. To compare strings, use = (-eq compares integers):

if [ $choice = "Yes" ]; then

I'm also not sure what you would want this to do:

user_input routers[@] router_number R

You may want

routers=( $(user_input "$router_number" R) )

or something similar (and modify your user_input function to match that).

You should also double quote the variable expansions. See "Security implications of forgetting to quote a variable in bash/POSIX shells" as to why.

Paste your script into ShellCheck to get a fuller list of possible problems: https://www.shellcheck.net/

As a general tip, you may want to test run your script as you develop it, just to make sure that what you've just written actually works. As opposed to writing a complete script before testing it at all.

Another tip to avoid lengthy if-then-elif bits is to use case ... esac:

case "$1" in
    Router)
        echo 'How many routers?'
        read router_number
        # etc.
        ;;
    Switch)
        # stuff for switches
        ;;
    Host)
        # stuff for hosts
        ;;
    *)
        echo 'Something is wrong' >&1
        exit 1
        ;;
esac

This also allows for easy pattern matching. R*) in a "case label" will match Router and any other string that starts with R. R*|H*) would match any string that starts with either R or H, etc.

for((i=1; i <= $2; ++i)); do would be OK even if not "canonical", ( can act as a token delimiter (like <, ; or | but contrary to [ or { for instance). So things like if(cmd) or while<foo cmd should work. — Stéphane Chazelas, Jun 02 '17 at 21:20

syntax error near unexpected token `then'

1 Answers1