CODEWITHSUNDEEP
bashshellrecursion
JohnnyF
JohnnyF

Reputation: 1101

bash - using recursion - counting the executable files in dir

i am not looking for some high level ans, with just one line, i am trying to make a script that will work and to make it work with recursion and to understand it :)

I am trying to learn bash scripting, and geting into some problems:

I am trying to count the number of executable files in a dir and in its sub dirs

this it what i was thinking about

file name countFiles:

#!/bin/bash
counter = 0
for FILE in ($1/*)  /// going over all the files in the input path
do
 if (-d $FILE); then        /// checking if dir
  ./countFiles $FILE  /// calling the script again with ned path
  counter= $($counter + $0) /// addint to the counter the number of exe files from FILE
 fi
 if (-f $FILE); then /// checking if file
  if (-x $FILE); then /// checking id exe file
   counter = $counter + 1 // adding counter by 1
  fi
 fi
done
exit($counter) // return the number for exe files

Upvotes: 1

Views: 143

Answers (1)

gniourf_gniourf
gniourf_gniourf

Reputation: 46903

If you really want to use recursion (which is a bad idea in Bash): first, don't call your script recursively. Instead, call a function recursively. This will be more efficient (no forking overhead). Trying to fix your syntax:

#!/bin/bash

shopt -s nullglob

count_x_files() {
   # Counts number of executable (by user, regular and non-hidden) files
   # in directory given in first argument
   # Return value in variable count_x_files_ret
   # This is a recursive function and will fail miserably if there are
   # too deeply nested directories
   count_x_files_ret=0
   local file counter=0
   for file in "$1"/*; do
      if [[ -d $file ]]; then
         count_x_files "$file"
         ((counter+=count_x_files_ret))
      elif [[ -f $file ]] && [[ -x $file ]]; then
         ((++counter))
      fi
   done
   count_x_files_ret=$counter
}

count_x_files "$1"
echo "$count_x_files_ret"

Upvotes: 1

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