So I want to copy some files from one directory to another.
Essentially, I want to capture the directory path to a variable, say, "pathname" and use
"cp-r $pathname ." to copy file1 and file2 to to a new folder in which I made using mkdir and have cd'ed into (hence the "." as the second command line argument).
source_to_copy_from:
home/folder1/folder2/file1
home/folder1/folder2/file2
destination_to_copy_to:
home/newfolder1/newfolder2/file1
home/newfolder1/newfolder2/file2
I did:
pathname=$"$(pwd)"
//code to make the newfolder1 here
cp -r $"$pathname" .
But there appears to be nothing in the new pathname that was supposed to be copied in.
Also am using Mac bash
also quite beginner to bash
Related
I have a text file called "list.txt" that contain all the directories of the files that need to be copied to a new folder (dir_newfolder). I wrote the code like below:
for file in $(cat list.txt); do cp ${file} dir_newfolder; done
I got list of errors: cp:"file_name":No such file or directory. The file_names are the lines pulled out from the "list.txt". But when I copy each file_names from the error message and use cp to copy to the new folder. There is no error.
I am using mac os terminal.
Thanks in advance.
Copy a file or folder locally
In the Terminal app on your Mac, use the cp command to make a copy of a file.
For example, to copy a folder named Expenses in your Documents folder to another volume named Data:
% cp -R ~/Documents/Expenses /Volumes/Data/Expenses
The -R flag causes cp to copy the folder and its contents. Note that the folder name does not end with a slash, which would change how cp copies the folder.
in your case:
make sure you are providing correct path list.txt and the correct path for destiny folder, also i mentioned how to access file variable in double quotes , try this code it's working for me
for file in $(cat ~/Documents/list.txt); do cp "$file" ~/dir_newfolder; done
I have two folder xyz and Bin. I want to move xyz directory and it's sub directory as well to Bin folder.
I want output like this:
Bin/xyz/q.html
/images/0.gif
How can do move files and subdirectory and I will get that type of output.
Move (mv) can be thought of as rename. If mv is used to move a file from one drive to another, then a copy will be made and the original will be deleted.
If the file is a directory (folder), then all of the contents of that directory are also moved.
$ mv -v xyz Bin
moves the file called xyz, no matter what kind of file it is, even if it is a directory, into the existing directory called Bin. If Bin does not exist, then the move command will rename xyz to Bin.
If you only want to move a few files, for example, Bin/1.html is to be moved to Bin/xyz/1.html, then file being moved is a file called 1.html (not a directory called bin or xyz).
$ mv -v Bin/1.html Bin/xyz/1.html
will perform the move if the Bin/xyz exists and is a directory. If directory Bin/xyz does not exist, then this mv command will error because mv does not create parent destination directories.
$ mkdir -v Bin/xyz
will create an empty directory called xyz in an existing directory called Bin. Then files can be moved into the existing Bin/xyz directory.
The -v (--verbose) tells mv and mkdir to report what happened. The output that they produce given a success is what -v does. The most common way to list the content of a directory is ls, but the indentation that you desire will require some programming.
$ man mv
$ man mkdir
will show all the features of the mv an mkdir commands.
The title not might be the best, I apologize, I'm rather new to scripting.
I'm trying to copy 2 files from each directory and place these compounds in a separate directory that only shares directory names.
for clarity;
/path/to/directory/all/variable_directories/
Inside this directory will be multiple files, I need 2 files which will have the same name in every individual variable directory.
I am trying to copy these 2 files from each individual variable directory and put them in variable directories based on the basename of /variable_directory/
the copy destination will be;
/path/to/magical/shit/subset/set_with_variable_name/variable_directories/
Only some of the destination directories are located in each /set_with_variable_name/
The script will need to be able to go through each /set_with_variable_name/ until it finds the directory that shares the basename of the directory that these files are originally being cp'd from
There's about 100 directories
to cp from and to and about 200 files total that need to be copied and sorted appropriately.
I can get it to cp ALL the files to the SAME directory using;
#! /usr/bin/env bash
for i in */;
do cd $i;
cp filename /path/to/destination/;
cp other_filename /path/to/destination/;
cd ..;
done;
It's the sorting the files to the correct destinations that I am completely lost at.
I appreciate any help, I'm a novice to this type of scripting
Looks like you need something like that.
# looping through all the variable directories
for var_dir in <path1>/all/*; do
# creating the destination directory if not exists
mkdir -p "<path2>/set_with_variable_name/$(basename ${var_dir})"
# copying the first file
cp "${var_dir}/filename1" "<path2>/set_with_variable_name/$(basename ${var_dir})/filename1"
# and the second
cp "${var_dir}/filename2" "<path2>/set_with_variable_name/$(basename ${var_dir})/filename2"
done
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"/* "${filepath}"/
done
I have a lot of files named the same, with a directory structure (simplified) like this:
../foo1/bar1/dir/file_1.ps
../foo1/bar2/dir/file_1.ps
../foo2/bar1/dir/file_1.ps
.... and many more
As it is extremely inefficient to view all of those ps files by going to the
respective directory, I'd like to copy all of them into another directory, but include
the name of the first two directories (which are those relevant to my purpose) in the
file name.
I have previously tried like this, but I cannot get which file is from where, as they
are all named consecutively:
#!/bin/bash -xv
cp -v --backup=numbered {} */*/dir/file* ../plots/;
Where ../plots is the folder where I copy them. However, they are now of the form file.ps.~x~ (x is a number) so I get rid of the ".ps.~*~" and leave only the ps extension with:
rename 's/\.ps.~*~//g' *;
rename 's/\~/.ps/g' *;
Then, as the ps files have hundreds of points sometimes and take a long time to open, I just transform them into jpg.
for file in * ; do convert -density 150 -quality 70 "$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