Question

What is the best way, using Bash, to rename files in the form:

(foo1, foo2, ..., foo1300, ..., fooN)

With zero-padded file names:

(foo00001, foo00002, ..., foo01300, ..., fooN)
Was it helpful?

Solution

In case N is not a priori fixed:

 for f in foo[0-9]*; do mv $f `printf foo%05d ${f#foo}`; done

OTHER TIPS

It's not pure bash, but much easier with the rename command:

rename 's/\d+/sprintf("%05d",$&)/e' foo*

I had a more complex case where the file names had a postfix as well as a prefix. I also needed to perform a subtraction on the number from the filename.

For example, I wanted foo56.png to become foo00000055.png.

I hope this helps if you're doing something more complex.

#!/bin/bash

prefix="foo"
postfix=".png"
targetDir="../newframes"
paddingLength=8

for file in ${prefix}[0-9]*${postfix}; do
  # strip the prefix off the file name
  postfile=${file#$prefix}
  # strip the postfix off the file name
  number=${postfile%$postfix}
  # subtract 1 from the resulting number
  i=$((number-1))
  # copy to a new name with padded zeros in a new folder
  cp ${file} "$targetDir"/$(printf $prefix%0${paddingLength}d$postfix $i)
done

The oneline command that I use is this:

ls * | cat -n | while read i f; do mv "$f" `printf "PATTERN" "$i"`; done

PATTERN can be for example:

  • rename with increment counter: %04d.${f#*.} (keep original file extension)
  • rename with increment counter with prefix: photo_%04d.${f#*.} (keep original extension)
  • rename with increment counter and change extension to jpg: %04d.jpg
  • rename with increment counter with prefix and file basename: photo_$(basename $f .${f#*.})_%04d.${f#*.}
  • ...

You can filter the file to rename with for example ls *.jpg | ...

You have available the variable f that is the file name and i that is the counter.

For your question the right command is:

ls * | cat -n | while read i f; do mv "$f" `printf "foo%d05" "$i"`; done

Pure Bash, no external processes other than 'mv':

for file in foo*; do
  newnumber='00000'${file#foo}      # get number, pack with zeros
  newnumber=${newnumber:(-5)}       # the last five characters
  mv $file foo$newnumber            # rename
done

The following will do it:

for ((i=1; i<=N; i++)) ; do mv foo$i `printf foo%05d $i` ; done

EDIT: changed to use ((i=1,...)), thanks mweerden!

Here's a quick solution that assumes a fixed length prefix (your "foo") and fixed length padding. If you need more flexibility, maybe this will at least be a helpful starting point.

#!/bin/bash

# some test data
files="foo1
foo2
foo100
foo200
foo9999"

for f in $files; do
    prefix=`echo "$f" | cut -c 1-3`        # chars 1-3 = "foo"
    number=`echo "$f" | cut -c 4-`         # chars 4-end = the number
    printf "%s%04d\n" "$prefix" "$number"
done

My solution replaces numbers, everywhere in a string

for f in * ; do
    number=`echo $f | sed 's/[^0-9]*//g'`
    padded=`printf "%04d" $number`
    echo $f | sed "s/${number}/${padded}/";
done

You can easily try it, since it just prints transformed file names (no filesystem operations are performed).

Explanation:

Looping through list of files

A loop: for f in * ; do ;done, lists all files and passes each filename as $f variable to loop body.

Grabbing the number from string

With echo $f | sed we pipe variable $f to sed program.

In command sed 's/[^0-9]*//g', part [^0-9]* with modifier ^ tells to match opposite from digit 0-9 (not a number) and then remove it it with empty replacement //. Why not just remove [a-z]? Because filename can contain dots, dashes etc. So, we strip everything, that is not a number and get a number.

Next, we assign the result to number variable. Remember to not put spaces in assignment, like number = …, because you get different behavior.

We assign execution result of a command to variable, wrapping the command with backtick symbols `.

Zero padding

Command printf "%04d" $number changes format of a number to 4 digits and adds zeros if our number contains less than 4 digits.

Replacing number to zero-padded number

We use sed again with replacement command like s/substring/replacement/. To interpret our variables, we use double quotes and substitute our variables in this way ${number}.


The script above just prints transformed names, so, let's do actual renaming job:

for f in *.js ; do
    number=`echo $f | sed 's/[^0-9]*//g'`
    padded=`printf "%04d" $number`
    new_name=`echo $f | sed "s/${number}/${padded}/"`
    mv $f $new_name;
done

Hope this helps someone.

I spent several hours to figure this out.

To left-pad numbers in filenames:

$ ls -l
total 0
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:24 010
-rw-r--r-- 1 victoria victoria 0 Mar 28 18:09 050
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:23 050.zzz
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:24 10
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:23 1.zzz

$ for f in [0-9]*.[a-z]*; do tmp=`echo $f | awk -F. '{printf "%04d.%s\n", $1, $2}'`; mv "$f" "$tmp"; done;

$ ls -l
total 0
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:23 0001.zzz
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:23 0050.zzz
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:24 010
-rw-r--r-- 1 victoria victoria 0 Mar 28 18:09 050
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:24 10

Explanation

for f in [0-9]*.[a-z]*; do tmp=`echo $f | \
awk -F. '{printf "%04d.%s\n", $1, $2}'`; mv "$f" "$tmp"; done;
  • note the backticks: `echo ... $2}\` (The backslash, \, immediately above just splits that one-liner over two lines for readability)
  • in a loop find files that are named as numbers with lowercase alphabet extensions: [0-9]*.[a-z]*
  • echo that filename ($f) to pass it to awk
  • -F. : awk field separator, a period (.): if matched, separates the file names as two fields ($1 = number; $2 = extension)
  • format with printf: print first field ($1, the number part) as 4 digits (%04d), then print the period, then print the second field ($2: the extension) as a string (%s). All of that is assigned to the $tmp variable
  • lastly, move the source file ($f) to the new filename ($tmp)
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