I want to copy a complete directory content from /home/private_html/userx/ into the /home/private_html/usery/, the problem is that the directory userx contains few symlinks, and when using the cp it just skip them (skip occurs, if symlinks directs into a file, in case if it points into the directory, it just copy WHOLE directory instead...).
The command I was using looks following:
# cp -iprv /home/private_html/userx/ /home/private_html/usery/
Has anyone a solution to copy the directory "just as it is" into other place?
On FreeBSD, cp doesn't have an -r option. It does have -R, which should do what you want:
-R If source_file designates a directory, cp copies the directory and
the entire subtree connected at that point. If the source_file
ends in a /, the contents of the directory are copied rather than
the directory itself. This option also causes symbolic links to be
copied, rather than indirected through, and for cp to create spe‐
cial files rather than copying them as normal files. Created
directories have the same mode as the corresponding source direc‐
tory, unmodified by the process' umask.
Roland is right about the -R flag. You could also use a pair of tar-processes, which would make your command a little bit more system-independent:
tar -C /home/private_html/userx/ -cpf - . | tar -C /home/private_html/usery/ -epf -
Related
Using any of the standard Robot libraries, is it possible to recursively copy the contents of a directory to an existing destination directory?
Basically, I'm looking for the equivalent of the following shell command: cp -r foo/. bar (note the trailing dot)
I tried Copy Directory but this creates a directory foo inside bar (as documented) and it doesn't stop doing that even when supplying the trailing dot. Copy Files chokes when it encounters a directory.
Is there anything I overlooked? Or do I need to just call cp -r myself?
As I only need this to work on Linux, I ended up implementing a custom keyword calling cp -r. If this is ever needed cross-platform, then I'll follow the suggestions to directly implement it in Python.
Copy Directory Contents
[Documentation] Recursively copies the contents of the source directory into the destination.
[Arguments] ${source} ${destination}
Directory Should Exist ${source}
Directory Should Exist ${destination}
${result} = Run Process cp -r ${source}/. ${destination}/
Should Be Equal As Integers ${result.rc} 0
I am on Mac OS.
I have a directory with round about 3000 files and several subdirectories (wordpress installation)
Now I have to find all the files in a similar directory (have to separate master and child installation) that are additional files and have to copy them away into another directory.
I use this command:
$ diff -rq dt-the7 dt-the7-master-from-Yana|grep 'Only in dt-the7'|awk {' print $3 $4 '}|sed 's/:/\//g'|xargs -J {} rsync -av {} neu/
but somehow a certain file 3d.png and a list of other that should be in a subdir of the destination dir are copied into the root dir of the destination.
Any idea why that might be?
It makes no difference whether I use cp, rsync or ditto
You need the -R relative option on your rsync command.
Without this rsync just copies the item referenced rather than the path referenced, so items at the root level are copied as you expected but items in sub-directories are also copied to the root, which is not what you wanted.
With the option rsync takes account of the relative path and recreates it at the destination.
An example with another command might help, consider:
cp A/B.txt C/
that will copy B.txt into C, it does not create a folder A in C which in turn contains the file B.txt. rsync without -R behaves like that cp command, with -R it creates the A directory in C.
HTH
I need to copy a directory and its content to a NEW directory.
Using: cp -r dir1/dir2 dir1/dir3
where dir3 is going to be created anew, it just copies the file(s) I have within dir2... It's like it copies dir2 naming it dir3.
Is there any way to copy dir2 into dir3?
I need to do it with a single command.
It's not really possible to do it in 1 command with cp. What you want is for dir3 to already exist when you do your copy. I'm not sure if there's a real reason why you have to do it in 1 command or not. You can certainly do it in one line.
mkdir dir/dir3;cp -r dir1/dir2 dir1/dir3
I tried to create a symbolic link to a non existing file
ln -s non_existing_file.txt $HOME/dir1/dir2/my_symbolic_link
Then I tried to write something in the non existing file using the symbolic link
vi $HOME/dir1/dir2/my_symbolic_link
Now after saving and exiting non_existing_file.txt is created under dir2.
Can someone explain why?
ln -s target linkpath
creates a symlink at linkpath which holds the name target. Operations on the symlink interpret the name target relative to the directory where the symlink resides, not the present working directory.
So, if you have a symlink holding, say, ../usr in /tmp/link-to-usr, then ls /tmp/link-to-usr will list the contents of /usr (which is /tmp/../usr) regardless of where the ls command is executed.
for example, my directory like this:
lib
|
+--foo.so
+--bar.so -> bar.so.1.0.0.0
+--bar.so.1.0.0.0
Are these both ways always same?
The reason I ask this question is that I found unix will copy to real so file when I
cp -r lib /path/to/
new directory like this:
/path/to/lib
|
+--foo.so
+--bar.so
+--bar.so.1.0.0.0
The difference between so and ls -s is the difference between a file and a symbolic link. Symbolic links are like aliases to other files and operations on them result in changes in the linked files. When you do cp, it copies the linked file to the target directory with the link name as the file name, i.e., it reads the linked file when it opens the symbolic link to copy it. So lose the link and instead get a copy of the linked file. If you use -P option of the cp command you can preserve the symbolic link information.
cp -P lib /path/to/