I have a list of files that a date has been added to the end.
ex: Chorus Left Octave (consolidated) (2020_10_14 20_27_18 UTC). The files will end with .wav or .mp3
I want to leave the (consolidated) but take out the date. I have come up with the regex and tested with regexr.com. It does format the text correctly there.
The regex is: /(\([0-9]+(.*)(?=.wav|.mp3))+/g
Now, I am trying to actually rename the files. In my terminal I have cd'ed into the folder with the files. Based on other answers here I have tried:
rename -n '/(\([0-9]+(.*)(?=.wav|.mp3))+/g' *.wav|*.mp3 - using rename installed with homebrew
sed '/(\([0-9]+(.*))+/g' *.wav|*.mp3
for f in *.wav|*.mp3; do mv "$f" "${f/(\([0-9]+(.*)(?=.wav|.mp3))+/g}” done
The first two do not throw any errors, but do not do any renames (I know that the -n after rename just prints out the files that will be changed, it doesn't actually change the files)
The last one starts a bash session.
I'd rather use the rename or sed, seems simpler to me. But, what am I doing wrong?.
In plain bash:
#!/bin/bash
pat='([0-9][0-9][0-9][0-9]_[0-9][0-9]_[0-9][0-9] [0-9][0-9]_[0-9][0-9]_[0-9][0-9] UTC)'
for f in *.mp3 *.wav; do echo mv "$f" "${f/$pat}"; done
Remove the echo preceding the mv after making sure it will work as intended. You may also consider adding the -i option to the mv in order to avoid clobbering an existing file unintentionally.
Related
I have a folder with many files where I only need some columns so I tried this to extract what I need:
mkdir ./raw_data/selection
doit() {
csvfix read_dsv -f 1,3,7 -s \; $1 > $1 | sed 's/raw_data/raw_data\/selection/'
}
export -f doit
Files_To_Parse=`ls ./raw_data/*csv`
parallel doit ::: $Files_To_Parse
This doesn't work.
But if I to this:
cd ./raw_data
doit() {
csvfix read_dsv -f 1,3,7 -s \; $1 > selection/$1
}
export -f doit
Files_To_Parse=`ls -1 *csv`
parallel doit ::: $Files_To_Parse
it works but I'd like to be able to run this from the top folder in this project (i.e to put this in a file named brief_csv.sh and call it from IDEs)
If you used Bash, you could:
for f in raw_data/*.csv
do
csvfix ... "$f" > raw_data/selection/"${f##*/}"
done
Also, instead of csvfix for extracting columns you could use cut:
$ cut -d \; -f 1,3,7 $f ...
I don't know the commands you are using, but this line:
csvfix read_dsv -f 1,3,7 -s \; $1 > $1 | sed ...
redirects the output in the same file you are reading; this can not work. In fact, you say that your modified code instead works. You could use temporary files to store intermediate results, don't be afraid to use many of them: debugging will be easier (you can see intermediate passages) and the system doesn't suffer. /tmp is a good place to put those intermediate files.
Use csvfix to do the first step, and redirect in /tmp/my-csvfix-intermediate; then use sed to read /tmp/my-csvfix-intermediate, and write in /tmp/my-grep-intermediate. After the last passage, you can take the last intermediate result and overwrite the original file, perhaps after having backed it up. You can move files everywhere you need, I don't see any problem in running your script from an IDE - just use as many passages as you need.
Avoid to parallelize when debugging, when the script will work, you can add parallelizing.
When two or more parallel processes will try to write in the same file (/tmp/my-...-intermediate), you will have one more problem. To overcome this you need to use different files for every process. The bash variable "$$" comes to help, just use file names like "/tmp/my-$$-blablabla", the $$ will be substituted with the PID of the process, and parallel processes can not have the same PID.
Hope it helps, regards.
How can I rename many files. Remove the digits at the beginning.
I have a Mac. All the files are in the same folder.
The pattern is:
1, 2 or 3 digits - any name.php
With Regular Expression, I think it would be:
\d*-(.*).php
For example:
1-marketing.php
2-3D.php
3-without.php
I want to remove the numbers and the dash at the beginning.
In the example it would be:
marketing.php
3D.php
without.php
What I have explored two ways:
Select the files > ctrl click > rename items. This is a fantastic method to change the name of files. But I think it cannot be used in this case. If I understand, it does not support Regex. Am I right?
Terminal. I am not very familiar with terminal. I tried mv 1-marketing.php marketing.php It works for 1 file, but how can I do the same for many? I am new with the terminal. If it can be done, please explain the basic.
Open the terminal app in Mac OS X and navigate to the folder containing the .php files
cd /my/path/to-php-files/
and run the below command on the command-line.
for file in *.php; do mv -v "$file" "${file#*-}"; done
The bash parameter expansion syntax ${file#*-} removes the characters before - from the beginning, so ideally 3-number-without.php becomes number-without.php
(or) use the perl rename utility not available by default in Mac OS, you can download and install it with homebrew 🍺:
brew install rename
and do
rename -n 's/^(\d+)-(.*)/$2/' *.php
The -n is just for a dry-run to see how the files are to be renamed, remove it as
rename 's/^(\d+)-(.*)/$2/' *.php
for the actual renaming.
I've been using the command line more frequently lately to increase my proficiency. I've created a .txt file containing URLs for libraries that I'd like to download. I batch-downloaded these files using
$ cat downloads.txt | xargs wget
When using the wget command I didn't specify a destination directory. I'd like to move each of the files that I've just downloaded into a directory called "vendor".
For the record, it has occurred to me that if I ran...
$ open .
...I could drag-and-drop these files into the desired directory. But in my opinion that would defeat the purpose of this exercise.
Now that I have the files in my cwd, I'd like to be able to target them and move them into the "vendor" directory.
As a side-question: Is there a useful way to print the most recently created files to STDOUT? Currently, I can grab the filenames from the URLs within downloads.txt pretty simply using the following pipeline and Perl script...
$ cat downloads.txt | perl -n -e 'if (/(?<=\/)([-.a-z]+)$/) { print $1 . "\n" }'
This will produce...
react.js
redux.js
react-dom.js
expect.js
...which is great as these are file that I intended on targeting. I'd like to transform each of these lines into a command within a pipeline that resembles this...
$ mv {./,./vendor/}<filename>
... where <filename> is "react.js" then "redux.js", and so forth.
I figure that I may be able to accomplish this using some combination of xargs, eval, and mv. This is where my bash skills drop-off.
Just to reiterate, I'm aware that the method in which I am approaching this problem is neither simple nor ideal. This is intentionally a convoluted exercise intended to stretch my bash knowledge.
Is there anyone who knows how I can use xargs, eval, and mv to accomplish this goal?
Thank you!
xargs -l -a downloads.txt basename | xargs -i mv {} ./vendor
How this works: The first instance of xargs reads the file names from downloads.txt and calls basename for each of these file names individually (alternatively, you could use basename -a). These basenames are then piped to another instance of xargs, which uses the arguments to call mv, replacing the string {} with the current argument.
mv $(basename -a $(<downloads.txt)) ./vendor
How this works: Since you want to move all the files into the same directory, you can use a single call to mv. The command substitution ("backticks") inserts the output of the command basename -a, which, in turn, reads its arguments from the file.
I have thousands of files named something like filename.gz.gz.gz.gz.gz.gz.gz.gz.gz.gz.gz
I am using the find command like this find . -name "*.gz*" to locate these files and either use -exec or pipe to xargs and have some magic command to clean this mess, so that I end up with filename.gz
Someone please help me come up with this magic command that would remove the unneeded instances of .gz. I had tried experimenting with sed 's/\.gz//' and sed 's/(\.gz)//' but they do not seem to work (or to be more honest, I am not very familiar with sed). I do not have to use sed by the way, any solution that would help solve this problem would be welcome :-)
one way with find and awk:
find $(pwd) -name '*.gz'|awk '{n=$0;sub(/(\.gz)+$/,".gz",n);print "mv",$0,n}'|sh
Note:
I assume there is no special chars (like spaces...) in your filename. If there were, you need quote the filename in mv command.
I added a $(pwd) to get the absolute path of found name.
you can remove the ending |sh to check generated mv ... .... cmd, if it is correct.
If everything looks good, add the |sh to execute the mv
see example here:
You may use
ls a.gz.gz.gz |sed -r 's/(\.gz)+/.gz/'
or without the regex flag
ls a.gz.gz.gz |sed 's/\(\.gz\)\+/.gz/'
ls *.gz | perl -ne '/((.*?.gz).*)/; print "mv $1 $2\n"'
It will print shell commands to rename your files, it won't execute those commands. It is safe. To execute it, you can save it to file and execute, or simply pipe to shell:
ls *.gz | ... | sh
sed is great for replacing text inside files.
You can do that with bash string substitution:
for file in *.gz.gz; do
mv "${file}" "${file%%.*}.gz"
done
This might work for you (GNU sed):
echo *.gz | sed -r 's/^([^.]*)(\.gz){2,}$/mv -v & \1\2/e'
find . -name "*.gz.gz" |
while read f; do echo mv "$f" "$(sed -r 's/(\.gz)+$/.gz/' <<<"$f")"; done
This only previews the renaming (mv) command; remove the echo to perform actual renaming.
Processes matching files in the current directory tree, as in the OP (and not just files located directly in the current directory).
Limits matching to files that end in at least 2 .gz extensions (so as not to needlessly process files that end in just one).
When determining the new name with sed, makes sure that substring .gz doesn't just match anywhere in the filename, but only as part of a contiguous sequence of .gz extensions at the end of the filename.
Handles filenames with special chars. such as embedded spaces correctly (with the exception of filenames with embedded newlines.)
Using bash string substitution:
for f in *.gz.gz; do
mv "$f" "${f%%.gz.gz*}.gz"
done
This is a slight modification of jaypal's nice answer (which would fail if any of your files had a period as part of its name, such as foo.c.gz.gz). (Mine is not perfect, either) Note the use of double-quotes, which protects against filenames with "bad" characters, such as spaces or stars.
If you wish to use find to process an entire directory tree, the variant is:
find . -name \*.gz.gz | \
while read f; do
mv "$f" "${f%%.gz.gz*}.gz"
done
And if you are fussy and need to handle filenames with embedded newlines, change the while read to while IFS= read -r -d $'\0', and add a -print0 to find; see How do I use a for-each loop to iterate over file paths output by the find utility in the shell / Bash?.
But is this renaming a good idea? How was your filename.gz.gz created? gzip has guards against accidentally doing so. If you circumvent these via something like gzip -c $1 > $1.gz, buried in some script, then renaming these files will give you grief.
Another way with rename:
find . -iname '*.gz.gz' -exec rename -n 's/(\.\w+)\1+$/$1/' {} +
When happy with the results remove -n (dry-run) option.
I've got a file structure that looks like:
A/
2098765.1ext
2098765.2ext
2098765.3ext
2098765.4ext
12345.1ext
12345.2ext
12345.3ext
12345.4ext
B/
2056789.1ext
2056789.2ext
2056789.3ext
2056789.4ext
54321.1ext
54321.2ext
54321.3ext
54321.4ext
I need to rename all the files that begin with 20 to start with 10; i.e., I need to rename B/2022222.1ext to B/1022222.1ext
I've seen many of the other questions regarding renaming multiple files, but couldn't seem to make it work for my case. Just to see if I can figure out what I'm doing before I actually try to do the copy/renaming I've done:
for file in "*/20?????.*"; do
echo "{$file/20/10}";
done
but all I get is
{*/20?????.*/20/10}
Can someone show me how to do this?
You just have a little bit of incorrect syntax is all:
for file in */20?????.*; do mv $file ${file/20/10}; done
Remove quotes from the argument to in. Otherwise, the filename expansion does not occur.
The $ in the substitution should go before the bracket
Here is a solution which use the find command:
find . -name '20*' | while read oldname; do echo mv "$oldname" "${oldname/20/10}"; done
This command does not actually do your bidding, it only prints out what should be done. Review the output and if you are happy, remove the echo command and run it for real.
Just wanna add to Explosion Pill's answer.
On OS X though, you must say
mv "${file}" "${file_expression}"
Or the mv command does not recognize it.
Brace expansions like :
{*/20?????.*/20/10}
can't be surrounded by quotes.
Instead, try doing (with Perl rename) :
rename 's/^10/^20/' */*.ext
You can do this using the Perl tool rename from the shell prompt. (There are other tools with the same name which may or may not be able to do this, so be careful.)
If you want to do a dry run to make sure you don't clobber any files, add the -n switch to the command.
note
If you run the following command (linux)
$ file $(readlink -f $(type -p rename))
and you have a result like
.../rename: Perl script, ASCII text executable
then this seems to be the right tool =)
This seems to be the default rename command on Ubuntu.
To make it the default on Debian and derivative like Ubuntu :
sudo update-alternatives --set rename /path/to/rename
The glob behavior of * is suppressed in double quotes. Try:
for file in */20?????.*; do
echo "${file/20/10}";
done