I have a large number of files and directories in one directory.
I need to sort them in terms of the permissions.
For example
drwx------
drwxr-xr-x
drwxr-x---
I am just wondering if we can sort the files and dirs using ls?
How to sort files by their permissions using ls
lssort
Related Solutions
With GNU sort and GNU du (which it appears you have, since you state you are using du's -h option):
du -sh -- * | sort -rh # Files and directories, or
du -sh -- */ | sort -rh # Directories only
The output looks something like this:
22G foo/
21G bar/
5.4G baz/
2.1G qux/
1021M wibble/
4.0K wobble/
You should decompose your goal into several steps easier to solve. This will have two advantages:
- It will be easier to solve,
- The resulting code will be clearer and more reusable.
The scripts below basically follows these steps:
- Generate raw statistic files. An easy way is to append the file size and the file name in a temporary file named after the original file extension. So, if you have the file
/path/to/foo.mp3which is 3000000 large, it will append3000000 /path/to/foo.mp3at the end of a temporary file namedmp3. - Handle specific cases. Here it will process the temporary file
m4aand create the two other filesm4a_aacandm4a_alacbased on the test you gave in the question. - Generate output. All the needed information being now available, it just has to:
- Count the number of line in each temporary file to determine the number of file of this type,
- Sum up each size to get the total size of files of this type.
Here is the script:
#!/bin/sh
# This script takes the searched directory as first parameter.
# For instance: ./this-script.sh ~/Music
: ${1:?"You must pass the search directory as first parameter."}
searchdir="$1"
# Create a temporary directory
statsdir=""
trap 'rm -rf $statsdir' EXIT
statsdir=$(mktemp -d "/tmp/tmp.XXXXXXXXXX") || exit 1
# Generate one listing file per extension
awkscript='/\.[[:alnum:]]+$/ {print $0 >statsdir"/"$(NF)}'
# For Linux: stat -c "%s %n"
# For Mac: stat -f "%z %N"
find "$searchdir" -type f -exec stat -f "%z %N" {} + | \
awk -F '.' -v statsdir="$statsdir" "$awkscript"
# Distinguish between m4a/AAC and m4a/ALAC
if [ -f "$statsdir/m4a" ]; then
input="$statsdir/m4a"
while IFS= read -r line; do
filename=${line#* }
if avprobe "$filename" 2>&1 | grep -q 'Audio: alac'; then
echo "$line" >> "$statsdir/m4a_alac"
else
echo "$line" >> "$statsdir/m4a_aac"
fi
done < "$input"
rm "$statsdir/m4a"
fi
# Generate and display result
{
printf "Type Count Size\n"
for extension in $(ls "$statsdir"); do
count=$(wc -l "$statsdir/$extension" | cut -d ' ' -f 1)
totalsize=$(awk '{s+=$1} END {print s}' "$statsdir/$extension")
printf "%s %d %d\n" "$extension" "$count" "$totalsize"
done
} | column -t
Best Answer
lsdoes not directly support sorting by permissions, but you can combine it with the sort command:You can use the
-koption to sort to start matching from a specific character, the format is-k FIELD.CHAR, the permissions are the first field in thelsoutput. So e.g.-k 1.2will start from the second character of the permission string, which will ignore any directory / device / link etc. flag, or-k 1.5for sorting by group permissions.If you don't want the additional output of
ls -l, you can remove it with awk:This will print only the first field (the permissions) and the last one (the filename).