Currently, I'm working on a project where I need to git ignore my local config files. This means that if the user clones the repository, he would see *.sample.js files inside the config dir.
If the user executes make config the following script is executed.
config: ###Misc Create config files
#cd src/backend/config
#cp cache.sample.js cache.js
#cp database.sample.js database.js
#cp steam.sample.js steam.js
#cp teamspeak.sample.js teamspeak.js
#cp website.sample.js website.js
#echo
This script basically removes the "sample" part from the file name. The above code needs to be modified when ever I add a new config file, which is not a practical thing.
I would like to convert this into a simple regex command like:
find -iname \*.sample.js -type f -exec rename -n 's/(.*sample\.js)/$1/' {} \;
This command doesn't work, at least on windows MINGW64 bash. I need a vanilla solution.
Thanks to #Jetchisel for a working solution:
find -iname '*.sample.js' -type f -exec bash -c 'for f; do cp "$f" "${f/.sample}"; done' sh {} +
Bonus: why that last + sign?
That will process as many files as possible while avoiding arg_max
as opposed to \; which will process one file per -exec call
Related
I am currently working on an auto grading script for a class project. It has to be able to search any number of given directories lets say
for example
usr/autograder/jdoe/
jdoe contains two files house.c and readme.txt.
I need to create a file in jdoe called jdoe.pdf
Currently i'm using this line of code below to get the path to where i need to create the file. Where $1 is user input of the path containing the projects the auto grader will grade.
find $1 -name "*.txt" -exec sh -c "dirname {}"
When I try adding /somename.pdf to the end of this statement I get readme.txt/somename.pdf
along with another -exec to get the name for the file.
\; -exec sh -c "dirname {} xargs -n 1 basename" \;
I'm having problems combining these two into one working statement.
I'm new to unix programming and would appreciate any advice or help even if it means re-writing the code using different unix tools.
The main question here is how do I create files in a path other than the directory I call my script from. Thanks in advance.
How about this?
find "$1" -name "*.txt" -exec bash -c 'd=$(dirname "$1"); touch $d"/"$(basename "$d").pdf' - {} \;
You can create files in another path using change directory command (cd).
If you start your script in usr/autograder/script and want to change to usr/autograder/jdoe you can change directory with shell command cd ../jdoe (relative) or cd usr/autograder/jdoe (absolute).
Now you are in the directory of usr/autograder/jdoe and you are able to create files in this directory, for example gedit readme.txt will open gedit and creates the file in usr/autograder/jdoe.
The simplest way is to loop over the files returned by find and then do whatever you need to do.
For example:
find "$1" -type f -name "*.txt" -print0 | while IFS= read -r -d $'\0' filename; do
dir=$(dirname "$filename")
# create pdf file
touch "$dir/${dir##*/}.pdf"
done
(Note the use of find -print0 to correctly handle filenames containing whitespace and newline characters.)
Is this what you are looking for?
function process_file {
dir=$(dirname "$1")
name=$(basename "$1")
echo name is $name and dir is $dir;
cd "$dir"
touch "${dir##*/}.pdf" # or anything else
}
# export the function, so that it is known in the child processes
export -f process_file
find . -name '*.txt' -exec bash -c "process_file '{}'" \;
I know there are dozen of questions about similar topcis but I still can't beat this up.
I need to copy all .svn directories recursively from /var/foo to /var/foo2 on a Debian machine:
/var/www/foo/.svn
/var/www/foo/bar/.svn
...
I tried these two commands without success:
find /var/foo -name ".svn" -type f -exec cp {} ./var/foo2 \;
find /var/foo -name ".svn" -type d -exec cp {} /var/foo2 \;
Once only the svn directory right inside foo is copied, while another time nothing is copied.
Given following file structure:
./
./a/
./a/test/
./a/test/2
./b/
./b/3
./test/
./test/1
Running following script in the directory to be copied:
find -type d -iname test -exec sh -c 'mkdir -p "$(dirname ~/tmp2/{})"; cp -r {}/ ~/tmp2/{}' \;
Should copy all test directories to ~/tmp2/.
Points of interest:
Directories are copied to the destination on a one-by-one basis
Parent directories are created in advance so that cp doesn't complain about target not existing
Rather than just cp, cp -r is used
The whole command is wrapped with sh -c so that operations on {} such as dirname can be performed (so that the shell expands it for each directory separately, rather than expanding it once during calling the find)
Resulting structure in ~/tmp2:
./
./a/
./a/test/
./a/test/2
./test/
./test/1
So all you should need to do is to replace test with .svn and ~/tmp2 with directory of choice. Just remember about running it in the source directory, instead of using absolute paths.
I find that using tar for such operations makes the code often much more readable:
$ mkdir /var/www/foo2
$ cd /var/www/foo2
$ find ../foo/ -type d -name .svn -exec tar c \{\} \+ | \
tar x --strip-components=1
find will list all directories named .svn, and call tar to create (c) an archive file (that is sent to stdout) with all these directories. the archive on stdout is then extracted (x) by another tar instance in the target directory. the relative path portion (../) is automatically removed by the archiving tar, but since we also want to remove the first path component (foo/) we need to add --strip-components.
Note: This will only work if you do not have very many .svn directories you want to copy (more than $(getconf ARG_MAX)-2, which on my system is more than 200000).
I'm making a bash script that should backup all files and dir structure to another dir.
I made the following code to do that:
find . -type f -exec cp {} $HOME/$bdir \; -o -type d -exec mkdir -p {} $HOME/$bdir \; ;
The problem is, is that this only copies the files and not the dir structure.
NOTE: I may not use cp -r, cp -R or something like it because this code is part of an assignment.
I hope somebody can put me in the right direction. ;)
Joeri
EDIT:
I changed it to:
find . -type d -exec mkdir -p $HOME/$bdir/{} \; ;
find . -type f -exec cp {} $HOME/$bdir/{} \; ;
And it works! Ty guys ;)
This sounds like a job for rsync.
You mention that this is an assignment. What are your restrictions? Are you limited to only using find? Does it have to be a single command?
One way to do this is to do it in two find calls. The first call only looks for directories. When a directory is found, mkdir the corresponding directory in the destination hierarchy. The second find call would look for files, and would use a cp command like you currently have.
You can also take each filename, transform the path manually, and use that with the cp command. Here's an example of how to generate the destination filename:
> find . -type f | sed -e "s|^\./|/new/dir/|"
/new/dir/file1.txt
/new/dir/file2.txt
/new/dir/dir1/file1_1.txt
/new/dir/dir1/file1_2.txt
For your purposes, you could write a short bash script that take the source file as input, uses sed to generate the destination filename, and then passes those two paths to cp. The dirname command will return the directory portion of a filename, so mkdir -p $(dirname $destination_path) will ensure that the destination directory exists before you call cp. Armed with a script like that, you can simply have find execute the script for every file it finds.
cd olddir; tar c . | (cd newdir; tar xp)
Can you do your find with "-type d" and exec a "mkdir -p" first, followed by your find that copies the files rather than having it all in one command? It should probably also be mkdir -p $HOME/$bdir/{}.
I was in the process of creating a User class where one of the methods was get_privileges();.
After hours of slamming my head into the keyboard, I finally discovered that the previous coder who I inherited this particular database spelled the word "privileges" as "privelages" in the MySQL database, and thus also everywhere in the hundreds of files that access these "privelages" it is spelled that way.
Is there a way in Linux (Ubuntu Server) that I can go through every place in the /var/www folder and replace "privelages" with "privileges", so that I don't have to deal with this typo and code around it?
A variation that takes into account subdirectories (untested):
find /var/www -type f -exec sed -i 's/privelages/privileges/g' {} \;
This will find all files (not directories, specified by -type f) under /var/www, and perform a sed command to replace "privelages" with "privileges" on each file it finds.
Check this out: http://www.cyberciti.biz/faq/unix-linux-replace-string-words-in-many-files/
cd /var/www
sed -i 's/privelages/privileges/g' *
I generally use this short script, which will rename a string in all files and all directory names and filenames. To use it, you can copy the text below into a file called replace_string, run sudo chmod u+x replace_string and then move it into your sudo mv replace_string /usr/local/bin folder to be able to execute it in any directory.
NOTE: this only works on linux (tested on ubuntu), and fails on MacOS. Also be careful with this because it can mess up things like git files. I haven't tested it on binaries either.
#!/usr/bin/env bash
# This will replace all instances of a string in folder names, filenames,
# and within files. Sometimes you have to run it twice, if directory names change.
# Example usage:
# replace_string apple banana
echo $1
echo $2
find ./ -type f -exec sed -i -e "s/$1/$2/g" {} \; # rename within files
find ./ -type d -exec rename "s/$1/$2/g" {} \; # rename directories
find ./ -type f -exec rename "s/$1/$2/g" {} \; # rename files
I'm searching for a config folder, and trying to change to that directory:
find . -name "config" -exec cd {} \;
There is one match, ./my-applications/config, but after I try this it says:
find: `cd': No such file or directory
What am I doing wrong?
The command cd is a shell built-in, not found in /bin or /usr/bin.
Of course, you can't change directory to a file and your search doesn't limit itself to directories. And the cd command would only affect the executed command, not the parent shell that executes the find command.
Use:
cd $(find . -name config -type d | sed 1q)
Note that if your directory is not found, you'll be back in your home directory when the command completes. (The sed 1q ensures you only pass one directory name to cd; the Korn shell cd takes two values on the command and does something fairly sensible, but Bash ignores the extras.)
In case you have more than one config directory:
select config in $(find . -name config -type d)
do
cd $config
break
done
find runs -exec programs as subprocesses and subprocesses cannot affect their parent process. So, it cannot be done. You may want to try
cd `find . -name "config"`