Let say the command be my_command
And this command has to be prepared specific files (file1, file2, and file3) in the current working directory.
Because I often use my_command in many different directories, I'd like to keep the certain files in a certain directory and execute my_command without those three files in the working directory.
I mean I don't want to copy those three files to every working directory.
For example:
Directory containing the three files /home/chest
Working directory: /home/wd
If I execute command my_command, it automatically recognizes the three files in /home/chest/
I've thought the way is similar to add $PATH and not the executable files but just files.
It seems like the files needs to be in the current working directory for the vasp_std command to work as expected, I am thinking that you could simply add all files in a include folder in you home directory and then create a symbolic link to this folder from your script. In the end of your script the symbolic link will then be deleted:
#!/bin/bash
# create a symbolic link to our resource folder
ln -s ~/include src
# execute other commands here
# finally remove the symbolic link from the current directory
unlink src
If the vasp_std command require that the files are placed directly under the current working directory you could instead create a symbolic link for each file:
#!/bin/bash
# create link for to all resource files
for file in ~/include/*
do
ln -s $file `basename $file`
done
# execute other commands here
# remove any previously created links
for file in ~/include/*
do
unlink `basename $file`
done
Related
So, I have a requirement to delete all files from specific folders within a directory. These are folders that end with "-outputs" in their names and I need to delete all files in those particular folders.
Is there a command in linux that lets you do that?
You can execute the following command:
rm $YOUR_PATH\*-outputs
Change $YOUR_PATH to the path where the files are located.
If you want to ignore nonexistent files and arguments, and avoid the prompt, you can use the -f option.
rm -f $YOUR_PATH\*-outputs
You can see the different rm options here:
rm Linual manual page
I'm new to linux and shell script in general. I'm using a distribution of Debian on the WSL (Windows Subsystem for Linux). I'm trying to write a very simple bash script that will do the following:
create a file in a directory (child-directory-a)
move to the directory it is in
move the file to another directory (child-directory-b)
move to that directory
move the file to the parent directory
This is what I have so far (trying to keep things extremely simple for now)
touch child-directory-a/test.txt
cd child-directory-a
mv child-directory-a/test.txt home/username/child-directory-b
The first two lines work, but I keep getting a 'no such directory exists' error with the last one. The directory exists and that is the correct path (checked with pwd). I have also tried using different paths (i.e. child-directory-b, username/child-directory-b etc.) but to no avail. I can't understand why it's not working.
I've looked around forums/documentation and it seems that these commands should work as they do in the command line, but I can't seem to do the same in the script.
If anyone could explain what I'm missing/not understanding that would be brilliant.
Thank you.
You could create the script like this:
#!/bin/bash
# Store both child directories on variables that can be loaded
# as environment variables.
CHILD_A=${CHILD_A:=/home/username/child-directory-a}
CHILD_B=${CHILD_B:=/home/username/child-directory-b}
# Create both child folders. If they already exist nothing will
# be done, and no error will be emitted.
mkdir -p $CHILD_A
mkdir -p $CHILD_B
# Create a file inside CHILD_A
touch $CHILD_A/test.txt
# Change directory into CHILD_A
cd $CHILD_A
# Move the file to CHILD_B
mv $CHILD_A/test.txt $CHILD_B/test.txt
# Move to CHILD_B
cd $CHILD_B
# Move the file to the parent folder
mv $CHILD_B/test.txt ../test.txt
Take into account the following:
We make sure that all the folders exists and are created.
Use variables to avoid typos, with the ability to load dynamic values from environment variables.
Use absolute paths to simplify the movement between folders.
Use relative paths to move files relatives to where we are.
Another command that might be of use is pwd. It will tell you the directory you are on.
with your second line, you change the current directory to child-directory-a
so, in your third line there is an error because there is no subdirectory child-directory-a into subdirectory child-directory-a
Your third line should be instead :
mv test.txt ../child-directory-b
The point #4 of your script should be:
cd ../child-directory-b
(before that command the current directory is home/username/child-directory-a and after this command it becomes home/username/child-directory-b)
Then the point #5 and final point of your script should be:
mv test.txt ..
NB: you can display the current directory at any line of your script by using the command pwd (print working directory) in your script, it that helps
#!/bin/sh
# Variables
WORKING_DIR="/home/username/example scripts"
FILE_NAME="test file.txt"
DIR_A="${WORKING_DIR}/child-directory-a"
DIR_B="${WORKING_DIR}/child-directory-b"
# create a file in a directory (child-directory-a)
touch "${DIR_A}/${FILE_NAME}"
# move to the directory it is in
cd "${DIR_A}"
# move the file to another directory (child-directory-b)
mv "${FILE_NAME}" "${DIR_B}/"
# move to that directory
cd "${DIR_B}"
# move the file to the parent directory
mv "${FILE_NAME}" ../
I need to use a shell script to move all files in a directory into another directory. I manually did this without a problem and now scripting it is giving me an error on the mv command.
Inside the directory I want to move files out of are 2 directories, php and php.tmp. The error I get is cd: /path/to/working/directory/php: No such file or directory. I'm confused because it is there to begin with and listed when I ls the working directory.
The error I get is here:
ls $PWD #ensure the files are there
mv $PWD/* /company/home/directory
ls /company/home/directory #ensure the files are moved
When I use ls $PWD I see the directories I want to move but the error afterward says it doesn't exist. Then when I ssh to the machine this is running on I see the files were moved correctly.
If it matters the directory I am moving files from is owned by a different user but the shell is executing as root.
I don't understand why I would get this error so, any help would be great.
Add a / after the path to specify you want to move the file, not rename the directory.
You should try this:
mv $PWD/\* /home/user/directory/
Are your variables properly quoted? You could try :
ls "$PWD" #ensure the files are there
mv "$PWD"/* "/company/home/directory"
ls "/company/home/directory" #ensure the files are moved
If any of your file or directory names contains characters such as spaces or tabs, your "mv" command may not be seeing the argument list you think it is seeing.
I have a directory containing a set of subdirectories and files. I need to recursively copy all the content of this directory to all the subdirectories of another directory, also recursively.
How do I achieve this, preferably without using a script and only with the cp command?
You can write this in a script but you don't have to. Just write it line by line in the terminal:
# $TARGET is the directory containing subdirectories where you want to STORE the copies
# $SOURCE is the directory containing the subdirectories you want to COPY
for dir in $(ls $TARGET); do
cp -r $SOURCE/* $TARGET/$dir
done
Only uses cp and runs on both bash and zsh.
You can't. cp can copy multiple sources but will only copy to a single destination. You need to arrange to invoke cp multiple times - once per destination - for what you want to do; using, as you say, a loop or some other tool.
The first part of the command before the pipe instruct tar to create an archive of everything in the current directory and write it to standard output (the – in place of a file-name frequently indicates stdout).
tar cf - * | ( cd /target; tar xfp -)
The commands within parentheses cause the shell to change directory to the target directory and untar data from standard input. Since the cd and tar commands are contained within parentheses, their actions are performed together.
The -p option in the tar extraction command directs tar to preserve permission and ownership information, if possible given the user executing the command. If you are running the command as superuser, this option is turned on by default and can be omitted.
Also you can use the following command, but it seems to be quite slower than tar;
cp -a * /target
I have a folder containing 100 pictures from a webcam. When the webcam sends a new picture, I want this one to replace number 0 and have all the other jpg's move up one number. I've set up a script where inotify monitors a directory. When a new file is put into this directory the script renumbers all the files in the picture directory, renames the new uploaded picture and puts it in the folder with the rest.
This script 'sort of' works. 'Sort of', because sometimes it does what it's supposed to do and sometimes it complains about missing files:
mv: cannot stat `webcam1.jpg': No such file or directory
Sometimes it complains about only one file, sometimes 4 or 5. Of course I made sure all 100 files were there, properly named before the script was run. After the script is run, the files it complains about are indeed missing.
This is the script, in the version I tested the full paths to the directories are used of course.
#!/bin/bash
dir1= /foo # directory to be watched
while inotifywait -qqre modify "$dir1"; do
cd /f002 #directory where the images are
for i in {99..1}
do
j=$(($i+1))
f1a=".jpg"
f1="webcam$i$f1a"
f2="test"
f2="webcam$j$f1a"
mv $f1 $f2
done
rm webcam100.jpg
mv dir1/*.jpg /f002/webcam0.jpg
done
I also need to implement some error checking, but for now I don't understand why it is missing files that are there.
You are executing the following mv commands:
mv webcam99.jpg webcam100.jpg
...
mv webcam1.jpg webcam2.jpg
The mv webcam0.jpg to webcam1.jpg is missing. With the first change to "$dir" you have the following files in /foo2:
webcam99.jp
...
webcam2.jpg
webcam0.jpg
With subsequent "$dir" change you will have the following:
webcam99.jp
...
webcam3.jpg
webcam0.jpg
In other words -- you are forgetting to move webcam0.jpg to webcam1.jpg. I would modify your script like this:
rm webcam99.jpg
for i in {98..0}
do
j=$(($i+1))
f1a=".jpg"
f1="webcam$i$f1a"
f2="test"
f2="webcam$j$f1a"
mv $f1 $f2
done
mv dir1/*.jpg /f002/webcam0.jpg