I have a script to move files of type .txt to a particular folder .It looks for the files in work folder and move it to completed folder.
I would like to make the script generic i.e to enhance the script so that the scripts works not for just one particular folder but other similar folders as well.
Example: If there is a .txt file in folder /tmp/swan/test/work and also in folder /tmp/swan/test11/work, the files should move to /tmp/swan/test/done and /tmp/swan/test11/done respectively.
EDIT:Also, if there is a .txt file in a sub folder like /tmp/swan/test11/work/APX that should also move to /tmp/swan/test11/done
Below is the current script.
#!/bin/bash
MY_DIR=/tmp/swan
cd $MY_DIR
find . -path "*work*" -iname "*.txt" -type f -execdir mv '{}' /tmp/swan/test/done \;
With -execdir, the mv command is executed in whatever directory the file is found in. Since you just want to move the file to a "sibling" directory, each command can use the same relative path ../done.
find . -path "*work*" -iname "*.txt" -type f -execdir mv '{}' ../done \;
One way to do it:
Background:
$ tree
.
├── a
│ └── work
└── b
└── work
Renaming:
find . -type f -name work -exec \
sh -c 'echo mv "$1" "$(dirname "$1")"/done' -- {} \;
Output:
mv ./a/work ./a/done
mv ./b/work ./b/done
You can remove the echo if it does what you want it to.
What about:
find . -path '*work/*.txt' -exec sh -c 'd=$(dirname $(dirname $1))/done; mkdir -p $d; mv $1 $d' _ {} \;
(also creates the target directory if it does not exist already).
Related
I want to write a shell script to do the following :
I have a folder with many subfolders. Each of these subfolders has a *.gz file and some other files which I don't need. I want to move all .gz files into a new subfolder called needed_files (I have already created this subfolder). So I did the following :
I went to the parent folder with all the subfolder and cp /.gz > needed_files/., but this did not work. Can you suggest what I should be doing?
grep is irrelevant here. Use find:
find . ! \( -type d -name needed_files -prune \) -type f -name '*.gz' \
-exec echo mv -t needed_files {} +
POSIX equivalent of that -exec is
-exec sh -c 'echo mv "$#" needed_files' _ {} +
If its output looks good, remove echo.
Btw I noticed that the title says copy but you also say I want to move, so decide on what you exactly want to do and let me know so I can edit my answer.
Hi I am working on a script that syncs content from a remote site using SFTP, then extracts any archives. However I am struggling to get the extracted files to the correct (source) directory.
I started with a for loop, but was thinking this could be more efficient.
list=`find $local_dir -type f -name *.rar`
for line in $list; do
DEST=${line%/*}
unrar x -o- $line $DEST
done
Requirements:
1. Find all RAR files
2. Extract all rar files to the directory each rar is located
/sync/testfile.rar --> /sync/testfile.file
/sync/testdir/testfile1.rar --> sync/testdir/testfile1.file
/sync/testdir/testfile2.rar --> sync/testdir/testfile2.file
Here is my current command. In the loop I used I specifically called out the destination, but am not sure how to do that here in one line.
find $local_dir -type f -name '*.rar' -exec unrar x -o- {} \;
I have also tried unrar xe -o- but that gives me the same result where the contents are extracted to the directory where the script was ran from.
Maybe:
find $local_dir -type f -name '*.rar' -exec sh -c 'echo unrar x -o- {} $(dirname {})' \;
I prepended unrar with echo so that it's possible to examine what
command is going to do. For example, for the directory structure like
this:
$ tree
.
├── a.rar
├── b
│ └── z.rar
└── b.rar
1 directory, 3 files
It would print:
$ find $local_dir -type f -name '*.rar' -exec sh -c 'echo unrar x -o- "{}" "$(dirname {})"' \;
unrar x -o- ./b.rar .
unrar x -o- ./a.rar .
unrar x -o- ./b/z.rar ./b
Basically in your example "-exec" should be "-execdir" (unrar switches not tested)
Executed from the folder above sync:
folderAboveSync$ find -name '*.rar' -execdir unrar e {} \;
http://man7.org/linux/man-pages/man1/find.1.html
How can I bulk-zip folders in subdirectories without including the parent folder in the zip archives? I have a folder structure like this:
folder01
folder02
file01
file02
When I run:
find . -type d -name "folder02" -exec zip -r '{}'.zip '{}' \;
I get "folder02.zip" which always extracts its contents into a parent folder "folder01". How can I prevent this? For me it creates useless parent folder structures when extracting these archives anywhere else.
Using some simple bash:
find . -type d -name "folder02" -exec bash -c 'cd "$(dirname "{}")"; zip -r "$(basename "{}")".zip "$(basename "{}")"' \;
I have a folder with tens of thousands of different file types. Id like to copy them all to a new folder (Copy1) but also rename them all to $RANDOM but keep the extension intact. I realize I can write a line specifying which extension to find and how to name it, but there is got to be a way to do it dynamically, because there are at least 100 file types and may be more in the future.
I have the following so far:
find ./ -name '*.*' -type f -exec bash -c 'cp "$1" "${1/\/123_//_$RANDOM}"' -- {} \;
but that puts the random number after the extension, and also it puts the all in the same folder. I cant figure out how to do the following 2 things:
1 - Keep all paths intact, but in a new root folder (Copy1)
2 - How to have name be $RANDOM.extension, instead of .extension.$RANDOM
PS - by $RANDOM i mean actual randomly generated number. I am interested in keeping folder structure, so we are dealing with a few hundred files at most per directory, but all directories/files need to be renamed to $RANDOM. Another way to look at what I need to do. Copy all contents or Folder1 with all subdirectories and files to Folder2 (where Fodler2 is a $RANDOM name), then rename all folders and files to random names but keep all extensions.
EDIT: Ok i figured out how to rename and keep extension. But I have a problem where its dumping all of the files into the root directory where script is run from. How do I keep them in their respective folders? Command Im using is:
find ./ -name '*.*' -type f -exec bash -c 'mv "$1" $RANDOM.${1##*.}' -- {} \;
Thanks!
Ok i figured out how to rename and keep extension. But I have a
problem where its dumping all of the files into the root directory
where script is run from. How do I keep them in their respective
folders? Command Im using is:
find ./ -name '*.*' -type f -exec bash -c 'mv "$1" $RANDOM.${1##*.}' -- {} \;
Change your command to:
PATH=/bin:/usr/bin find . -name '*.*' -type f -execdir bash -c 'mv "$1" $RANDOM.${1##*.}' -- {} \;
Or alternatively using uuids instead of random numbers:
PATH=/bin:/usr/bin find . -name '*.*' -type f -execdir bash -c 'mv "$1" $(uuidgen).${1##*.}' -- {} \;
Here's what I came up with :
i=1
random="whatever"
find . -name "*.*" -type f | while read f
do
newbase=${f/*./$random$i.} //added counter to filename
cp $f /Path/Name/"$newbase"
((i++))
done
I had to add a counter to random (i), otherwise, if the extensions are similar, your files would overwrite themselves when copied.
In your new folder, your files should look like this :
whatever1.txt
whatever2.txt
etc etc
I hope this is what you were looking for.
Here is the command that worked for me.
find . -name '*.pdf' -type f -exec bash -c 'echo "{}" && cp "$1" ./$RANDOM.${1##*.}' -- {} \;
I have a folder with a bunch of log files. Each set of log files is in a folder detailing the time and date that the program was run. Inside these log folders, I've got some video files that I want to extract. All I want is the video files, nothing else. I tried using this command to only copy the video files, but it didn't work because a directory didn't exist.
.rmv is the file extension of the files I want.
$ find . -regex ".*\.rmv" -type f -exec cp '{}' /copy/to/here/'{}'
If I have a folder structure such as:
|--root
|
|--folder1
| |
| |--file.rmv
|
|--folder2
|
|--file2.rmv
How can I get it to copy to copy/to/here with it copying the structure of folder1 and folder2 in the destination directory?
cp has argument --parents
so the shortest way to do what you want is:
find root -name '*.rmv' -type f -exec cp --parents "{}" /copy/to/here \;
I would just use rsync.
The {} represents the full path of the found file, so your cp command evaluate to this sort of thing:
cp /root/folder1/file.rmv /copy/to/here/root/folder1/file.rmv
If you just drop the second {} it will instead be
cp /root/folder1/file.rmv /copy/to/here
the copy-file-to-directory form of cp, which should do the trick.
Also, instead of -regex, yor could just use the -name operand:
find root -name '*.rmv' -type f -exec cp {} /copy/to/here \;
Assuming src is your root and dst is your /copy/to/here
#!/bin/sh
find . -name *.rmv | while read f
do
path=$(dirname "$f" | sed -re 's/src(\/)?/dst\1/')
echo "$f -> $path"
mkdir -p "$path"
cp "$f" "$path"
done
putting this in cp.sh and running ./cp.sh from the directory over root
Output:
./src/folder1/file.rmv -> ./dst/folder1
./src/My File.rmv -> ./dst
./src/folder2/file2.rmv -> ./dst/folder2
EDIT: improved script version (thanks for the comment)