Bash Scripting explanation

mijohnst · 08-28-2009, 04:55 PM

Howdy all... I'm writing a simple script to change some security settings and I need someone to explain the -z option in a if statement. The way I understand it, if I have an empty variable then do something... for example:

Code:

for variable in '/bin/grep -i ^banner /tmp/sshd_config'; do
  if [ -z "$variable" ]; then
   sed -i '109i\banner /etc/motd' /tmp/sshd_config
  elif [ -n "$variable" ]; then
   sed -i '/^[Bb]anner/ d' /tmp/sshd_config
   sed -i '108i\Banner /etc/motd' /tmp/sshd_config
  fi
done

So the way I understand it, if grep comes up with nothing, then the string is empty...so in the empty "variable" add it at line 109. If there IS a line that begins with banner anywhere, then delete it and added it on line 108.

Am I misunderstanding something here because it will only run the option with -n. If I change the first option to -n then it run that option. Is the string not really empty? Thanks for the help!

David1357 · 08-28-2009, 06:14 PM

Quote:

Originally Posted by mijohnst

Is the string not really empty?

This script

Code:

#!/bin/bash
BLAH=$(grep -i ^banner blah.sh)
if [ -z "$BLAH" ] ; then
        echo "BLAH has a length of zero"
fi

if [ -n "$BLAH" ] ; then
        echo "BLAH has a non-zero length"
fi

produces this output

Code:

[user@machine:~]:./blah.sh
BLAH has a length of zero

On the other hand, this script

Code:

#!/bin/bash
for variable in $(grep -i ^banner bleh.sh) ; do
        if [ -z "$variable" ] ; then
                echo "variable has a length of zero"
        elif [ -n "$variable" ] ; then
                echo "variable has a non-zero length"
        else
                echo "variable failed all tests"
        fi
done

produces no output

Code:

[user@machine:~]:./bleh.sh
[user@machine:~]:

So, I am going to go with, "The string is not really empty."

David1357 · 08-28-2009, 06:16 PM

Instead of this

Code:

for variable in '/bin/grep -i ^banner /tmp/sshd_config'; do

did you really want this

Code:

for variable in `/bin/grep -i ^banner /tmp/sshd_config`; do

The first one is definitely a non-empty string. The second one executes the command in a sub-shell (note the backticks instead of the single quotes).

mijohnst · 08-28-2009, 11:17 PM

Thanks David1357... your first example worked. I will have to call out a variable instead of using a for loop. For curiosities sake, is it possible to do what I wanted with the for loop? I tried changing the ticks like you recommended but that didn't help. Thanks again for your help! Believe or not, I'm learning...

i92guboj · 08-29-2009, 12:02 AM

Quote:

Originally Posted by David1357

This script

Code:

#!/bin/bash
BLAH=$(grep -i ^banner blah.sh)
if [ -z "$BLAH" ] ; then
        echo "BLAH has a length of zero"
fi

if [ -n "$BLAH" ] ; then
        echo "BLAH has a non-zero length"
fi

produces this output

Code:

[user@machine:~]:./blah.sh
BLAH has a length of zero

On the other hand, this script

Code:

#!/bin/bash
for variable in $(grep -i ^banner bleh.sh) ; do
        if [ -z "$variable" ] ; then
                echo "variable has a length of zero"
        elif [ -n "$variable" ] ; then
                echo "variable has a non-zero length"
        else
                echo "variable failed all tests"
        fi
done

produces no output

Code:

[user@machine:~]:./bleh.sh
[user@machine:~]:

So, I am going to go with, "The string is not really empty."

Nope. The problem with that script is that the contents of for is never run. "for" runs an interation for each element, since there are no elements, there's no looping at all.

David the H. · 08-29-2009, 12:59 AM

"test" is actually a separate command, the output of which can used by the if statement. 'test foo' and '[ foo ]' are equivalent commands.

test evaluates the expression given to it, and outputs a true-or-false exit value. "[ -z $variable ]" is really saying "the variable is a null string: T/F?", so if it's null the answer is "true", and false otherwise.

The if statement then reads that exit value to decide what to do next.

Another useful way you can test whether a variable is set or not is to use parameter substitution to define a default value.

Code:

if [ ${variable:-novalue}" = "novalue" ]; then
   echo "The variable is unset."
else
   echo "The variable is $variable."

If the variable is empty or unset here, then it will default to "novalue", and test will echo true. If the variable contains something, it will echo false, and if will branch to the alternate action.

This won't work correctly, however, if the variable actually contains the string "novalue" (it will test true), so there's also an operator that works the opposite way:

Code:

if [ "${variable:+setvalue}" = "setvalue" ]; then 
   echo "The variable is $variable."
else
   echo "No variable defined."
fi

In this case, if the variable contains something, then the substitute value will be output in place of it, and test will echo true. Otherwise the test reads false.

mijohnst · 08-29-2009, 01:06 AM

Thank you all very much for your help. It makes sense now and also explains why I haven't been able to find any "for" examples the way I tried to do it. Thanks again for taking the time! Once again this forum and it's users don't disappoint!

David1357 · 08-29-2009, 10:29 AM

Quote:

Originally Posted by mijohnst

For curiosities sake, is it possible to do what I wanted with the for loop?

Can you post your modified script here? I would like to see your changes before trying to answer this question.

David1357 · 08-29-2009, 10:32 AM

Quote:

Originally Posted by i92guboj

The problem with that script is that the contents of for is never run.

That was implicit in what I was saying. If the string in the original post was empty, the "for" loop would not even get executed. I was merely providing what you might call a proof.