Cannot increment a variable in bash

Question

I'm trying to create a script to count the number of hidden and non-hidden files in a folder where the script is run. However, I run into an issue where I cannot increment variables.

#!/bin/bash

#A simple script to count the number of hidden and non-hidden files in the folder this script is run in

#Variables to store the number of hidden and non-hidden files and folders
#Variables with a 'h' at the end represent hidden items

files=0
fileh=0

#List all files and folders
#Use grep to folder entries beginning with '-', which are files
#Return the 9th word in the string which is the filename
#Read the filename into the variable 'fls'
ls -al | grep ^- | awk '{print $9}' | while read fls

#If the filename begins, with a dot, it is a hidden file
do
    if [[ $fls == .* ]]
    then
        #Therefore increment the number of hidden files by one
        let fileh++
    else
        #Else, increment the number if non-hidden files by one
        let files++
    fi
done

#Print out the two numbers
echo $files 'non-hidden files'
echo $fileh 'hidden files'

#When I run this script, the output is always zero for both variables
#I don't know why this doesn't work?!

The output of this script is as follows:

jai@L502X~$ ./script.sh 
0 non-hidden files
0 hidden files

quote your increments like let "fileh++"

SMA
– SMA

2014-12-28 15:26:48 +00:00
Commented Dec 28, 2014 at 15:26 — SMA
– SMA, Commented Dec 28, 2014 at 15:26
It did not make any difference.

jai_
– jai_

2014-12-28 15:31:13 +00:00
Commented Dec 28, 2014 at 15:31 — jai_
– jai_, Commented Dec 28, 2014 at 15:31
I feel like there is probably an easier way:

Daniel
– Daniel

2014-12-28 15:35:10 +00:00
Commented Dec 28, 2014 at 15:35 — Daniel
– Daniel, Commented Dec 28, 2014 at 15:35

Richard-Degenne · Accepted Answer · 2014-12-28 15:31:28Z

3

If you want to increment a variable using let, you have to quote your expression, like in

let "i++"

However, I personally prefer to use the double-parentheses syntax, which is

((i++))
# or, if you want a pre-fixed increment
((++i))

Plus, you can use a way shorter syntax for your if statement, using && and ||:

[[ $fls == .* ]] && ((++fileh)) || ((++files))

answered Dec 28, 2014 at 15:31

Richard-Degenne

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Comments

choroba · Accepted Answer · 2014-12-28 15:31:11Z

1

What happens on the right hand side of a | happens in a subshell. Changes to variables in a subshell don't propagate back to the parent shell.

Common workaround: don't use a pipe, use Process substitution:

while read fls ; do
   ...
done < <(ls -al | grep ^- | awk '{print $9}')

answered Dec 28, 2014 at 15:31

choroba

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1 Comment

jai_ Over a year ago

Thanks, you solution works! I do understand the explanation that you gave, however, I don't understand the syntax used.

Daniel · Accepted Answer · 2014-12-28 15:41:51Z

0

Not the answer to the "increment" question, but a much easier script to do what you're trying to do:

files=`find . -type f`
echo "non-hidden files " `echo "$files" | egrep -v "[/]\.[^/]+$" | wc -l`
echo "hidden files " `echo "$files" | egrep "[/]\.[^/]+$" | wc -l`

answered Dec 28, 2014 at 15:41

Daniel

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1 Comment

jai_ Over a year ago

Thanks for the suggestion but I'm doing a challenge where I can't use the commands 'find' 'du' 'locate' or any recursive function.

fejese · Accepted Answer · 2014-12-28 15:50:21Z

0

Quote the variable increment

let "fileh++"

edited Dec 28, 2014 at 15:50

fejese

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answered Dec 28, 2014 at 15:30

Arnab Nandy

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