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.
Related
I have 200,000 files I want to send to different folders based on key words in the file name
in English if a file name has (shtf or prepper or prepping or survival) in the name send(move) it to folder shtf
if a file has (cookbook or gluten or recipe) move to food folder
*cookbook* *GLUTEN* *RECIPE*
example
(file name)
more shtf tips.epub move to folder shtf
ifshtfbeready.epub move to folder shtf
oldworldcookbook.epub move to folder food
i'm old retired ibmer small basic sas dos commands or ????
Here is a bash command, you may be able to adapt it into dos etc. I'm posting this because others may it useful as well.
find . | grep -E "(cookbook|gluten|recipe)" | while read name; do mv $name directory; done;
Where directory is the name of the directory you want to move the file. You can replace . with whatever starting directory you want, of course.
You can use wildcard in the source filename list and use a directory as the target to move multiple files with one command.
move c:\dir1\*cookbook*.* c:\food
move c:\dir1\*gluten*.* c:\food
Using puttys pscp -r folder\to\copy\* user#server:/path/to/copy/folder/to it only copies the content of path\to\copy\folder\* and does not include the "main" folder which the subfiles and subdirectories are in.
What I need is that the folder itself is also copied such that I get a folder with the same name as the one I copied with the content inside.
I know I just can create a parent-folder for the one I want to copy and parse that as the path\to\copy\folder\* but that is not the case
Just use pscp -r folder\to\copy user#server:/path/to/copy/folder/to.
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.
file1 = fullfile(pwd,'folder','toto.txt');
file2 = fullfile(pwd,'folder','toto2.txt.sav');
movefile(file1,file2)
=>this creates a folder named toto2.txt.sav
whereas I'm looking for renaming the file in the same directory
any idea ?
Check the documentation:
Renaming a File in the Current Folder
In the current folder, rename myfunction.m to oldfunction.m:
movefile('myfunction.m','oldfunction.m')
Thus:
o=pwd
cd('folder')
movefile('toto.txt','toto2.txt.sav')
cd(o);
Was looking for this as well. Now on a windows machine if I want to rename all files in a folder, I do:
system('rename *.* *.sav')
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 *