Renaming files in a folder to sequential numbers

Question

I want to rename the files in a directory to sequential numbers. Based on creation date of the files.

For Example sadf.jpg to 0001.jpg, wrjr3.jpg to 0002.jpg and so on, the number of leading zeroes depending on the total amount of files (no need for extra zeroes if not needed).

Answer 1

If your rename doesn't support -N, you can do something like this:

ls -1 -c | xargs rename -n 's/.*/our $i; sprintf("%04d.jpg", $i++)/e'

Edit To start with a given number, you can use the (somewhat ugly-looking) code below, just replace 123 with the number you want:

ls -1 -c | xargs rename -n 's/.*/our $i; if(!$i) { $i=123; } sprintf("%04d.jpg", $i++)/e'

This lists files in order by creation time (newest first, add -r to ls to reverse sort), then sends this list of files to rename. Rename uses perl code in the regex to format and increment counter.

However, if you're dealing with JPEG images with EXIF information, I'd recommend exiftool

This is from the exiftool documentation, under "Renaming Examples"

   exiftool '-FileName<CreateDate' -d %Y%m%d_%H%M%S%%-c.%%e dir

   Rename all images in "dir" according to the "CreateDate" date and time, adding a copy number with leading '-' if the file already exists ("%-c"), and
   preserving the original file extension (%e).  Note the extra '%' necessary to escape the filename codes (%c and %e) in the date format string.

Answer 2

You can also use ls

ls *.JPG| awk 'BEGIN{ a=0 }{ printf "mv %s gopro_%04d.jpg\n", $0, a++ }' | bash

Answer 3

Let us assume we have these files in a directory, listed in order of creation, the first being the oldest:

a.jpg
b.JPG
c.jpeg
d.tar.gz
e

then ls -1cr outputs exactly the list above. You can then use rename:

ls -1cr | xargs rename -n 's/^[^\.]*(\..*)?$/our $i; sprintf("%03d$1", $i++)/e'

which outputs

rename(a.jpg, 000.jpg)
rename(b.JPG, 001.JPG)
rename(c.jpeg, 002.jpeg)
rename(d.tar.gz, 003.tar.gz)
Use of uninitialized value $1 in concatenation (.) or string at (eval 4) line 1.
rename(e, 004)

The warning ”use of uninitialized value […]” is displayed for files without an extension; you can ignore it.

Remove -n from the rename command to actually apply the renaming.

This answer is inspired by Luke’s answer of April 2014. It ignores Gnutt’s requirement of setting the number of leading zeroes depending on the total amount of files.

Answer 4

To me this combination of answers worked perfectly:

ls -v | gawk 'BEGIN{ a=1 }{ printf "mv %s %04d.jpg\n", $0, a++ }' | bash

• ls -v helps with ordering 1 10 9 in correct: 1 9 10 order, avoiding filename extension problems with jpg JPG jpeg
• gawk 'BEGIN{ a=1 }{ printf "mv %s %04d.jpg\n", $0, a++ }' renumbers with 4 characters and leading zeros. By avoiding mv I do not accidentally try to overwrite anything that is there already by accidentally having the same number.
• bash executes

Be aware of what @xhienne said, piping unknown content to bash is a security risk. But this was not the case for me as I was using my scanned photos.

Answer 5

Follow command rename all files to sequence and also lowercase extension:

rename --counter-format 000001 --lower-case --keep-extension --expr='$_ = "$N" if @EXT' *

Answer 6

To renumber 6000, files in one folder you could use the 'Rename' option of the ACDsee program.

For defining a prefix use this format: ####"*"

Then set the start number and press Rename and the program will rename all 6000 files with sequential numbers.