Question 39392 in Ubuntu Question posted in
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Bash script ,which puts all the files in order of size

I need help with a bash script. The problem is that I want to sort all the files in order of size, but I only need files, not folders,and to show me their size as well.
I have this code but folders also appear:

read -p "Enter the size of the top: " MARIMETOP
du  -a | sort -n -r | head -n $MARIMETOP | /usr/bin/awk 'BEGIN{ pref[1]="K";  pref[2]="M"; pref[3]="G";} { total = total + $1; x = $1; y = 1; while( x  > 1024 ) { x = (x + 1023)/1024; y++; }  printf("%g%st%sn",int(x*10)/10,pref[y],$2); } END { y = 1; while(  total > 1024 ) { total = (total + 1023)/1024; y++; } ; }'
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3

Answers


  1. 0

    This will print all regular files in the current directory and subdirectories sorted by size:

    find . -type f -print0 | xargs -0 -n100000 ls -Sl

    or if you want only size and filenames:

    find . -type f -print0 | xargs -0 -n100000 stat -f "%z %N" | sort -n -k1 -r

    With the -n100000 flag this will handle at most 100000 files found by find.

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  2. 0

    find from GNU findutils can print the file size

    find . -type f -printf '%st%pn' | sort -k1,1nr
    • %s = file size
    • %p = file path (use %f for just the file’s basename)
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  3. 0

    While the OP tagged the question by bash, he said in a comment that a zsh solution would also be OK. In zsh you could write it as

    du *(.OL)

    Explanation:

    .   : Only consider plain files
O   : Sort descending
L   : Sorting criterium is the length of the file
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