How can I compare two directories to update a log file in BASH?

Question

I am attempting to compare two directories and then update the log file after determining what files are different between the two, but I keep getting error messages saying that it expects an integer result in the line [ if $userinput -eq $userinput.log ]. Is there a way to compare these two that I am missing? Here is the script I have written.

userinput=$1
if [ ! -d "$userinput" ];
then
        echo "Usage: dirlog.sh directory_name" 1>&2
        exit 0
else
        if [ ! -f "$userinput.log" ];
        then
                ls -l > $userinput.log
                echo ".logfile created for $userinput"
        elif [ -f "$userinput.log" ];
        then
                if [ $userinput -eq $userinput.log  ]; #if they are different
                then
                        differences=diff -rq $userinput $userinput.log
                        for file in $differences
                        do
                                echo "$file missing from directory $userinput"
                        done
                else
                        echo "no files missing from directory $userinput"
                fi
                #Update this no matter what
                ls -l > $userinput.log
                echo "logfile updated for directory $userinput"
        fi
fi

Answer 1

You can compare two directories with the diff -r command. This will show you all files that differ, which files are missing from one directory to another. The --brief option or --summary option will simply list the differences and not the content of the files.

Now, about your if [ $userinput -eq $userinput.log ]; ...

In shell, a variable could hold numeric or could hold alphanumeric data. Let's take something like this:

foo="0456"
bar="456"

In the above, $foo and $bar could be strings or numbers. If I think of them as numbers, they're equal. However, if they are strings, they're not equal.

Here's another example:

foo="123"
bar="45"

Which is greater? Again, it depends whether $foo and $bar are being treated as strings or numbers. If I treat them numerically $foo is greater. However, if I treat them as text strings, $bar is greater because the first character of $bar which is 4 comes after the first character of $foo , 1 .

So, how does bash know whether I want numeric or string comparison? Simple, Bash provides me with two different sets of comparison operators. One is for strings and one is for text:

                    String   Numeric
                    ------   -------
Equals                =       -eq
Greater Than          >       -gt
Less Than             <       -lt
Not equals           !=       -ne
Greater Than/Equals  >=       -ge
Less Than/Equals     <=       -le

In your case, you want to compare two strings and you're using -eq . This is why you're getting the error that the -eq operator expects numeric results. By the way, if you're using Bash or Kornshell, you should go ahead and use the doubled square brackets [[ ... ]] . These are less error prone and have a few more nice features such as pattern matching:

if [[ $userinput == "$userinput.log" ]]

However, I really recommend you to use he diff -r --brief command for comparing two directories.