Bash: handling “[[ statement ]] || echo problem found ; exit 1” logic

All these are equivalent:

[ $(vagrant --version > /dev/null) ]
[ ]
test

shells are not like other programming languages. They have 2 outputs: The return value, and stdout. You are passing stdout to test, but stdout is the empty string, as it has been redirected.

What you were looking for is.

[ $(vagrant --version > /dev/null; echo $? ) ]

However [ 0 ] and [ 1 ] etc all return true because [ … ] with a single word inside tests if the word is non-empty. So you need

[ $(vagrant --version > /dev/null; echo $? ) = 0 ]

or more simply

vagrant --version > /dev/null

You don't need grep here at all, awk can do the pattern match on the port number.

awk can also keep track of whether the port was found or not and exit with an appropriate exit code.

For example:

$ port=23
$ awk  '$2 ~ /^'"$port"'\// {print $1 ; found=1} END {exit !found}' /etc/services 
telnet
$ echo $?
0

$ port=99999
$ awk  '$2 ~ /^'"$port"'\// {print $1 ; found=1} END {exit !found}' /etc/services 
$ echo $?
1

The exit !found works because awk variables default to zero (or true) if they haven't previously been defined - exit !0 is exit 1. So if we set found=1 when we match then exit !found in the END block is exit 0.

Here's how to use that awk script with your if/then/else.

#!/bin/bash

awk  '$2 ~ /^'"$1"'\// {print $1 ; found=1} END {exit !found}' /etc/services 
if [ $? -eq 0 ]; then 
    echo "Service(s) is(are) found correctly!"
else
    echo "There is(are) no such service(s)!"
fi

You can also do it like this:

if awk  '$2 ~ /^'"$1"'\// {print $1;found=1} END{exit !found}' /etc/services ; then 
    echo "Service(s) is(are) found correctly!"
else
    echo "There is(are) no such service(s)!"
fi

Or even like this:

awk  '$2 ~ /^'"$1"'\// {print $1 ; found=1} END {exit !found}' /etc/services \
  && echo "Service(s) is(are) found correctly!" \
  || echo "There is(are) no such service(s)!"