Find Files with the Same Name as the Directory

bashdiff()findgrepshell-script

I have the next problem:

Directory Example1 has three files: Example1, Things and Pictures.

Directory Example2 has three files: Example2, Example3 and Pictures.

I need a list showing only the files that match the directory name, this is: Example1 and Example2. I have tried with diff, find, locate and ls… but I have not achieved anything.

Best Answer

Since there are generally fewer directories than files, let's look for all directories and then test whether they contain the required filename.

find . -type d -exec sh -c '
    for dirpath do
        filepath="$dirpath/${dirpath##*/}"
        [ -f "$filepath" ] && printf "%s\n" "$filepath"
    done' sh {} +

This would print the pathnames of all regular files (and symbolic links to regular files) that is located in a directory that has the same name as the file.

The test is done in a short in-line sh -c script that will get a number of directory pathnames as arguments. It iterates over each directory pathname and constructs a file pathname with the name that we're looking for. The ${dirpath##*/} in the code could be replaced by $(basename "$dirpath").

For the given example directory structure, this would output

./Example1/Example1
./Example2/Example2

To just test for any name, not just regular files, change the -f test to a -e test.

Related Solutions

Shell – Recursively compare directory contents by name, ignoring file extensions

To see a listing, here are two variants, one that recurses into subdirectories and one that doesn't. All use syntax specific to bash, ksh and zsh.

comm -3 <(cd source && find -type f | sed 's/\.[^.]*$//' | sort) \
        <(cd dest && find -type f | sed 's/\.[^.]*$//' | sort)
comm -3 <(cd source && for x in *; do printf '%s\n' "${x%.*}"; done | sort) \
        <(cd dest && for x in *; do printf '%s\n' "${x%.*}"; done | sort)

Shorter, in zsh:

comm -3 <(cd source && print -lr **/*(:r)) <(cd dest && print -lr **/*(:r))
comm -3 <(print -lr source/*(:t:r)) <(print -lr dest/*(:t:r))

The comm command lists the lines that are common to two files (comm -12), that are only in the first file (comm -23) or that are only in the second file (comm -13). The numbers indicate what is subtracted from the output¹. The two input files must be sorted.

Here, the files are in fact the output of a command. The shell evaluates the <(…) construct by providing a “fake” file (a FIFO or a /dev/fd/ named file descriptor) as the argument to the command.

¹ _{So here the minus sayers are fully justified.}

If you want to perform actions on the files, you'll probably want to iterate over the source files.

cd source
for x in *; do
  set -- "…/dest/${x%.*}".*
  if [ $# -eq 1 ] && ! [ -e "$1" ]; then
    echo "$x has not been converted"
  elif [ $# -gt 1 ]; then
    echo "$x has been converted to more than one output file: " "$@"
  else
    echo "$x has been converted to $1"
  fi
done

Bash – Find files with same name but different content

Update: fixed a typo in the script: changed print $NF to print $3; also tidied things up, and added some comments.

Assuming file names do not contain \n, the following prints out a sorted list which breaks (as in: section control breaks) at unique file name, unique md5sum, and shows the corresponding group of file paths.

#!/bin/bash

# Choose which script to use for the final awk step 
out_script=out_all

# Print all duplicated file names, even when md5sum is the same 
out_all='{ if( p1 != $1 ) { print nl $1; print I $2 }
      else if( p2 != $2 ) { print I $2 }
      print I I $3; p1=$1; p2=$2; nl="\n" }
   END { printf nl}'

# Print only duplicated file names which have multiple md5sums.
out_only='{ if( p1 != $1 ) { if( multi ) { print pend }
                             multi=0; pend=$1 "\n" I $2 "\n" }
       else if( p2 != $2 ) { multi++; pend=pend I $2 "\n" } 
       pend=pend I I $3 "\n"; p1=$1; p2=$2 } 
   END { if( multi ) print pend }'

# The main pipeline 
find "${1:-.}" -type f -name '*' |  # awk for duplicate names
awk -F/ '{ if( name[$NF] ) { dname[$NF]++ }
           name[$NF]=name[$NF] $0 "\n" } 
     END { for( d in dname ) { printf name[d] } 
   }' |                             # standard md5sum output 
xargs -d'\n' md5sum |               # " "==text, "*"==binary
sed 's/ [ *]/\x00/' |               # prefix with file name  
awk -F/ '{ print $3 "\x00" $0 }' |  # sort by name. md5sum, path 
sort |                              # awk to print result
awk -F"\x00" -v"I=   " "${!out_script}"

Output showing only file names with multiple md5s

afile.html
   53232474d80cf50b606069a821374a0a
      ./test/afile.html
      ./test/dir.svn/afile.html
   6b1b4b5b7aa12cdbcc72a16215990417
      ./test/dir.svn/dir.show/afile.html

Output showing all files with the same name.

afile.html
   53232474d80cf50b606069a821374a0a
      ./test/afile.html
      ./test/dir.svn/afile.html
   6b1b4b5b7aa12cdbcc72a16215990417
      ./test/dir.svn/dir.show/afile.html

fi    le.html
   53232474d80cf50b606069a821374a0a
      ./test/dir.svn/dir.show/fi    le.html
      ./test/dir.svn/dir.svn/fi    le.html

file.html
   53232474d80cf50b606069a821374a0a
      ./test/dir.show/dir.show/file.html
      ./test/dir.show/dir.svn/file.html

file.svn
   53232474d80cf50b606069a821374a0a
      ./test/dir.show/dir.show/file.svn
      ./test/dir.show/dir.svn/file.svn
      ./test/dir.svn/dir.show/file.svn
      ./test/dir.svn/dir.svn/file.svn

file.txt
   53232474d80cf50b606069a821374a0a
      ./test/dir.show/dir.show/file.txt
      ./test/dir.show/dir.svn/file.txt
      ./test/dir.svn/dir.show/file.txt
      ./test/dir.svn/dir.svn/file.txt

Best Answer

Related Solutions

Shell – Recursively compare directory contents by name, ignoring file extensions

Bash – Find files with same name but different content

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