How to find different files in UNIX based on . extension?

Question

How to find different files in UNIX based extension. Lets say suppose if I have 100 files for each vamsi.log-vamsi100.log, krishna.txt-krishna100.txt, super.csv..., rajat.jpeg..., kiran.xml-kiran100.xml and etc. My requirement is if I give command it'd only give how many different patterns do I have in one folder. It should also deliver one for each extension. How can I achieve this?

ls -ltr gives all files. But I need to know how many different type of files there in one folder?

Answer 1

Bash command: for i in *; do echo "${i##*.}"; done | uniq -c

Answer 2

One way:

find . -maxdepth 1 -type f | sort -u -t. -k3 | wc -l

Explanation:

• find . -maxdepth 1 -type f: find all normal files in current directory
• sort -u -t. -k3: sort the output on 3 filed with . as a field separator and print only first of equal lines
• wc -l: count the lines

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Edit

If you are using zsh shell then you can run

echo ${(u)$(echo *(:e))} | wc -w

The result is probably better then in find case - at least multi-dot files are properly taken into account, if you are also interested in files which names start from . then

echo ${(u)$(echo {.,}*(:e))} | wc -w

Answer 3

I think a simple ls *.extension should do the trick :)

To count them, you can use ls *.extension -1 | wc -l

Answer 4

ls -1 -a | ls *.foo | wc -l

Where foo is your extension.

Answer 5

find . -maxdepth 1 -type f | awk -F. '{print $NF}' | sort | uniq -c

awk being used to grab whatever comes after the last . (the "extension")

sort to get all occurrences of each "extension" grouped together

uniq -c to show the number of duplicates of each row/extension.

Answer 6

Some script with bash-builtins and assotiative array

declare -A F_TYPE
unset F_TYPE
while IFS=" ," read b c d f
do
  n="$b${c:+ $c}${d:+ $d}"
  F_TYPE[$n]=$[${F_TYPE[$n]}+1]
done < <(file -b *)
for i in "${!F_TYPE[@]}"
do
  echo "$i = ${F_TYPE[$i]}"
done
echo "Totally ${#F_TYPE[@]} types"