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I am trying to write a shell script which will count the number of lines, words and characters in Bash. 

    echo "Enter file name"
    read file
    if [ -f $file ]
    then
        echo "The number of lines in $file are "
        echo $(wc -l $file | cut -d " " -f1 )
    fi

The program does take the output but the cut part is not formatting the output. I checked the syntax for cut as well as wc. How do I get the number of lines without the filename at the end which is the default characteristic of the wc command?

This is the output I am getting now.

    Enter file name
    pow.sh
    The number of lines in pow.sh are

This is what is required.

    Enter file name
    pow.sh
    The number of lines in pow.sh are 3.
Kumar Divya Rajat
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  • Add a number after `cut`'s option `-f`, e.g. `1`. – Cyrus May 01 '15 at 18:38
  • First, you're missing an argument to your `cut` command; `-f` takes a number to tell it what field to display. Second, the output of `wc -l` starts with a number of space characters, each one of which is treated as a delimiter by `cut`. You probably want to use `awk` instead. – Mark Reed May 01 '15 at 18:38
  • Oh. I actually had the argument while executing. Probably while pasting here, I missed it somewhere. I will edit the question. I need to use shell script specifically for this purpose. I can't be using awk. – Kumar Divya Rajat May 01 '15 at 18:43

1 Answers1

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The typical way omit the filename is to avoid giving it to wc in the first place:

wc -l < $file

So you end up with:

printf "The number of lines in $file is "
wc -l < $file
William Pursell
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