How to print entire line in BASH

Question

I'm new to bash scripting and I've been trying to print out the entire line but couldn't find a way to work.

This is my code

#!/bin/bash
MOTD=`cat /etc/motd | awk '{print $1}'`
if [ "$MOTD" = "WARNING" ]
then
    echo "Audit Criteria: Warning banner exist."
    echo "Vulnerability: No."
    echo "Details: $MOTD "
else
    echo "Audit Criteria: Warning banners does not exist."
    echo "Vulnerability: Yes."
    echo "Details: $MOTD "
fi

my output is:

Audit Criteria: Warning banner exist.
Vulnerability: No.
Details: WARNING:

instead of the WARNING:Authorized uses only All activity may be monitored and reported. , only "WARNING" appeared in the Details:

I believe the problem lies on the

MOTD=`cat /etc/motd | awk '{print $1}'` 

and

if [ "$MOTD" = "WARNING" ] parts, I've tried {print$0} but still could not get it to work.

Answer 1

I guess you want to get the first line of /etc/motd , not the first word. If so, use the following:

MOTD=$(head -1 /etc/motd)

and then do the string comparison with

if [[ $MOTD == WARNING* ]; then

You can check String contains in bash for more information about check if a string contains a specific substring in bash.

Answer 2

Perhaps it would be simpler to do the whole thing in awk:

awk 'NR==1{
    if($1=="WARNING") {
        print "Audit Criteria: Warning banner exists."
        print "Vulnerability: No."
    }
    else { 
        print "Audit Criteria: Warning banner does not exist."
        print "Vulnerability: Yes."
    }
    print "Details: " $0
    exit
}' /etc/motd

The condition NR==1 and the exit at the end of the block mean that only the first line of the file is processed.

The code above is the most similar to your bash script but you could make it a lot shorter using variables:

awk 'NR==1{if($1=="WARNING"){b="exists";v="No"}else{b="does not exist";v="Yes"}
printf "Audit Criteria: Warning banner %s.\nVulnerability: %s.\nDetails: %s\n",b,v,$0
exit}' /etc/motd

Answer 3

You are using only using variable MOTD and it is having only value WARNING.

#!/bin/bash

MOTD=`cat /etc/motd | awk '{print $1}'`    
if [ "$MOTD" = "WARNING" ]    
then
    echo "Audit Criteria: Warning banner exist."        
    echo "Vulnerability: No."       
    echo "Details: `cat /etc/motd` "    
else        
    echo "Audit Criteria: Warning banners does not exist."      
    echo "Vulnerability: Yes."      
    echo "Details: `cat /etc/motd`"    
fi

Or in case if you have multiple lines in /etc/motd and you need to print only one line then.

#!/bin/bash

MOTDL=`grep WARNING /etc/motd`
MOTD=`cat /etc/motd | awk '{print $1}'`    
if [ "$MOTD" = "WARNING" ]    
then
    echo "Audit Criteria: Warning banner exist."        
    echo "Vulnerability: No."       
    echo "Details: $MOTDL "    
else        
    echo "Audit Criteria: Warning banners does not exist."      
    echo "Vulnerability: Yes."      
    echo "Details: $MOTDL"    
fi