Find Directory if Number of files is over X

Question

Am wondering if someone can advise on the best way to do this. In my Web Dir is around 100 folders; each folder is a clientID.

Within each ClientID folder is a Sub-Set of folders. In each subset of folders is a set of structured folders that are the same.

Example:

httpdocs/client1234/amazon/
httpdocs/client1234/ebay/

Within the Folder of Amazon and Ebay we have a directory structure like:

In/
Out/
Backup/

What I am trying to do is scan the whole Web dir and locate In folders that have more than 50 files in the IN directory.

**edited Included in the IN Directory. Only interested in File count on the IN dir.

Answer 1

This one-liner should work:

find httpdocs -mindepth 2 -name In | xargs ls -1 | awk '$0 ~ /httpdocs.*In/{name=$0; start=NR} (NR-start)==50{print name;}'

It will first find all In directories, then list all its files, and, finally, filter those with more than 50 files.

Adjust -mindepth according to your start folder.

Answer 2

This will loop through the requested directories, counting the files (and subdirectories' files) to see if there are more than 50 files; if so, it echoes the matching directory/directories:

for dir in httpdocs/*/*/In
do
  count=$(find $dir | wc -l)
  if [ $count -gt 50 ]
  then
    echo $dir has more than 50 files
  fi
done

Note I used strictly "greater-than" 50 because find's output includes that top-level "In" directory in its output to wc -l.

Answer 3

#!/bin/bash
 for dir in `find ./httpdocs/*/*/ -name In -type d` ; do
  count=`find $dir -type f | wc -l`
  if [[ $count -gt 50 ]]
     echo $dir
  fi
done