I want to move a couple of files from point a to point b
but I have to manually specify
mv /full/path/from/a /full/path/to/b
but some times there are 20 files which I have to move manually. Instead of /full/path/form/a, can't I just enter the a function which returns all the files which I want to move in my case;
/full/path/to/b is a directory, it's the target directory which all the files with extenstions mp3, exe and mp4 must go to:
mv ls *.{mp3,exe,mp4} /full/path/to/b
If I have to move a couple of files and I don't want to do it one by one, how can I optimize the problem?
The command mv ls *.{mp3,exe,mp4} /full/path/to/b in your question is not correct.
As pointed out in comments by #janos, the correct command is
mv *.{mp3,exe,mp4} /full/path/to/b
mv can complain about missing file if the file is really missing and/or the path is not accessible or is not valid.
As i can understand by your question description, if you go manually to the source path you can move the file to the desired directory.
Thus it seems that path is valid, and file exists.
In order mv to keeps complaining about *.mp3 not found (having a valid path and file) the only reason that pops up in my head is the Bash Pathname Expansion feature (enabled by default in my Debian).
Maybe for some reason this pathname expansion bash feature is disabled in your machine.
Try to enable this feature using command bellow and provide the correct command to mv and you should be fine.
$ set +f
PS: Check man bash about pathname expansion.
Related
When I copy something, I always forget the -R, then I have go all the way back to add it right after cp.
I want to add this to bash config files.
alias cp="cp -R"
I have not seen anything bad happen. Is it safe to do this?
The only thing I can think of that would cause unexpected behavior with the -R flag is that it doesn't work with wildcards.
What I mean is... for example you want to copy all mp3 Files in a directory and every subdirectory with: cp -R /path/*.mp3. Although -R is given it will not copy mp3s in the subdirectories of path - if there are any.
I wouldn't use aliases for changing the behaviour of normal commands. When you are in a different shell / another computer, the alias will be missing. When a friend wants to help you, he will not know what you did.
Once I had an alias rm="rm -i" and I performed rm *, while I just had changed shell with a su.
And sometimes you want to use cp without the -R option, will you remember to use /bin/cp in these cases (copy all files in the current dir to another location, but do not cp the subdirs)?
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 have a lastaction script I'm trying to run in my "log" folder, as I want to move all files and folders in the log folder inside the log/archive folder. So I simply added
mv log/* log/archive/2014
Obviously enough, I get an error saying archive folder cannot be moved to a subdirectory of itself, so I tried adding the extra parameter to the move command to move everything except the archive folder.
mv !(archive) log/* log/archive/2014
This exact command, if executed from cli, works just fine, but when added inside the lastaction/endscript block, it throws the following message
logrotate_script: 2: logrotate_script: Syntax error: "(" unexpected
Anybody has any clue on why this happens?
You are using bash as your shell. You also have the extglob setting enabled.
When logrotate is executing that shell script one or the other of those is not true.
Also that mv command looks odd to me. If archive is under log then I don't see why !(archive) doesn't have log/ as a prefix. Also the log/* should still be matching log/archive regardless of the extglob glob before it. (That is I would think you wanted mv log/!(archive) log/archive/2014, assuming you don't feel like just ignoring the warning from mv in the first place.)
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
To create a playlist for all of the music in a folder, I am using the following command in bash:
ls > list.txt
I would like to use the result of the pwd command for the name of the playlist.
Something like:
ls > ${pwd}.txt
That doesn't work though - can anyone tell me what syntax I need to use to do something like this?
Edit: As mentioned in the comments pwd will end up giving an absolute path, so my playlist will end up being named .txt in some directory - d'oh! So I'll have to trim the path. Thanks for spotting that - I would probably have spent ages wondering where my files went!
The best way to do this is with "$(command substitution)" (thanks, Landon):
ls > "$(pwd).txt"
You will sometimes also see people use the older backtick notation, but this has several drawbacks in terms of nesting and escaping:
ls > "`pwd`.txt"
Note that the unprocessed substitution of pwd is an absolute path, so the above command creates a file with the same name in the same directory as the working directory, but with a .txt extension. Thomas Kammeyer pointed out that the basename command strips the leading directory, so this would create a text file in the current directory with the name of that directory:
ls > "$(basename "$(pwd)").txt"
Also thanks to erichui for bringing up the problem of spaces in the path.
This is equivalent to the backtick solution:
ls > $(pwd).txt
To do literally what you said, you could try:
ls > `pwd`.txt
which will use the full pathname, which should be fine.
Note that if you do this in your home directory, which might
be in /home/hoboben, you will be trying the create /home/hoboben.txt,
a text file in the directory above.
Is this what you wanted?
If you wanted the directory to contain a file named after it, you would get
the basename of the current directory and append that with .txt to the pwd.
Now, rather than use the pwd command... why not use the PWD environment variable?
For example:
ls > $PWD.txt
or
ls > ${PWD}.txt
is probably what you were trying to remember with your second example.
If you're in /home/hoboben and you want to create /home/hoboben/hoboben.txt, try:
ls > ${PWD}/${PWD##*/}.txt
If you do this, the file will contain its own name, so most often, you would remedy this in one of a few ways. You could redirect to somewhere else and move the file or name the file beginning with a dot to hide it from the ls command as long as you don't use the -a flag (and then optionally rename the resulting file).
I write my own scripts to manage a directory hierarchy of music files and I use subdirectories named ".info", for example, to contain track data in some spare files (basically, I "hide" metadata this way). It works out okay because my needs are simple and my collection small.
I suspect the problem may be that there are spaces in one of the directory names. For example, if your working directory is "/home/user/music/artist name". Bash will be confused thinking that you are trying to redirect to /home/user/music/artist and name.txt. You can fix this with double quotes
ls > "$(pwd).txt"
Also, you may not want to redirect to $(pwd).txt. In the example above, you would be redirecting the output to the file "/home/user/music/artist name.txt"
The syntax is:
ls > `pwd`.txt
That is the '`' character up underneath the '~', not the regular single quote.
Using the above method will create the files one level above your current directory. If you want the play lists to all go to one directory you'd need to do something like:
#!/bin/sh
MYVAR=`pwd | sed "s|/|_|g"`
ls > /playlistdir/$MYVAR-list.txt
to strip all but the directory name
ls >/playlistdir/${PWD##/*}.txt
this is probably not what you want because then you don't know where the files are (unless you change the ls command)
to replace "/" with "_"
ls >/playlistdir/${PWD//\//_}.txt
but then the playlist would look ugly and maybe not even fit in the selection window
So this will give you both a short readable name and usable paths inside the file
ext=.mp3 #leave blank for all files
for FILE in "$PWD/*$ext"; do echo "$FILE";done >/playlistdir/${PWD##/*}.txt