Simple Bash Programming

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.

Here's one very good source ibm.com/developerworks/library/l-bash.html — helpermethod
– helpermethod, Commented Jun 16, 2010 at 10:13

unbeli · Accepted Answer · 2010-06-16 09:08:57Z

3

Use stat(1)

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

1 Comment

Hai Vu Over a year ago

I like this solution the best: It employs only one external tool (stat), unlike other solutions which use two.

sepp2k · Accepted Answer · 2010-06-16 09:11:49Z

1

filesize=`du -b $1 | cut -f 1`

cut -f 1 splits its input by tabs and then outputs the first field (i.e. it returns each line of du's output up to the fist tab character.).

answered Jun 16, 2010 at 9:11

sepp2k

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Comments

paxdiablo · Accepted Answer · 2010-06-16 09:17:37Z

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 :-)

Matthew Mott · Accepted Answer · 2010-06-16 09:10:49Z

0

You could use awk to split the output of du at the first space, as follows:

filesize=`du -b $1 | awk '{ print $1; }'`

answered Jun 16, 2010 at 9:10

Matthew Mott

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Comments

user44556 · Accepted Answer · 2010-06-16 09:12:52Z

0

Instead of using du -b $1 use du -b $1 | cut -f 1. It will only include the first field of output of du command, therefore only reporting the size of the file.

answered Jun 16, 2010 at 9:12

user44556

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Comments

kasperjj · Accepted Answer · 2010-06-16 09:14:25Z

0

You can use awk to pick out the first column of data.

filesize=`du -b $1 | awk '{print $1}'`

answered Jun 16, 2010 at 9:14

kasperjj

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