Count number of lines in each file within current directory using a for loop?

Question

• If a parameter is provided as an input, script should only check number of lines for that input file, otherwise script should display number of lines for each file within that directory.
• The script file should be ignored i.e. the count for the script file should not be displayed.

I tried:

#!/bin/bash
if [ $# -eq 0 ]
then
        for l in *.txt
        do
                echo $(wc -l $l)
        done
else
        for l in $*
        do
                echo $(wc -l $l)
        done
fi

But I am not allowed to explicitly mention the file type i.e *.txt - it must check all files except script file.

Solution

I think you almost got it. You can use basename "$0" to find the name of the script from within the script, and print the line count of everything except that

#!/bin/bash
if [ $# -eq 0 ]
then
    for k in *
    do
        if [[ ! -d "$k" && "$k" != `basename "$0"` ]]
        then
            wc -l "$k"
        fi
    done
else
    for k in $*
    do
         wc -l "$k"
    done
fi

I took the liberty of

• Using k instead of l (not recommended for variable names I'm pretty sure although I can't find a link (it can be confused with 1, I))
• Using 4 spaces for indentation instead of 8. In the end that is of course your decision, but I'd say 4 are more readable than 8.
• Quoting your variables. Highly recommended to avoid splitting variable names with spaces.
• Dropping the unnecessary echo like @Serg pointed out.

EDIT I added double brackets and a test for directories to ignore them as well to the if conditional.

Comments

[Sergiy Kolodyazhnyy]

First of all, you don't need echo $(), just call wc directly

[Sergiy Kolodyazhnyy]

Second, use $0 variable to skip the script itself.

[Koshur]

How do we ignore directories?

[Matthias]

if [ ! -d "$k" ], -d stands for directory. I'll modify my answer. I needed to change it to double brackets to get this to work