I'm developing a simple launchdaemon that copies files from one directory to another. I've gotten the files to transfer over fine.
I just want the files in the directory to be .mp3's instead of .dat's
Some of the files look like this:
6546785.8786.dat
3678685.9834.dat
4658679.4375.dat
I want them to look like this:
6546785.8786.mp3
3678685.9834.mp3
4658679.4375.mp3
This is what I have at the end of the bash script to rename the file extensions.
cd $mp3_dir
mv *.dat *.mp3
exit 0
Problem is the file comes out as *.mp3 instead of 6546785.8786.mp3
and when another 6546785.8786.dat file is imported to $mp3_dir, the *.mp3 is overwritten with the new .mp3
I need to rename just the .dat file extensions to .mp3 and keep the filename.
Ideas? Suggestions?
Try:
for file in *.dat; do mv "$file" "${file%dat}mp3"; done
Or, if your shell has it:
rename .dat .mp3 *.dat
Now, why your command didn't work: first of all, it is more than certain that you only had one file in your directory when it was renamed to *.mp3, otherwise mv would have failed with *.mp3: not a directory.
And mv does NOT do any magic with file globs, it is the shell which expands globs. Which means, if you had this file in the directory:
t.dat
and you typed:
mv *.dat *.mp3
the shell would have expanded *.dat to t.dat. However, as nothing would match *.mp3, the shell would have left it as is, meaning the fully expanded command is:
mv t.dat *.mp3
Which will create a file named, literally, *.mp3.
If, on the other hand, you had several files named *.dat, as in:
t1.dat t2.dat
the command would have expanded to:
mv t1.dat t2.dat *.mp3
But this will fail: if there are more than two arguments to mv, it expects the last argument (ie, *.mp3) to be a directory.
For anyone on a mac, this is quite easy if you have BREW, if you don't have brew then my advice is get it. then when installed just simply do this
$ brew install rename
then once rename is installed just type (in the directory where the files are)
$ rename -s dat mp3 *
Related
I'm unable to use the rm command to remove files remotely from another directory. I'm a beginner so I apologise for my inability to elaborate properly.
Here's what I'm trying to do:
I'm trying to delete all .srt files from a sub directory. It works when I cd into the specific directory like so:
Command 1:
cd /users/jakubdonovan/library/cloudstorage/iCloud\ drive/the-modern-python3-bootcamp/target_folder
Command 2:
rm *.srt
However, let's say I want to quickly delete a specific file type from a folder without first using the "cd" command, like so:
rm *.srt /users/jakubdonovan/library/cloudstorage/iCloud\ drive/the-modern-python3-bootcamp/target_folder
It returns with "No matches for wildcard '*.srt'. See help expand."
Which is strange because I can use the touch, cp and and all the other commands remotely without a problem.
Is there a way to make the command "rm *.filetype" remove all the files with that specific filetype from a folder and all its subfolders in one swoop?
If you would like to rm in a sub-directory you just have to specify that sub-directory in the command.
rm /path/to/folder/*.filetype
or if you know that the folder is inside your current directory you can try...
rm ./folder/*.filetype
Okay so I want to know how I would go about doing this, using grep to locate .txt files named "cocacola1", "cocacola2", "cocacola3" & then copying them to another directory. So searching for files named "cocacola" &/even if it contains other characters within the file name to then copy them to another directory/location.
You can just use unix find. Assuming the files you're searching for are in 'source' and you want to copy to 'destination':
find source -name '*cocacola*' -exec cp {} destination \;
I put the wildcard '*' before and after cocacola since you said other characters might exist in the file name.
I am trying to move a file from Downloads to a folder on my desktop.
I keep getting this:
Why does the usage message appear afterwards?
It looks like your file has a space in it, so it needs to be escaped. Otherwise, mv tries to find a file named "Tres", one named "Beijos_C.pdf" and move them into that directory.
# Either...
mv Tres\ Beijos_C.pdf ~/Desktop/choro/
# ...or...
mv "Tres Beijos_C.pdf" ~/Desktop/choro/
If your file name contains spaces, you should surround it with " like this:
mv "my file name.txt" /home/user/Desktop
Suppose I have a directory structure like
C:\Users\Desktop\abc\d
I want to rar archive the abc folder so that the structure of rar is:
abc\d
When I try using powershell to archive, winrar replicates the full path inside the archive, like:
\Users\Desktop\abc\d
I dont want the full path to be created inside the archive
Here's the script:
https://gist.github.com/saurabhwahile/50f1091fb29c2bb327b7
What am I doing wrong?
Use the command line:
Rar.exe a -r -ep1 Test.rar "C:\Users\Desktop\abc"
Rar.exe is the console version of WinRAR stored in same directory as WinRAR.exe. You can use this command line also with WinRAR.exe if you want to see the compression process in a graphic window.
a is the command and means add files to archive.
-r is a switch to recursively add all files and subdirectories including empty subdirectories to the archive.
-ep1 is another switch which results in execluding base directory.
For this command line the base directory is "C:\Users\Desktop\" and therefore the created archive Test.rar contains only abc and all files and subdirectories in directory abc which is what you want.
One more hint: Using the command line
Rar.exe a -r -ep1 Test.rar "C:\Users\Desktop\abc\"
results in adding all files and subdirectories of directory abc to the archive, but without directory name abc being also stored in the archive. The backslash at end makes this difference.
In other words: On using switch -ep1 everything up to last backslash in file/directory specification is not added to the archive.
For more information about available switches see the text file Rar.txt in the program files directory of WinRAR.
I am using 7z command line executable to zip files, but I see that while adding to an archive the path of the files is preserved in the archive.
So if I do
7z a -tzip myzip.zip dir1\dir2\*
the archive myzip.zip will contain the path dir1\dir2. I do not want this, rather I want only the files to be added to the zip file without the paths being preserved.
I searched quite a bit but do not seem to find any way of doing this, maybe I am missing something obvious?
Thanks
Just add a dot before the path, i.e.
7z a -tzip -r myzip.zip .\Relative\Dir\*
Give the full path. That should work. Not the relative path from the current location.
For example, I give the below, where I want the files in the man5 folder to be archived.
$ 7z a -tzip myzip.zip /home/pradeeban/Desktop/man4/man5/*
The zip contained only the files, without the directories.
Then I gave only the relative path. It had the directories, inside the zip.
$ 7z a -tzip myzip.zip Desktop/man4/man5/*
Tried with Linux (Ubuntu 12.04). Not sure whether that differs from Windows.
I discovered a way to do this by using a relative path:
7z a -tzip myzip.zip %CD%\dir1\dir2\*
%CD% is how you get the current path in a Windows batch file, but it also works from the command line. More info about Capturing the current directory from a batch file.
As explained in related question in 7-zip user FAQ, 7z stores paths relative to working directory, so you will need to first cd to desired top-level directory for archive and run 7-zip from here.
cd dir1\dir2\
7z a -tzip myzip.zip *
If you run it from script and don't want to affect it with changed directory, use directory push/pop facilities available in your shell of choice or run cd+7-zip in spawned process to avoid affecting your entire script with changed directory. For example, using Windows' start that would be:
start /D dir1\dir2\ /wait 7z a -tzip myzip.zip *
This worked for me
Consider folder structure like C:\Parent\SubFolders..... And you want to create parent.zip which will contain all files and folders C:\Parent without parent folder [i.e it will start from SubFolders.....]
cd /D "C:\Parent"
"7z.exe" a Parent.zip "*.*" -r
This will create Parent.zip in C:\Parent