Question

I am a newbie to BASH so please dont mind my stupid questions because I am not able to get any good sources to learn that.

I want to create a script to display filename and its size. This is what the code is like

filename=$1
if [ -f $filename ]; then
    filesize=`du -b $1`
    echo "The name of file is $1"
    echo "Its size is $filesize"
else
    echo "The file specified doesnot exists"
fi

The output is like this

$ ./filesize.sh aa
The name of file is aa
Its size is 88  aa

But in the last line I don’t want to show the name of the file. How do I do that ?
I want to do the same thing using wc as well.

Answer 1

All levels of question are accepted here.

You can use awk to only get the first field:

du -b /path/to/file | awk '{print $1}'

Your script would then be something like:

filename=$1
if [ -f ${filename} ]; then
    filesize=$(du -b ${filename} | awk '{print $1}')
    echo "The name of file is ${filename}"
    echo "Its size is ${filesize}"
else
    echo "The file specified does not exists"
fi

You’ll notice I’ve changed a couple of other things as well:

• I prefer $(X) to `X` for capturing output of command simply because they’re easier to nest.
• If you’ve already put $1 into $filename, you may as well use $filename from that point on (the $1 in the awk is a different beast).
• I prefer to put all my variables inside ${} so that it’s obvious to both the reader and the shell that ${var}X is $var appended with X, rather than $varX being the non-existent ${varX}.

These are really personal preferences but, as with all my personal preferences, I consider them best practices for the entire IT industry 🙂