I have a lot of files named the same, with a directory structure (simplified) like this:
../foo1/bar1/dir/file_1.ps
../foo1/bar2/dir/file_1.ps
../foo2/bar1/dir/file_1.ps
.... and many more
As it is extremely inefficient to view all of those ps files by going to the
respective directory, I'd like to copy all of them into another directory, but include
the name of the first two directories (which are those relevant to my purpose) in the
file name.
I have previously tried like this, but I cannot get which file is from where, as they
are all named consecutively:
#!/bin/bash -xv
cp -v --backup=numbered {} */*/dir/file* ../plots/;
Where ../plots is the folder where I copy them. However, they are now of the form file.ps.~x~ (x is a number) so I get rid of the ".ps.~*~" and leave only the ps extension with:
rename 's/\.ps.~*~//g' *;
rename 's/\~/.ps/g' *;
Then, as the ps files have hundreds of points sometimes and take a long time to open, I just transform them into jpg.
for file in * ; do convert -density 150 -quality 70 "$file" "${file/.ps/}".jpg; done;
This is not really a working bash script as I have to change the directory manually.
I guess the best way to do it is to copy the files form the beginning with the names
of the first two directories incorporated in the copied filename.
How can I do this last thing?
If you just have two levels of directories, you can use
for file in */*/*.ps
do
ln "$file" "${file//\//_}"
done
This goes over each ps file, and hard links them to the current directory with the /s replaced by _. Use cp instead of ln if you intend to edit the files but don't want to update the originals.
For arbitrary directory levels, you can use the bash specific
shopt -s globstar
for file in **/*.ps
do
ln "$file" "${file//\//_}"
done
But are you sure you need to copy them all to one directory? You might be able to open them all with yourreader */*/*.ps, which depending on your reader may let browse through them one by one while still seeing the full path.
You should run a find command and print the names first like
find . -name "file_1.ps" -print
Then iterate over each of them and do a string replacement of / to '-' or any other character like
${filename/\//-}
The general syntax is ${string/substring/replacement}. Then you can copy it to the required directory. The complete script can be written as follows. Haven't tested it (not on linux at the moment), so you might need to tweak the code if you get any syntax error ;)
for filename in `find . -name "file_1.ps" -print`
do
newFileName=${filename/\//-}
cp $filename YourNewDirectory/$newFileName
done
You will need to place the script in the same root directory or change the find command to look for the particular directory if you are placing the above script in some other directory.
References
string manipulation in bash
find man page
Related
I don't have much experience with the command line, but essentially I have a list of files in a single folder as follows:
file1_a_1
file1_a_2
file2_b_1
file2_b_2
file3_c_1
file3_c_2
And I also have a text file with the files I want. However, this list does not have the full file path, instead, it looks like this:
file1_a file3_c
because I want to move all files that start with 30 or so specific codes (i.e. everything that starts with file1_a and file1_c for all the files that start with this).
I have tried:
cp file1_a* file3_c* 'dir/dest'
but this does not work. I have also tried the find command. I think I have to use a loop to do this but I cannot find any help on looping through files with a wildcard on the end.
Thanks in advance! I am working on a linux machine in bash.
you can use the xargs command with find command and a pipe
find / -name xxxxx | xargs cp /..
I am trying to loop through all of the files in a directory and
move them to a workspace (I need to do this to do this because the workspace doesn't have much storage).
Run a program which produces an output directory that contains all the files I will want to work with in the future
Delete the original file from the workspace (to save space in the workspace)and
Move the output directory out of the workspace and back to the storage space
I am able to do this for each file singly (i.e. each line works if I actually use the name of the files), but I can't get the for loop to work. I am quite new to this, so I probably did something simple wrong.
Can anyone see where I am going wrong?
for i in path_to_files; do
#copy to home directory (from scratch)
cp $i .
#Run IDBA
idba_ud -l $i -o '$i'_out
#remove file from work directory (limited space)
rm $i
#copy out directory back to scratch
cp -r '$i'_out path_to_files
done
I keep getting an error that says
syntax error near unexpected token `cp'.
I have also tried replacing cp with copy and i/$i with file/$file with no luck.
If this is indeed POSIX compatible shell (your code looks suspiciously like that, but you haven't specified the actually used shell), then:
You should always quote filenames, in case it contains spaces or other weird characters:
But you should not use single-quotes, as this will prevent shell from expanding your variables.
when appending text to substituted variables, use ${} notation (e.g. if $i expands to "murgel", then ${i}foo will expand to "murgelfoo", whereas $ifoo will expand to "" (an empty string) if there is no variable ifoo)
Thus try:
filepath=/path/to/files
for i in "${filepath}"/*; do
#copy to home directory (from scratch)
cp "${i}" .
#Run IDBA
idba_ud -l "${i}" -o "${i}_out"
#remove file from work directory (limited space)
rm $i
#copy out directory back to scratch
cp "${i}_out"/* "${filepath}"/
done
I have following directory structure :
/home/dir1/abc.jpg
/home/dir1/abc.pdf
/home/dir1/dir2/abc.jpg
/home/dir1/dir2/abc1.jpg
/home/dir1/dir2/dir3/abc.jpg
and I want to copy jpg files from them to a new folder which will have same directory structure, for eg.:
/home/newdir1/abc.jpg
/home/newdir1/dir2/abc.jpg
/home/newdir1/dir2/abc1.jpg
/home/newdir1/dir2/dir3/abc.jpg
How to achieve it using rsync or any other software ?
Please help, Many Thanks !!
From the looks of what you've included in your question, there are a couple of things you might try.
You've specified that you want to "move" files. That means you either use the mv command, or use rsync's --remove-source-files option. For example:
mv /source1/* /source2/* /path/to/targetdir/
or
rsync -a /source1/ /source2/ /path/to/targetdir/
You've no doubt already read the part of rsync's man page that explains the difference between source dirs with and without their trailing slash. If not, read up, because it's important.
If your "thousands of source files [with] similar names" need to be matched from within your source directories, leaving some other files behind, you need to determine whether your "similar names" can be differentiated using pathname expansion or if you should use a regular expression. If the former, then adding the pathname expansion to your sources with either mv or rsync should be sufficient. If you need to use a regex, then find may be a better option:
find /source1/ /source2/ -regex ".*/file[A-F][0-9][0-9].txt" -exec mv "{}" /targetdir/ \;
If these don't solve the problem, then you'll need to supply more detail in your question.
I would try a little shell script like this:
#!/bin/sh
cd /home/dir1
JPEGS=`find . -name "*.jpg"`
tar cf - $JPEGS | (cd /home/newdir1 ; tar xf -)
This first gets the list of all your jpg files with their relative paths, then writes a tar file of them to a pipe into a subshell which changes to the new directory, and then extracts the tar from its stdin.
I am trying to write a script that finds all my files that are .jpg, and copies them do a new directory. It currently looks like this:
find ~/Pictures -iname \*.jpg -exec cp {} ...newDirectory \;
The problem is that some of my older files have the same name as newer files, when the IMG_#### reset back to 0001 and started counting again.
Is there a way to find the .jpgs and copy without overwriting the files? Ideally giving them a new name in the process.
EDIT
I ended up learning about rsync, which in its own way does exactly what I was looking for. Thanks for the help!
Use -n parameter for cp, that means: do not overwrite an existing file.
To prevent identical names, you could just name all of them unique.
Example:
$ touch screenshot.jpg
$ cp screenshot.jpg screenshot-$(date "+%s").jpg
So basically, mass rename the new files you want to copy to the same name+date.
That will make them different from what's already there, sice the older ones are unnamed or (if you repeat this later) will have different dates.
Is anybody able to point me in the right direction for writing a batch script for a UNIX shell to move files into a zip one at at time and then delete the original.
I cant use the standard zip function because i don't have enough space to fit the zip being created.
So any suggestions please
Try this:
zip -r -m source.zip *
Not a great solution but simple, i ended up finding a python script that recursively zips a folder and just added a line to delete the file after it is added to the zip
You can achieve this using find as
find . -type f -print0 | xargs -0 -n1 zip -m archive
This will move every file into the zip preserving the directory structure. You are then left with empty directories that you can easily remove. Moreover using find gives you a lot of freedom on what files you want to compress.
I use :
zip --move destination.zip src_file1 src_file2
Here the detail of "--move" option from the man pages
--move
Move the specified files into the zip archive; actually, this
deletes the target directories/files after making the specified zip
archive. If a directory becomes empty after removal of the files, the
directory is also removed. No deletions are done until zip has
created the archive without error. This is useful for conserving disk
space, but is potentially dangerous so it is recommended to use it in
combination with -T to test the archive before removing all input
files.