Identifying folder with name as largest number in the directory - bash

Question

there is a directory which contains folders named with numbers, i've to find the folder with largest number in that directory.
This is the script i've written to find that folder:
files='ls path/'
var=0
for file in $files
do
echo $file
tmp=$((file-"0"))
if [ $tmp -gt $var ]
then
var=$tmp
fi
done
echo $var
But it's not working. It gives below error after invoking the script using command sudo ./restore2.sh.
ls
path/
./restore2.sh: line 6: path/: syntax error: operand expected (error token is "/")
0

Answer 1

Try this:
#!/bin/bash
files=`ls path/`
var=0
for file in $files
do
echo $file
tmp=$((file-"0"))
if [ $tmp -gt $var ]
then
var=$tmp
fi
done
echo $var
there's a backtick here: ls path/ instead of single or double-quotes.
I've only corrected this statement and it worked. and notice to add #!/bin/bash at the top of the script. This will tell your system to run the script in a bash shell.

Answer 2

You're using single quotes instead of backticks files='ls path/'. It's trying to use it as a literal string instead of evaluating it.
Also, for that specific task, you can just do:
ls test | awk '{if($1 > largest){largest = $1}} END{print largest}'
To have it a bit simpler.

Answer 3

Use find instead:
find . -maxdepth 1 -type d -regextype "posix-extended" -regex "^.*[[:digit:]]+.*$" | sort -n | tail -1
Set the maxdepth to 1 to check for directories within this directory only and no deeper. Set the regular expression type to posix-extended and search for all directories that have one or more digits. Print the result and order through sort before taking the largest one with tail -1.

Answer 4

Does path/ have any files in it? It looks like it's empty.
You should be getting a completely different complaint...
You don't want the path info in the filename. Rather than strip it with ${file##*/}, just go there and use non-path'd names.
An adaptation using your own logic as its base -
cd /whatever/path/ # go where the files are
var=-1 # initialize comparator
for file in [0-9]* # each entry that starts with a digit
do [[ "$file" =~ [^0-9] ]] && continue # skip any file with nondigit contents
[[ -f "$file" ]] || continue # only process plain files
(( file > var )) && var=$file # remember largest seen
done
echo $var # report largest
If you are sure there will be no negative numbered filenames, this should do it.
If there can be valid negatives, then your initialization needs to be appropriately lower, and the exclusion of nondigits should include the minus sign, as well as the list of files to select.
Note that this doesn't parse ls and doesn't require piping through a sort or spawning any other processes -- it's all handled in the bash interpreter and should be pretty efficient.
If you are sure of your data, and know there aren't any negatives or files named just 0 or non-plain-file entries in the directory that match the [0-9]* pattern, you can simplify it to just
cd /whatever/path/ # go where the files are
for file in [0-9]*; do (( file > var )) && var=$file; done
echo $var # report largest
As an aside, if you wanted to preserve the "make a list first" logic, you should still NOT use ls. Use an array.
cd /wherever/your/files/are/
files=( [0-9]* )
for file in "${files[#]}"
do : ...