bash shell：更改多个文件名以添加前导零

Question

I have several files with this pattern: prefix.1.* , prefix.2.* , prefix.3.* , etc... and I want their name to be changed, respectively, to prefix.01.* , prefix.02.* , prefix.03.* , etc. With that, they will be properly sorted by name as there are filenames with already two digits (eg prefix.27.* ) in the set.

How can I do that using commands available in the bash shell?

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Note: Just after this issue, I had to cope with a list of files like prefix.1.* , prefix.2.* , prefix.17.* , prefix.157.* (arbitrary, not sequentially numbered), with the aim to convert to prefix.001.* , prefix.002.* , prefix.017.* , prefix.157.* , with a generic way of inserting the right number of leading zeros. If you face this situation, I strongly recommend you to follow the general solution provided by mklement0 answer ( More generic renaming solutions ) below.

Answer 1

for file in prefix.[0-9].*
do
   mv "$file" "${file/prefix./prefix.0}"
done

Answer 2

More generic renaming solutions , prompted by later comments by the OP:

Uses the Perl-based rename utility , which comes standard on some Linux distros (eg, Ubuntu). Caveat : some distros have a different utility of the same name from the util-linux package.
On OS X, you can install via Homebrew using brew install rename .

Both solutions determine the required padding width dynamically .

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Simpler solution, in case files are already numbered sequentially and only padding must be performed [does NOT meet the OP's requirements - see other solution below] :

printf '%s\n' prefix.[0-9]*.* |                # enumerate files of interest
  sort -t. -n -k2,2 |                          # sort them by the embedded number
    rename -n -N '...01' -e 's/\.\d+\./.$N./'  # rename with appropriate padding

• -n performs a dry-run only, showing only what would happen; remove it to perform actual renaming
• -N '...01' specifies the format for the special counter variable $N in a way that auto-determines appropriate zero-padding based on the number of input files .
• 's/\\.\\d+\\./.$N./' is a Perl expression that replaces the existing number component with the appropriately padded counter variable $N .

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More sophisticated solution that preserves existing numbers and merely pads them.

In this case, the padding width is derived from the largest existing number embedded in the filenames :

# Determine the longest (largest) number contained in the matching filenames...
maxNum=$(printf '%s\n' prefix.[0-9]*.* | sort -nrt. -k2,2 | head -1 | cut -d. -f2)
# ... and derive the padding length from it.
padLength=${#maxNum}

# Perform renaming by zero-padding the *existing* numbers in the filenames.
rename -n -e 's/\.(\d+)\./sprintf(".%0'${padLength}'s.", $1)/e' prefix.[0-9]*.*

• -n performs a dry-run only, showing only what would happen; remove it to perform actual renaming
• Note the use of Perl's e flag in the s/.../.../ command, which allows use of an arbitrary Perl expression in the replacement string; in this case, an sprintf expression is used to zero-pad the input number; credit for this approach belongs to @Adrian Frühwirth's answer here .

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On a general note, matching filenames could be made more robust with shopt -s extglob and prefix.+([0-9]).* ; similarly, shopt -s nullglob would make it easier to determine whether any files matched.

Answer 3

Note a pure bash solution, but a little more concise(*):

printf '%s\0' prefix.[0-9].* | 
  xargs -0 -I % bash -c 'f="%"; mv "$f" "${f/\./.0}"'

This assumes that glob prefix.[0-9].* expands to at least 1 actual filename.
If it's possible that NO files match, execute shopt -s nullglob first (and, if required, restore that shell option's value later).

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Performance note :

While this is more concise than @R Sahu's solution , it will perform worse , due to spawning an additional child shell ( bash -c ) for every input filename - so as to be able to assign the filename to a variable and use expansion on it.

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(*) In terms of character count, @R Sahu's solution is actually shorter, but the absence of a loop (and fewer lines) may result in this answer being perceived as shorter. At the end of the day, this solution doesn't add much in itself - except perhaps as an advanced example of using xargs :

• printf '%s\\0' ensures that all matching filenames are passed through the pipeline with NUL chars. as separators.
• xargs -0 then ensures that the filenames are recognized individually, even if they contain embedded whitespace.
• -I % ensures that each input filename results in its own invocation of the command that follows, with % instances replaced with the filename at hand.
• bash -c then invokes a child shell (a child process that happens to be another bash instance) and evaluates the specified string.

Answer 4

我发现的最简单的方法是在 bash shell 中输入这一行：

for ITER in 1 2 3 4 5 6 7 8 9; do for FILE in prefix."$ITER".*; do mv "$FILE" $(echo "$FILE" | sed 's/prefix.'$ITER'/prefix.0'$ITER'/'); done; done

Answer 5

One more way>>

#!/bin/bash
count=1
for i in $(ls prefix*)
do
Newname=prefix\.$(printf "%02d" $count)\.txt
mv $i $Newname
count=$(($count + 1))
done

Answer 6

Great answers so far. If the extension is same of all the files. Just use following. Here the extension is .jpg you can change prefix_ to your choice of prefix. Also if you want more leading zeros just change %02 to %03 and so on.

rename -n -e 's/\d+/sprintf("prefix_%02d",$&)/e' -- *.jpg