I have a bunch of files with different extensions and i would like to add a suffix at the end of all of their names:
Fe2-K-D4.rac
Fe2-K-D4.plo
Fe2-K-D4_iso.xy
...
to
Fe2-K-D4-4cc8.rac
Fe2-K-D4-4cc8.plo
Fe2-K-D4-4cc8_iso.xy
...
I read a few posts about changing the extension using the rename tool but i don't know how to change the name and keep the same extension (I'm a recent linus user).
Thanks for any help
Using Extract filename and extension in Bash, I would say:
for file in *
do
extension="${file##*.}"
filename="${file%.*}"
mv "$file" "${filename}-4cc8.${extension}"
done
This loops through all the files, gets its name and extension and then moves it (that is, renames it) to the given name with an extra -4cc8 value before the extension.
Using rename:
rename 's/[.]([^.]+)$/-4cc8.$1/' *
s/[.]([^.]+)$/-4cc8.$1/ is a perl expression of the form s/PATTERN/REPLACEMENT/
which tells rename to do a global substition.
[.]([^.]+)$ is a regex pattern with the following meaning:
[.] match a literal period
( followed by a group
[ containing a character class
^. composed of anything except a literal period
]+ match 1-or-more characters from the character class
) end group
$ match the end of the string.
The replacement pattern, -4cc8.$1, tells rename to replace the matched text with a literal -4cc8. followed by text in the first group matched, i.e. whatever followed the literal period.
Related
I've got a ton of files as follows
audiofile_drums_1-ktpcwybsh5c.wav
soundsample_drums_2-fghlkjy57sa.wav
noise_snippet_guitar_5-mxjtgqta3o1.wav
louder_flute_9-mdlsiqpfj6c.wav
I want to remove everything between and including the "-" and the .wav file extension, to be left with
audiofile_drums_1.wav
soundsample_drums_2.wav
noise_snippet_guitar_5.wav
louder_flute_9.wav
I've tried to do delete everything following and including the character "-" using
rename 's/-.*//' *
Which gives me
audiofile_drums_1
soundsample_drums_2
noise_snippet_guitar_5
louder_flute_9
And for lack of finding an easy way to rename all the files again, adding .wav the extension, I am hoping there is a slicker way to do this in one nifty command in one stage instead of 2.
Any suggestions?
Thanks
You can use rename 's/-[^\.]*\.wav$/\.wav/' *
The first part -[^\.]*\.wav$ searchs for a - followed by n chars that are not . followed by .wav and the end of filename. The end of filename and .wav is not strictly needed but it helps avoid renaming files you don't want to rename.
The /\.wav/ preserves the extension.
Please not that rename is not a standard utility, and is part of perl, so rename may not be available on every linux system.
This works in my specific case, but should work for any file extension.
rename -n 's/-.*(?=\.wav$)//' *
The command looks for all characters after and inclusive of the - symbol in the filename, then, using a positive lookahead** (?=\.wav$) to search for the characters (the file extension in this case) at the end of the filename (denoted by $, and replaces them with no characters (removing them).
** NOTE: A positive look ahead is a zero width assertion.
It will affect the match but it will not be included
in the replacement. (The '.wav' part will not be
erased)
In this example (?=\.wav$) is the positive lookahead. The dollar sign $, as in regex, denotes at the end of the line, so perfect for a file extension.
I need to rename a lot of files with mmv. I know how to do that but I have a problem with wildcard indexes followed by numbers in the filename.
Basically I need to have an output filename which contains a wildcard followed by numbers.
mmv -n ``\*2\\.3_\*'' ``#11.6#2''
Here, as you can see, I'd like to have an output filename which contains the first wildcard followed by 1.6.
Unfortunately, this way I have #11.6 and the code is interpreted as if I want the 11th wildcard, which of course do not exist.
By reading the documentation you should have been able to find a solution.
Citation from man mmv, see https://ss64.com/bash/mmv.html
To strip any character (e.g. ’*’, ’?’, or ’#’) of its special meaning to mmv, as when the actual replacement name must contain the character ’#’, precede the special character with a ´\’ (and enclose the argument in quotes because of the shell). This also works to terminate a wildcard index when it has to be followed by a digit in the filename, e.g. "a#1\1".
I have a folder that was created automatically. The user unintentionally provided smart (curly) quotes as part of the name, and the process that sanitizes the inputs did not catch these. As a result, the folder name contains the smart quotes. For example:
this-is-my-folder’s-name-“Bob”
I'm now trying to rename/remove said folder on the command line, and none of the standard tricks for dealing with files/folders with special characters (enclosing in quotes, escaping the characters, trying to rename it by inode, etc.) are working. All result in:
mv: cannot move this-is-my-folder’s-name-“Bob” to this-is-my-folders-name-BOB: No such file or directory
Can anyone provide some advice as to how I can achieve this?
To get the name in a format you can copy-and-paste into your shell:
printf '%q\n' this*
...will print out the filename in a manner the shell will accept as valid input. This might look something like:
$'this-is-my-folder200\231s-name-200\234Bob200\235'
...which you can then use as an argument to mv:
mv $'this-is-my-folder200\231s-name-200\234Bob200\235' this-is-my-folders-name-BOB
Incidentally, if your operating system works the same way mine does (when running the test above), this would explain why using single-character globs such as ? for those characters didn't work: They're actually more than one byte long each!
You can use shell globbing token ? to match any single character, so matching the smart quotes using ? should do:
mv this-is-my-folder?s-name-?Bob? new_name
Here replacing the smart quotes with ? to match the file name.
There are several possibilities.
If an initial substring of the file name ending before the first quote is unique within the directory, then you can use filename completion to help you type an appropriate command. Type "mv" (without the quotes) and the unique initial substring, then press the TAB key to request filename completion. Bash will complete the filename with the correct characters, correctly escaped.
Use a graphical file browser. Then you can select the file to rename by clicking on it. (Details of how to proceed from there depend on the browser.) If you don't have a graphical terminal and can't get one, then you may be able to do the same with a text-mode browser such as Midnight Commander.
A simple glob built with the ? or * wildcard should be able to match the filename
Use a more complex glob to select the filename, and perhaps others with the same problem. Maybe something like *[^a-zA-Z0-9-]* would do. Use a pattern substitution to assign a new name. Something like this:
for f in *[^a-zA-Z0-9-]*; do
mv "$f" "${f//[^a-zA-Z0-9-]/}"
done
The substitution replaces all appearances of a characters that are not decimal digits, appercase or lowercase Latin letters, or hyphens with nothing (i.e. it strips them). Do take care before you use this, though, to make sure you're not going to make more changes than you intend to do.
I would like to rename many files in this format
abc.123.fits
abcd.1234.fits
efg.12.fits
to this format
abc.fits
abcd.fits
efg.fits
I tried the rename function, but since the part I'm trying to replace is not the same in all files, it did not work. I am using Linux.
for f in *; do mv "$f" "${f%%.*}.${f##*.}"; done`
${f%%.*} removes everything after the first period, including the period. ${f##*.} removes everything before the last period, including the period (i.e. it gets the file extension). Concatenating these two, with a period between them, gives you the desired result.
You can change the * to a more restrictive pattern such as *.fits if you don't want to rename all files in the current directory. The quotes around the parameters to mv are necessary if any filenames contain whitespace.
Many other variable substitution expressions are available in bash; see a reference such as TLDP's Bash Parameter Substitution for more information.
I have a list of files with extension .elf like this
file1.elf
file2.elf
file3.elf
I am trying to run them in shell with run command like run file1.elf >file1.log and get the result in a log file with file name with .log addition.
My list of file is very big. I am trying out a vim regular expression so it will match the file name eg file1 in file1.elf and use it to create name for the log file. I am trying out like this
s/\(\(\<\w\+\)\#<=\.elf\)/\1 >\2\.log/
Here i try to match a text which is proceeded by .elf and keep it in \1 , i expect the entrire file name to be in it and \2 i was hoping would just contain the file name minus extension. but this gives me
run file1 >file1.run i.e \1 dose not take the full file name, it has some how missed .elf extension. I can do \1\.elf to get proper result but i was wondering why the expression is not working as i expected?
You use \#<= in your match pattern. This is the positiv lookahead assertion. As per documentation (:help /\#<=1),
Matches with zero width if the preceding atom matches just before what follows
The important part is that it matches with zero width, this is what you are experiancing, the .elf (which follows) is matched but with zero widht, so that \1 does not contain the suffix .elf.
Instead, it would be easier to go with a
%s/\v(.*)\.elf$/run \1.elf > \1.log/
Here, I've used \v to turn on very magic (:help magic). With this turned on, you don't need al those backslashes when you use grouping parantheses.
Then there is (.*) to match and store the filename up until
\.elf$ which seems to be each files suffix.
In the substitution part, after the / I add the literal run followed by \1. \1 will be replaced by the stored filename (without .elf suffix).
The \#<= seems pointless and unneeded. Removing it gets you the desired behavior.