使用regex重命名linux中的文件

Question

This is not actually a question, I solved this myself. But I wanted to post my solution here to save other people in the same situation time and effort.

So I came into the situation where I had to rename a lot (+3000) matching a certain pattern. In my case, the files were automatic backups from syncthing, so a file would be renamed like this:

foo.bar -> foo~20150221-1330.bar

Answer 1

After a lot of searching trough forums and man pages, I created the following one-liner which restores the original filename with the find, sed , xargs and mv commands in linux:

find . -type f  | sed -e 'p;s/\(.*\)~20[0-9]\{6\}-[0-9]\{6\}\(.*\)/\1\2/' | xargs -n2 -d'\n' mv

If you can replace the sed -part with your own pattern if you like. This command can handle whitespaces by the way (thanks to the -d'\\n' flag in xargs ), but not newlines. I hope some of you find this command useful.

Ok so I'll give some more information about what each command does:

• find : give all regular files (not directories) in the current directory

• sed : p will print every line from stdin, the s/regex/regex/ will print the same lines, but substituted. So you get each file followed by the fixed filename:

   ./foo/bar~20150221-172703.txt ./foo/bar.txt` 

• xargs : -n2 will take two lines and send them to mv as parameter, -d'\\n' will fix issues with whitespaces in folder names (the delimiter is set to newline instead of whitespace)

Answer 2

Your answer is good, but if you also want to address filenames that also contain newlines, if you don't want to use sed and if you want to make sure that only files with basename that matches this pattern are taken into account (your method fails, eg, with dir~20150221-172703/file , and your method makes mv complain a lot about file and file being the same file when the filename doesn't match your pattern) you need to proceed slightly differently.

---

One possibility, if your find supports the -print0 option (GNU find does): have find spit all the filenames, and use a while loop in which Bash (not sed ) will perform the substitution. Something like this:

find . -type f -print0 | while IFS= read -r -d '' file; do
    dirname=${file%/*}
    basename=${file##*/}
    # Perform the substitution only on basename
    # Since you like regex, you can use them
    if [[ $basename =~ ^(.*)~20[0-9]{6}-[0-9]{6}(.*)$ ]]; then
        new_basename=${BASH_REMATCH[1]}${BASH_REMATCH[2]}
        echo mv "$dirname/$basename" "$dirname/$new_basename"
    fi
done

You could also make sure find only spits matching files by using the -regex filter (not POSIX, but GNU find supports it):

find . -regextype posix-basic -regex '.*~20[0-9]\{6\}-[0-9]\{6\}[^/]*' -type f

Not sure this fully answers your question, but at least it fixes the flaws your method has.

Without GNU goodies, it's more difficult to have something robust and clean...

Answer 3

You can use rnm :

rnm -rs '/(.*)~20[0-9]{6}-[0-9]{6}(.*)/\1\2/' -fo -dp -1 *

1. -fo : file only mode
2. -dp : depth of directory (-1 means unlimited depth ie goes to all sub-directories).

You could use your own regex pattern exactly though, in that case you would have to change the regex mode to basic (BRE):

rnm --regex basic -rs '/\(.*\)~20[0-9]\{6\}-[0-9]\{6\}\(.*\)/\1\2/' -fo -dp -1 *

Note:

1. Only invalid characters are the null character and the path delimiter ( / ).
2. Default regex mode is javascript.