I'm relatively new to bash and I have tried multiples solutions that I could find here but none of them seem to be working in my case. It's pretty simple, I have a folder that looks like this:
- images/
- 0_image_1.jpg
- 0_image_2.jpg
- 0_image_3.jpg
- 1_image_1.jpg
- 1_image_2.jpg
- 1_image_3.jpg
and I would like to move these jpg files into subfolders based on the prefix number like so:
- images_0/
- 0_image_1.jpg
- 0_image_2.jpg
- 0_image_3.jpg
- images_1/
- 1_image_1.jpg
- 1_image_2.jpg
- 1_image_3.jpg
Is there a bash command that could do that in a simple way ?
Thank you
for src in *_*.jpg; do
dest=images_${src%%_*}/
echo mkdir -p "$dest"
echo mv -- "$src" "$dest"
done
Remove both echos if the output looks good.
I would do this with rename a.k.a. Perl rename. It is extremely powerful and performant. Here's a command for your use case:
rename --dry-run -p '$_="images_" . substr($_,0,1) . "/" . $_' ?_*jpg
Let's dissect that. At the right end, we specify we only want to work on files that start with a single character/digit before an underscore so we don't do damage trying to apply the command to files it wasn't meant for. Then --dry-run means it doesn't actually do anything, it just shows you what it would do - this is a very useful feature. Then -p which handily means "create any necessary directories for me as you go". Then the meat of the command. It passes you the current filename in a variable called $_ and we then need to create a new variable called $_ to say what we want the file to be called. In this case we just want the word images_ followed by the first digit of the existing filename and then a slash and the original name. Simples!
Sample Output
'0_image_1.jpg' would be renamed to 'images_0/0_image_1.jpg'
'0_image_2.jpg' would be renamed to 'images_0/0_image_2.jpg'
'1_image_3.jpg' would be renamed to 'images_1/1_image_3.jpg'
Remove the --dry-run and run again for real, if the output looks good.
Using rename has several benefits:
that it will warn and avoid any conflicts if two files rename to the same thing,
that it can rename across directories, creating any necessary intermediate directories on the way,
that you can do a dry run first to test it,
that you can use arbitrarily complex Perl code to specify the new name.
Note: On macOS, you can install rename using homebrew:
brew install rename
Note: On some Ones, rename is referred to as prename for Perl rename.
Related
I have a directory containing files following the following naming convention:
Label_0000_AA.gz
Label_0001_BB.gz
Label_0002_CC.gz
...
All I want to do is to rename these files so that the _#### number pattern is removed, resulting in:
Label_AA.gz
Label_BB.gz
Label_CC.gz
...
but only up to a certain number. E.g.: I may have 10000 files but might only want to remove the pattern in the first 3000. Would this be possible using something like bash?
If you don't have prename or rename -
(assuming the names are consistent)
for f in Label_[0-9][0-9][0-9][0-9]_[A-Z][A-Z].gz
do mv "$f" "${f//_[0-9][0-9][0-9][0-9]/}"
done
To just do a certain range -
for n in {0000..2999}
do for f in Label_${n}_??.gz
do mv $f ${f//_$n/}
done
done
You're sure there are not collisions?
If you can name the pattern you want to change/remove in a regex you can use the command prename:
prename 's/_[0-3][[:digit:]]{3}_/_/g' Label_*.gz
This regex would only remove numbers 0000-3999.
Using the flag -n does a "dry-run" and shows what it would do.
Edit: Thanks #KamilCuk to remind me about two renames. I made it clear and changed the name to prename.
How can I pass each one of my repository files and to do something with them?
For instance, I want to make a script:
#!/bin/bash
cd /myself
#for-loop that will select one by one all the files in /myself
#for each X file I will do this:
tar -cvfz X.tar.gz /myself2
So a for loop in bash is similar to python's model (or maybe the other way around?).
The model goes "for instance in list":
for some_instance in "${MY_ARRAY[#]}"; do
echo "doing something with $some_instance"
done
To get a list of files in a directory, the quick and dirty way is to parse the output of ls and slurp it into an array, a-la array=($(ls))
To quick explain what's going on here to the best of my knowledge, assigning a variable to a space-delimited string surrounded with parens splits the string and turns it into a list.
Downside of parsing ls is that it doesn't take into account files with spaces in their names. For that, I'll leave you with a link to turning a directory's contents into an array, the same place I lovingly :) ripped off the original array=($(ls -d */)) command.
you can use while loop, as it will take care of whole lines that include spaces as well:
#!/bin/bash
cd /myself
ls|while read f
do
tar -cvfz "$f.tar.gz" "$f"
done
you can try this way also.
for i in $(ls /myself/*)
do
tar -cvfz $f.tar.gz /myfile2
done
I need to copy all files from
/dirA/[NAME].20151231.txt
to
/dirB/20151231.[NAME].txt
and
/dirC/20151231/[NAME].txt
i.e. I need to copy the files, but change the name.
You can assume that I know the "date" string before hand, so we can assume 20151231 is a supplied argument.
if I have a list of names, I can do something like
for n in $names; do; cp /dirA/$n.$date.txt /dirB/$date.$n.txt; done;
But what if I dont have a list of names? I am looking for an elegant solution as extracting them from dirA sounds a bit cumbersome.
Thanks!
A reasonably reliable way of processing this material is:
date=20151231
cd /dirA || exit 1
mkdir -p "/dirC/$date" || exit 1
for file in *."$date.txt"
do
name="${file%.$date.txt}"
cp "$file" "/dirB/$date.$name.txt"
cp "$file" "/dirC/$date/$name.txt"
done
The cd operation is checked; if it fails, there is no point in continuing. Likewise, the mkdir -p operation ensures that the dated directory under /dirC exists or exits. The relevant error messages were already generated by cd and mkdir.
Using the shell globbing to generate the file names is best; it avoids issues with 'what happens if the file name contains spaces (or newlines, or other unexpected characters)'.
The assignment extracts the '[NAME]' portion of the file name. This is then used to copy the file from /dirA to the relevant locations under /dirB and /dirC. It would be feasible to check that /dirB and /dirC also exist if you thought that was necessary.
Maybe I am just awful at asking questions. What I was looking for was a "sed for file names". And I found the answer -- that's rename.
I would like to rearrange and rename files.
I have this tree structure of files :
ada/rda/0.05/alpha1_freeSurface.md
ada/rda/0.05/p_freeSurface.md
ada/rda/0.05/U_freeSurface.md
ada/rda/0.1/alpha1_freeSurface.md
ada/rda/0.1/p_freeSurface.md
ada/rda/0.1/U_freeSurface.md
I want that files will be renamed and rearranged like this structure below:
ada/rda/ada-0.05-alpha1.md
ada/rda/ada-0.05-p.md
ada/rda/ada-0.05-U.md
ada/rda/ada-0.1-alpha1.md
ada/rda/ada-0.1-p.md
ada/rda/ada-0.1-U.md
Using the perl rename (sometimes called prename) utility:
rename 's|ada/rda/([^/]*)/([^_]*).*|ada/rda/ada-$1-$2.md|' ada/rda/*/*
(Note: by default, some distributions install a rename command from the util-linux package. This command is incompatible. If you have such a distribution, see if the perl version is available under the name prename.)
How it works
rename takes a perl commands as an argument. Here the argument consists of a single substitute command. The new name for the file is found from applying the substitute command to the old name. This allows us not only to give the file a new name but also a new directory as above.
In more detail, the substitute command looks like s|old|new|. In our case, old is ada/rda/([^/]*)/([^_]*).*. This captures the number in group 1 and the beginning of the filename (the part before the first _) in group 2. The new part is ada/rda/ada-$1-$2.md. This creates the new file name using the two captured groups.
You can use basename and dirname functions to reconstruct the new filename:
get_new_name()
{
oldname=$1
prefix=$(basename $oldname _freeSurface.md)
dname=$(dirname $oldname)
basedir=$(dirname $dname)
dname=$(basename $dname)
echo "$basedir/ada-$dname-$prefix.md"
}
e.g. get_new_name("ada/rda/0.05/alpha1_freeSurface.md") will show ada/rda/ada-0.05-alpha1.md in console.
Then, you can loop through all your files and use mv command to rename the files.
I need to rename hundreds of files in Linux to change the unique identifier of each from the command line. For sake of examples, I have a file containing:
old_name1 new_name1
old_name2 new_name2
and need to change the names from new to old IDs. The file names contain the IDs, but have extra characters as well. My plan is therefore to end up with:
abcd_old_name1_1234.txt ==> abcd_new_name1_1234.txt
abcd_old_name2_1234.txt ==> abcd_new_name2_1234.txt
Use of rename is obviously fairly helpful here, but I am struggling to work out how to iterate through the file of the desired name changes and pass this as input into rename?
Edit: To clarify, I am looking to make hundreds of different rename commands, the different changes that need to be made are listed in a text file.
Apologies if this is already answered, I've has a good hunt, but can't find a similar case.
rename 's/^(abcd_)old_name(\d+_1234\.txt)$/$1new_name$2/' *.txt
Should work, depending on whether you have that package installed. Also have a look at qmv (rename-utils)
If you want more options, use e.g.
shopt -s globstart
rename 's/^(abcd_)old_name(\d+_1234\.txt)$/$1new_name$2/' folder/**/*.txt
(finds all txt files in subdirectories of folder), or
find folder -type f -iname '*.txt' -exec rename 's/^(abcd_)old_name(\d+_1234\.txt)$/$1new_name$2/' {} \+
To do then same using GNU find
while read -r old_name new_name; do
rename "s/$old_name/$new_name/" *$old_name*.txt
done < file_with_names
In this way, you read the IDs from file_with_names and rename the files replacing $old_name with $new_name leaving the rest of the filename untouched.
I was about to write a php function to do this for myself, but I came upon a faster method:
ls and copy & paste the directory contents into excel from the terminal window. Perhaps you may need to use on online line break removal or addition tool. Assume that your file names are in column A In excel, use the following formula in another column:
="mv "&A1&" prefix"&A1&"suffix"
or
="mv "&A1&" "&substitute(A1,"jpeg","jpg")&"suffix"
or
="mv olddirectory/"&A1&" newdirectory/"&A1
back in Linux, create a new file with
nano rename.txt and paste in the values from excel. They should look something like this:
mv oldname1.jpg newname1.jpg
mv oldname1.jpg newname2.jpg
then close out of nano and run the following command:
bash rename.txt. Bash just runs every line in the file as if you had typed it.
and you are done! This method gives verbose output on errors, which is handy.