Break down a large folder into multiple subfolders by sequential names

I have a folder with 20000 files in it. I need to split it into subfolders with increasing directory name (dir_1, dir_2....) with each folder having 500 files each in such a way that for example if files are from file_1.png to file_20000, the first folder should contain first 500 files i.e., file_1 to file 500 and immediate folder should contain files in a continued manner i.e, file_501 to file_1000 and so on. I have tried many scripts on internet but none helped me. Can anyone please help me in making this possible..
enter image description here

edited 23 mins ago

asked 31 mins ago

Praveen Kumar

New contributor

1

I'd love to see a more informative title to this question!
– Jeff Schaller
29 mins ago

2

please please do not post images of text. Or, if you must, at least don't post such sparse images, and make them more helpful in the context of the question.
– DopeGhoti
14 mins ago

add a comment |

edited 23 mins ago

asked 31 mins ago

Praveen Kumar

New contributor

1

I'd love to see a more informative title to this question!
– Jeff Schaller
29 mins ago

2

please please do not post images of text. Or, if you must, at least don't post such sparse images, and make them more helpful in the context of the question.
– DopeGhoti
14 mins ago

add a comment |

edited 23 mins ago

asked 31 mins ago

Praveen Kumar

New contributor

shell-script

edited 23 mins ago

asked 31 mins ago

Praveen Kumar

New contributor

edited 23 mins ago

asked 31 mins ago

Praveen Kumar

New contributor

edited 23 mins ago

asked 31 mins ago

Praveen Kumar

New contributor

asked 31 mins ago

Praveen Kumar

asked 31 mins ago

Praveen Kumar

New contributor

Praveen Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

1

I'd love to see a more informative title to this question!
– Jeff Schaller
29 mins ago

2

please please do not post images of text. Or, if you must, at least don't post such sparse images, and make them more helpful in the context of the question.
– DopeGhoti
14 mins ago

add a comment |

1

I'd love to see a more informative title to this question!
– Jeff Schaller
29 mins ago

2

please please do not post images of text. Or, if you must, at least don't post such sparse images, and make them more helpful in the context of the question.
– DopeGhoti
14 mins ago

I'd love to see a more informative title to this question!
– Jeff Schaller
29 mins ago

please please do not post images of text. Or, if you must, at least don't post such sparse images, and make them more helpful in the context of the question.
– DopeGhoti
14 mins ago

add a comment |

3 Answers
3

active

oldest

votes

#!/bin/sh



files_per_dir=500



set -- file_*



printf 'There are %d filesn' "$#"

printf 'Putting %d files in each new directoryn' "$files_per_dir"



N=0 # directory counter

n=0 # file counter



for filename do



    if [ "$(( n % files_per_dir ))" -eq 0 ]; then

        N=$(( N + 1 ))

        dir="dir_$N"

        printf 'Creating directory %sn' "$dir"

        # mkdir "$dir"

    fi



    n=$(( n + 1 ))



    printf 'Moving %s to %sn' "$filename" "$dir"

    # mv -i -- "$filename" "$dir"



done

The above would put 500 files into each directory (the actual command that changes things have been commented out for safety). The filenames are assumed to match file_* and each new directory will be called dir_N where N is a positive integer.

If you want to move the files based on their numerical names (the above sorts the files in lexicographical order), then use

printf 'Moving %s to %dn' "file_$n" "$dir"

# mv -i -- "file_$n" "$dir"

instead (i.e. replace the corresponding two lines in the first script with these, in this order).

answered 10 mins ago

Kusalananda

122k16230375

add a comment |

cd lotsafiles

for i in {1..20000}; do

  dirnum="$(((i/500)+1))"

  mkdir "dir_${dirnum}"

  mv "file${i}.png" "dir_${dirnum}"/

done

edited 6 mins ago

Kusalananda

122k16230375

answered 14 mins ago

DopeGhoti

43.5k55382

add a comment |

-1

Something simple along the lines of:

#!/bin/bash

typeset -i i

typeset -i j

typeset -i k



i=0

j=0

k=0

mkdir dir_$j

for file in * ; do

   i=$i+1

   k=$k+1

   if [ $i = 500 ] ; then

       j=$j+1

       i=0

       mkdir  dir_$j

   fi

   mv "$file" dir_$j/file_$k

done

Not tested, so you might put an echo in front of the mv "$file" dir_$j to see if it does what you want. Anyway it will give you an idea of how to do what you want.

answered 13 mins ago

Ljm Dullaart

58817

Yes it works. That's why the typeset -i is there. Yes, I use string comparison, but that also works. Would you prefer I put #!/bin/sh as first line?
– Ljm Dullaart
6 mins ago

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

#!/bin/sh



files_per_dir=500



set -- file_*



printf 'There are %d filesn' "$#"

printf 'Putting %d files in each new directoryn' "$files_per_dir"



N=0 # directory counter

n=0 # file counter



for filename do



    if [ "$(( n % files_per_dir ))" -eq 0 ]; then

        N=$(( N + 1 ))

        dir="dir_$N"

        printf 'Creating directory %sn' "$dir"

        # mkdir "$dir"

    fi



    n=$(( n + 1 ))



    printf 'Moving %s to %sn' "$filename" "$dir"

    # mv -i -- "$filename" "$dir"



done

If you want to move the files based on their numerical names (the above sorts the files in lexicographical order), then use

printf 'Moving %s to %dn' "file_$n" "$dir"

# mv -i -- "file_$n" "$dir"

instead (i.e. replace the corresponding two lines in the first script with these, in this order).

answered 10 mins ago

Kusalananda

122k16230375

add a comment |

#!/bin/sh



files_per_dir=500



set -- file_*



printf 'There are %d filesn' "$#"

printf 'Putting %d files in each new directoryn' "$files_per_dir"



N=0 # directory counter

n=0 # file counter



for filename do



    if [ "$(( n % files_per_dir ))" -eq 0 ]; then

        N=$(( N + 1 ))

        dir="dir_$N"

        printf 'Creating directory %sn' "$dir"

        # mkdir "$dir"

    fi



    n=$(( n + 1 ))



    printf 'Moving %s to %sn' "$filename" "$dir"

    # mv -i -- "$filename" "$dir"



done

If you want to move the files based on their numerical names (the above sorts the files in lexicographical order), then use

printf 'Moving %s to %dn' "file_$n" "$dir"

# mv -i -- "file_$n" "$dir"

instead (i.e. replace the corresponding two lines in the first script with these, in this order).

answered 10 mins ago

Kusalananda

122k16230375

add a comment |

#!/bin/sh



files_per_dir=500



set -- file_*



printf 'There are %d filesn' "$#"

printf 'Putting %d files in each new directoryn' "$files_per_dir"



N=0 # directory counter

n=0 # file counter



for filename do



    if [ "$(( n % files_per_dir ))" -eq 0 ]; then

        N=$(( N + 1 ))

        dir="dir_$N"

        printf 'Creating directory %sn' "$dir"

        # mkdir "$dir"

    fi



    n=$(( n + 1 ))



    printf 'Moving %s to %sn' "$filename" "$dir"

    # mv -i -- "$filename" "$dir"



done

If you want to move the files based on their numerical names (the above sorts the files in lexicographical order), then use

printf 'Moving %s to %dn' "file_$n" "$dir"

# mv -i -- "file_$n" "$dir"

instead (i.e. replace the corresponding two lines in the first script with these, in this order).

answered 10 mins ago

Kusalananda

122k16230375

#!/bin/sh



files_per_dir=500



set -- file_*



printf 'There are %d filesn' "$#"

printf 'Putting %d files in each new directoryn' "$files_per_dir"



N=0 # directory counter

n=0 # file counter



for filename do



    if [ "$(( n % files_per_dir ))" -eq 0 ]; then

        N=$(( N + 1 ))

        dir="dir_$N"

        printf 'Creating directory %sn' "$dir"

        # mkdir "$dir"

    fi



    n=$(( n + 1 ))



    printf 'Moving %s to %sn' "$filename" "$dir"

    # mv -i -- "$filename" "$dir"



done

If you want to move the files based on their numerical names (the above sorts the files in lexicographical order), then use

printf 'Moving %s to %dn' "file_$n" "$dir"

# mv -i -- "file_$n" "$dir"

instead (i.e. replace the corresponding two lines in the first script with these, in this order).

answered 10 mins ago

Kusalananda

122k16230375

answered 10 mins ago

Kusalananda

122k16230375

answered 10 mins ago

Kusalananda

122k16230375

answered 10 mins ago

Kusalananda

122k16230375

add a comment |

cd lotsafiles

for i in {1..20000}; do

  dirnum="$(((i/500)+1))"

  mkdir "dir_${dirnum}"

  mv "file${i}.png" "dir_${dirnum}"/

done

edited 6 mins ago

Kusalananda

122k16230375

answered 14 mins ago

DopeGhoti

43.5k55382

add a comment |

cd lotsafiles

for i in {1..20000}; do

  dirnum="$(((i/500)+1))"

  mkdir "dir_${dirnum}"

  mv "file${i}.png" "dir_${dirnum}"/

done

edited 6 mins ago

Kusalananda

122k16230375

answered 14 mins ago

DopeGhoti

43.5k55382

add a comment |

cd lotsafiles

for i in {1..20000}; do

  dirnum="$(((i/500)+1))"

  mkdir "dir_${dirnum}"

  mv "file${i}.png" "dir_${dirnum}"/

done

edited 6 mins ago

Kusalananda

122k16230375

answered 14 mins ago

DopeGhoti

43.5k55382

cd lotsafiles

for i in {1..20000}; do

  dirnum="$(((i/500)+1))"

  mkdir "dir_${dirnum}"

  mv "file${i}.png" "dir_${dirnum}"/

done

edited 6 mins ago

Kusalananda

122k16230375

answered 14 mins ago

DopeGhoti

43.5k55382

edited 6 mins ago

Kusalananda

122k16230375

edited 6 mins ago

Kusalananda

122k16230375

edited 6 mins ago

Kusalananda

122k16230375

answered 14 mins ago

DopeGhoti

43.5k55382

answered 14 mins ago

DopeGhoti

43.5k55382

answered 14 mins ago

DopeGhoti

43.5k55382

add a comment |

-1

Something simple along the lines of:

#!/bin/bash

typeset -i i

typeset -i j

typeset -i k



i=0

j=0

k=0

mkdir dir_$j

for file in * ; do

   i=$i+1

   k=$k+1

   if [ $i = 500 ] ; then

       j=$j+1

       i=0

       mkdir  dir_$j

   fi

   mv "$file" dir_$j/file_$k

done

Not tested, so you might put an echo in front of the mv "$file" dir_$j to see if it does what you want. Anyway it will give you an idea of how to do what you want.

answered 13 mins ago

Ljm Dullaart

58817

Yes it works. That's why the typeset -i is there. Yes, I use string comparison, but that also works. Would you prefer I put #!/bin/sh as first line?
– Ljm Dullaart
6 mins ago

add a comment |

-1

Something simple along the lines of:

#!/bin/bash

typeset -i i

typeset -i j

typeset -i k



i=0

j=0

k=0

mkdir dir_$j

for file in * ; do

   i=$i+1

   k=$k+1

   if [ $i = 500 ] ; then

       j=$j+1

       i=0

       mkdir  dir_$j

   fi

   mv "$file" dir_$j/file_$k

done

Not tested, so you might put an echo in front of the mv "$file" dir_$j to see if it does what you want. Anyway it will give you an idea of how to do what you want.

answered 13 mins ago

Ljm Dullaart

58817

Yes it works. That's why the typeset -i is there. Yes, I use string comparison, but that also works. Would you prefer I put #!/bin/sh as first line?
– Ljm Dullaart
6 mins ago

add a comment |

-1

Something simple along the lines of:

#!/bin/bash

typeset -i i

typeset -i j

typeset -i k



i=0

j=0

k=0

mkdir dir_$j

for file in * ; do

   i=$i+1

   k=$k+1

   if [ $i = 500 ] ; then

       j=$j+1

       i=0

       mkdir  dir_$j

   fi

   mv "$file" dir_$j/file_$k

done

Not tested, so you might put an echo in front of the mv "$file" dir_$j to see if it does what you want. Anyway it will give you an idea of how to do what you want.

answered 13 mins ago

Ljm Dullaart

58817

Something simple along the lines of:

#!/bin/bash

typeset -i i

typeset -i j

typeset -i k



i=0

j=0

k=0

mkdir dir_$j

for file in * ; do

   i=$i+1

   k=$k+1

   if [ $i = 500 ] ; then

       j=$j+1

       i=0

       mkdir  dir_$j

   fi

   mv "$file" dir_$j/file_$k

done

Not tested, so you might put an echo in front of the mv "$file" dir_$j to see if it does what you want. Anyway it will give you an idea of how to do what you want.

answered 13 mins ago

Ljm Dullaart

58817

answered 13 mins ago

Ljm Dullaart

58817

answered 13 mins ago

Ljm Dullaart

58817

answered 13 mins ago

Ljm Dullaart

58817

Yes it works. That's why the typeset -i is there. Yes, I use string comparison, but that also works. Would you prefer I put #!/bin/sh as first line?
– Ljm Dullaart
6 mins ago

add a comment |

Yes it works. That's why the typeset -i is there. Yes, I use string comparison, but that also works. Would you prefer I put #!/bin/sh as first line?
– Ljm Dullaart
6 mins ago

Yes it works. That's why the typeset -i is there. Yes, I use string comparison, but that also works. Would you prefer I put #!/bin/sh as first line?
– Ljm Dullaart
6 mins ago

add a comment |

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