Find files of specified size using bash in Unix [closed]

Question

I am looking for a Unix command to print the files with its size. I used this but it didn't work.

find . -size +10000k -print. 

I want to print the size of the file along with the filename/directory.

Answer 1

find . -size +10000k -exec ls -sd {} + 

If your version of find won't accept the + notation (which acts rather like xargs does), then you might use (GNU find and xargs, so find probably supports + anyway):

find . -size +10000k -print0 | xargs -0 ls -sd 

or you might replace the + with \; (and live with the relative inefficiency of this), or you might live with problems caused by spaces in names and use the portable:

find . -size +10000k -print | xargs ls -sd 

The -d on the ls commands ensures that if a directory is ever found (unlikely, but...), then the directory information will be printed, not the files in the directory. And, if you're looking for files more than 1 MB (as a now-deleted comment suggested), you need to adjust the +10000k to 1000k or maybe +1024k, or +2048 (for 512-byte blocks, the default unit for -size). This will list the size and then the file name. You could avoid the need for -d by adding -type f to the find command, of course.

Answer 2

Find can be used to print out the file-size in bytes with %s as a printf. %h/%f prints the directory prefix and filename respectively. \n forces a newline.

Example

find . -size +10000k -printf "%h/%f,%s\n" 

Output

./DOTT/extract/DOTT/TENTACLE.001,11358470 ./DOTT/Day Of The Tentacle.nrg,297308316 ./DOTT/foo.iso,297001116