How could I move all .txt files from a folder and all included folders into a target directory .
And preferably rename them to the folder they where included in, although thats not that important. I'm not exactly familiar with bash.
To recursively move files, combine find with mv.
find src/dir/ -name '*.txt' -exec mv -t target/dir/ -- {} +
Or if on a UNIX system without GNU's version of find, such as macOS, use:
find src/dir/ -name '*.txt' -exec mv -- {} target/dir/ ';'
To rename the files when you move them it's trickier. One way is to have a loop that uses "${var//from/to}" to replace all occurrences of from with to in $var.
find src/dir/ -name '*.txt' -print0 | while IFS= read -rd $'\0' file; do
mv -- "$file" target/dir/"${file//\//_}"
done
This is ugly because from is a slash, which needs to be escaped as \/.
See also:
Unix.SE: Understanding IFS= read -r line
BashFAQ: How can I read a file (data stream, variable) line-by-line (and/or field-by-field)?
Try this:
find source -name '*.txt' | xargs -I files mv files target
This will work faster than any option with -exec, since it will not invoke a singe mv process for every file which needs to be moved.
If it's just one level:
mv *.txt */*.txt target/directory/somewhere/.
Related
The scenario is that I want to convert all of my music files from .mp3 to .ogg. They are in a folder called "Music". In this folder there are folders and files. The files are .mp3s. The directories may contain .mp3s or directories which further contain .mp3s or directories, and so on. This is because some artists have albums which have parts and some do not, etc.
I want to write a script that converts each file using avconv.
Basically, what I am going to do is manually cd into every directory and run the following:
for file in $(ls); do avconv -i $file `echo \`basename $file .mp3\`.ogg`; done
This successfully gets me what I want. However, this is not great as I have a lot of folders, and manually going into each of them and executing this is slow.
My question, then, is how do I write a script that runs this in any directory that has .mp3s, and then goes into any subdirectory it finds and recursively calls itself? My intuition tells me to use Perl or Python because of the complex nature of this.
Thanks for any suggestions!
I'm not familiar with avconv but assuming your command is:
avconv -i inputname outputname
And you want to convert all inputname.mp3 to inputname.ogg in their original directories below Music, then the following should work in bash:
#!/bin/bash
while read -r fname; do
avconv -i "$fname" "${fname%.mp3}.ogg"
done < <(find /path/to/Music -type f -name "*.mp3")
Note: this does not remove the original .mp3, and the space between < < is required. Also note, for file in $(ls) is filled with potential for errors.
You can do it with bash in one liner:
First you find all files (of type file (-type f) ) that match next pattern "*.mp3". To read each one you use 'while' and invoke avconf.
For exchange extension I prefer 'sed' command, that keep folder so you don't need the 'cd' command.
Notice that you must put quotes on $FN variable because it can contain spaces.
find -type f -iname "*.mp3" | while read "FN" ; do avconf -i "$FN" $(echo "$FN" | sed 's/\.mp3/\.ogg/g') ; done
find <music-folder> -type f -name '*.mp3' | \
xargs -I{} bash -c 'mp3="$0"; ogg="${mp3%.mp3}.ogg"; avconv -i "$mp3" "$ogg";' {}
This should survive in cases of "weird" filenames with spaces, quotes and other strange symbols within.
You can list directories with absolute paths and recursively cd into every directory using find $PWD -type d syntax:
Just inside from Music directory run:
for d in $(find $PWD -type d)
do
cd $d
for file in $(find . -maxdepth 1 -type f)
do
echo $file
avconv -i $file `echo \`basename $file .mp3\`.ogg`
done
done
I am not sure how to word this question to find the solution easily online, so after much searching I thought I would ask here.
I access my website's files using bitvise ssh client and I use command lines for various grep and sed functions that I've been recently taught, but I can't seem to find a simple way to do this:
What is the command line to make a backup copy (.bak) of EVERY file that ends in .php? I am looking for the command to instantly make a backup of every php file at once, so when I go into my files I see things like...
index.php
index.php.bak
For every php file.
Also, what is the command line to do this for EVERY file at once, regardless of extension?
Would be awesome to see a solution that uses xargs or find's -exec.
But here is how can do this with a shell loop and find:
Note, this recursively backs up files in sub directories.
For .php files:
find . -iname '*.php' -type f -print0 | while read -d $'\0' file; do cp "$file" "$file.bak"; done
For all files:
find . -type f -print0 | while read -d $'\0' file; do cp "$file" "$file.bak"; done
For all files that have an extension:
find . -iname '*.*' -type f -print0 | while read -d $'\0' file; do cp "$file" "$file.bak"; done
You can just use the cp command
enter the dir that you have all the .php files then type
cp *.php temp/ where temp is a directory in the current directory. The * means all
if you just want to copy the whole folder you could
cp -R foldername destinationArea
files=`find C:/PATH/TO/DIRECTORY -name *.txt`
for file in $files; do
#need code to rename $file, by moving it into the same directory
eg. $file was found in C:/PATH/TO/DIRECTORY/2014-05-08.
how do I rename $file to back to that directory and not to C:/PATH/TO/DIRECTORY?
You can use -execdir option in find:
find C:/PATH/TO/DIRECTORY -name '*.txt' -execdir mv '{}' '{}'-new \;
As per man find:
-execdir utility [argument ...] ;
The -execdir primary is identical to the -exec primary with the exception that utility will be executed
from the directory that holds the current file.
You would be better served by using this structure:
while read fname
do
....
done < <(find ...)
Or, if you're not using bash:
find ... | while read fname
do
....
done
The problem with storing the output of find in a variable, or even doing for fname in $(find ...), is with word splitting on whitespace. The above structures still fail if you have a file name with a newline in it, since they assume that you have one file name per line, but they're better than what you have now.
An even better solution would be something like this:
find ... -print0 | xargs -0 -n1 -Ixxx somescript.sh "xxx"
But even that might have issues if filenames have quotes or other things in them.
The bottom line is that parsing arbitrary data (which filenames can be) is hard...
find C:/PATH/TO/DIRECTORY -name \*.txt | while read file
do
dir=$(dirname "$file")
base=$(basename "$file")
mv "$file" "$dir/new_file_name"
done
So I have some folder
|-Folder1
||-SubFolder1
||-SubFolder2
|-Folder2
||-SubFolder3
||-SubFolder4
Each subfolder contains several jpg I want to zip to the root folder...
I'm a little bit stuck on "How to enter each folder"
Here is my code:
find ./ -type f -name '*.jpg' | while IFS= read i
do
foldName=${PWD##*/}
zip ../../foldName *
done
The better would be to store FolderName+SubFolderName and give it to the zip command as name...
Zipping JPEGs (for Compression) is Usually Wasted Effort
First of all, attempting to compress already-compressed formats like JPEG files is usually a waste of time, and can sometimes result in archives that are larger than the original files. However, it is sometimes useful to do so for the convenience of having a bunch of files in a single package.
Just something to keep in mind. YMMV.
Use Find's -execdir Flag
What you need is the find utility's -execdir flag. The GNU find man page says:
-execdir command {} +
Like -exec, but the specified command is run from the subdirec‐
tory containing the matched file, which is not normally the
directory in which you started find.
For example, given the following test corpus:
cd /tmp
mkdir -p foo/bar/baz
touch foo/bar/1.jpg
touch foo/bar/baz/2.jpg
you can zip the entire set of files with find while excluding the path information with a single invocation. For example:
find /tmp/foo -name \*jpg -execdir zip /tmp/my.zip {} +
Use Zip's --junk-paths Flag
The zip utility on many systems supports a --junk-paths flag. The man page for zip says:
--junk-paths
Store just the name of a saved file (junk the path), and do not
store directory names.
So, if your find utility doesn't support -execdir, but you do have a zip that supports junking paths, you could do this instead:
find /tmp/foo -name \*jpg -print0 | xargs -0 zip --junk-paths /tmp/my.zip
You can use dirname to get the directory name of a file/directory it is located in.
You can also simplify the find command to search only for directories by using -type d. Then you should use basename to get only the name of the subdirs:
find ./*/* -type d | while read line; do
zip --junk-paths "$(basename $line)" $line/*.jpg
done
Explanation
find ./*/* -type d
will print out all directories located in ./*/* which will result in all subdirs of directories located in the current dir
while read line reads each line from the stream and stores it in the variable "line". Thus $line will be the relative path to the subdir, e.g. "Folder1/Subdir2"
"$(basename $line)" returns the only the name of the subdir, e.g. "Subdir2"
Update: add --junk-paths to the zip command if you do not want the directy paths to be stored in the zip filde
So a little check, I finally got something working:
find ./*/* -type d | while read line; do
#printf '%s\n' "$line"
zip ./"$line" "$line"/*.jpg
done
But this create un archive containing:
Subfolder.zip
Folder
|-Subfolder
||-File1.jpg
||-File2.jpg
||-File3.jpg
Instead I fold like it to be:
Subfolder.zip
|-File1.jpg
|-File2.jpg
|-File3.jpg
So I tried using basename and dirname in differnet combination...Always got some error...
And just to learn how to: what if I would like the new archive to be created in the same root directory as "Folder"?
Ok finally got it!
find ./* -name \*.zip -type f -print0 | xargs -0 rm -rf
find ./*/* -type d | while read line; do
#printf '%s\n' "$line"
zip --junk-paths ./"$line" "$line"/*.jpg
done
find . -name \*.zip -type f -mindepth 2 -exec mv -- '{}' . \;
In first row I simply remove all .zip files,
Then I zip all and in the final row I move all zip to the root directory!
Thanks everbody for your help!
I know this is probably elementary to unix people, but I haven't found a straightforward answer online.
I have a directory with sub-directories. Some of these sub-dirs have .mov files in them. I want to consolidate all the movs to a single directory. I don't need to worry about file naming conflicts because the files are from a digital camera and it names the files incrementally, but divides them into daily folders.
What is the Unix-fu for grabbing all these files and copying (or even better, moving them) to a directory in my home folder?
Thanks.
How about this?
find "$SOURCE_DIRECTORY" -type f -name '*.mov' -exec mv '{}' "$TARGET_DIRECTORY" ';'
If the source and target directories do not overlap this should work fine.
EDIT:
BTW, if you have mixed-case extensions (x.mov, y.Mov, Z.MOV) as is the case with many cameras, this would be better. It uses -iname which is case-insensitive when matching:
find "$SOURCE_DIRECTORY" -type f -iname '*.mov' -exec mv '{}' "$TARGET_DIRECTORY" ';'
Make sure to replace the $SOURCE_DIRECTORY and $TARGET_DIRECTORY variables with the actual directories and that they do not overlap (i.e. the target being somewhere under the source)
EDIT 2:
PS: I just noticed that khachik caught this one with his edit
mv `find . -name "*.mov" | xargs` OUTPUTDIR/
Update after thkala's comment:
find . -iname "*.mov" | while read line; do mv "$line" OUTPUTDIR/; done
If you need to cope with weird filenames (spaces, special characters), try this:
$ cd <source parent directory>
$ find -name '*.mov' -print0 | xargs -0 echo mv -v -t <target directory>
Remove the "echo" above to actually do the move, rather than print what would happen.
"mv -v" gives verbose output, "mv -t ..." specifies the target directory (possibly GNU-specific).
"-print0" and "-0" are extensions to cope with weird filenames. On non-GNU systems you might need to remove those options, which will result in newline-separated data. This will still work on filenames with spaces, but not filenames with newlines (yes, it's possible).