UNIX for Beginners Questions & Answers

hi,
i would like to create a bash script that check which port in my Linux server are closed (not in use) from a specific range, port range (3000-3010).
the print output need to be only 1 port, and it will be nice if the output will be saved as a variable or in same file.

my code is:

Code:

#!/bin/bash
IP=$1
first_port=$2
last_port=$3
function scanner

{
for ((port=$first_port; port<=$last_port; port++))
        do
                (echo >/dev/tcp/$IP/$port)> /dev/null 2>&1 && echo $port open || echo "$port closed"
        done
}

scanner

than to run it i use:
./demo2.sh 127.0.0.1 3000 3010

the output for this code is:
the result that i am trying to get is only:

Hi,

In terms of your script itself as a solution to your problem, a change like this should work:

Code:

#!/bin/bash
IP=$1
first_port=$2
last_port=$3
function scanner

{
for ((port=$first_port; port<=$last_port; port++))
        do
                if ! (echo >/dev/tcp/$IP/$port)2>/dev/null
                then
                        echo $port
                        return
                fi
        done
}

scanner

This will cause the first TCP port that cannot have a blank line written to it by the echo command to be echoed out to standard output. The key is the return statement, which will cause the function to end as soon as it is executed. So once we find the first port we can't write a blank line to, we print it out and return control to the next line of the script (which doesn't exist, and so the script ends at that point).

Now, a better question: what is it you actually need to do here, and why ? This script might well work for certain ports in certain circumstances, but a far better bet here would be to use a utility that is dedicated to the purpose of port scanning (like nmap, for example) rather than trying to re-invent the wheel with a shell script.

Couldn't you take your output, grep for the word 'closed' and then get the top line head -1