Need to count files & create log of that.

[paragnehete]

Hi Friends,
I need some help. First look at my files hierchachy.

/<level-1>/<level-2>/<level-3>/*.tif

eg. :

/2010-07-01/AFFIDAVIT-OF-SERVICE---FOR-SC/001/Babylon2_20100701012049_1278004849.49892_000.tif
./2010-07-01/AFFIDAVIT-OF-SERVICE---FOR-SC/002/Babylon2_20100701012049_1278004849.49892_016.tif
./2010-07-01/AFFIDAVIT-OF-SERVICE---FOR-SC/003/Babylon2_20100701012049_1278004849.49892_035.tif
./2010-07-01/AFFIDAVIT-OF-SERVICE---FOR-SC/004/Babylon2_20100701012049_1278004849.49892_046.tif
./2010-07-01/AFFIDAVIT-OF-SERVICE---FOR-SC/005/Babylon2_20100701012049_1278004849.49892_058.tif
./2010-07-01/AFFIDAVIT-OF-SERVICE---FOR-SC/006/Babylon2_20100701012049_1278004849.49892_078.tif
./2010-07-01/AFFIDAVIT-OF-SERVICE---FOR-SC/007/Babylon2_20100701012049_1278004849.49892_096.tif
./2010-07-01/AFFIDAVIT-OF-SERVICE---FOR-SC/008/Babylon2_20100701012049_1278004849.49892_105.tif
./2010-07-01/AFFIDAVIT-OF-SERVICE---FOR-SC/009/Babylon2_20100701012049_1278004849.49892_117.tif

I want to find out the list of level-3 folders, which contain more that 2 TIF images. I want list of full path from date to that tif image.

[atreyu]

This assumes that the level-1 directory is 2010-07-01, but you could modify that, too...

#!/bin/bash
dirs=$(find /2010-07-01/*/* -type d)
for dir in $dirs; do
  echo -e "\nDir:   $dir"
  declare -a files
  files=($(find $dir -type f -name '*.tif'))
  echo Files: ${#files[*]}
  if [ ${#files[*]} -gt 2 ]; then
    for file in ${files[*]}; do
      echo $file
    done
  fi
done