for i in `seq 1 4` issue

[DeadBeet]

So, I'm currently writing a script for IP addresses and what not.

Currently it looks like this:

#!/bin/bash
echo -n "Enter Network: "
read IP;

for i in `seq 1 4`
do

GATEWAY=$(echo $IP | awk -F "." '{print $1 "." $2 "." $3 "." ($4+$i+2)}')

done

echo $GATEWAY;

Very very basic.. However, this is what happens:

# ./usable
# Enter Network: 127.0.0.1
127.0.0.130

Why does it return .130?

[Cabhan]

I can't tell you exactly why, but it is because there is no such variable as $i. Remember, the code that awk is executing is awk code, but you have defined i as a Bash variable. Bash variables and awk variables are completely different.

The first thing to be aware of is that the loop in your program does nothing. Your problem would operate the same if you wrote:

#!/bin/bash
echo -n "Enter Network: "
read IP;

GATEWAY=$(echo $IP | awk -F "." '{print $1 "." $2 "." $3 "." ($4+4+2)}')

echo $GATEWAY

In your loop, you just overwrite GATEWAY, so only the last value is remembered.

In any case, if you actually did want to use a loop, you can turn a Bash variable into an awk variable by doing:

#!/bin/bash
echo -n "Enter Network: "
read IP;

for i in `seq 1 4`
do

GATEWAY=$(echo $IP | awk -v i="$i" -F "." '{print $1 "." $2 "." $3 "." ($4+i+2)}')

done

echo $GATEWAY

Note the -v i="$i" bit. This assigns the Bash variable $i to the awk variable i. Don't use $i for awk variables: $ variables are always fields in awk, and you just want the variable (if i was defined as 4, for instance, $i would return the fourth field of the line of input, while i would return the number 4).

Does this make sense? I ran it as follows and it worked:

$ for i in $(seq 1 4); do echo '127.0.0.1' | awk -v i=$i -F. '{print $1 "." $2 "." $3 "." ($4+i+2)}'; done
127.0.0.4
127.0.0.5
127.0.0.6
127.0.0.7

[DeadBeet]

IT workeD!! Thanks so much!!