Hi I just created a file by mistake, doing a tar actually, anyway the problem I have is that I can't remove that file. It is called --exclude-tag-under=hey.txt
I am trying to use rm -rf command but it doesn't do the trick. this is the output
[root]# rm -rf '--exclude-tag-under\=hey.txt'
rm: unrecognized option '--exclude-tag-under\=hey.txt'
Try 'rm --help' for more information.
the problem here is that the command rm is recognizing the file as a flag and thats a problem, I've tried also
rm -rf *hey.txt
but it doesnt work neither
I've also tried to change the name of the file but its the same problem
Prepend ./ like this: rm ./--exclude-tag-under\=hey.txt
When in doubt, check the man pages.
Running man rm will give you the rm man page, which, on Linux and OpenBSD (the ones I have tested) at least, will have a section saying:
To remove a file whose name starts with a '-', for example '-foo', use
one of these commands:
rm -- -foo
rm ./-foo
Use rm -- --exclude-tag-under=hey.txt
$ ls
--exclude-tag-under=hey.txt
test
$ rm -- --exclude-tag-under=hey.txt
$ ls
test
Given
A modern Linux/UNIX/OSX (w/ realpath)
bash 4+ (even on OSX)
Is
"$PWD" == "$(realpath .)"
Always true?
It's pretty easy to test that this is not always the case.
$ mkdir /tmp/realdir
$ cd /tmp/realdir
$ echo $PWD
/tmp/realdir
$ ln -s realdir /tmp/fakedir
$ cd /tmp/fakedir
$ echo $PWD
/tmp/fakedir
$ realpath .
/tmp/realdir
so no, $PWD is not always the same as $(realpath .).
The bash manual indicates that the PWD variable is set by the built-in cd command. the default behaviour of cd is:
symbolic links are followed by default or with the -L option
This means that if you cd into a symlink the variable gets resolved relative to the symlink, not relative to the physical path. You can change this behavior for a cd command by using the -P option. This will cause it to report the physical directory in the PWD variable:
$ cd -P /tmp/fakedir
$ echo $PWD
/tmp/realdir
You can change the default behavior of bash using the -P option:
$ set -P
$ cd /tmp/fakedir
$ echo $PWD
/tmp/realdir
$ set +P
$ cd /tmp/fakedir
$ echo $PWD
/tmp/fakedir
This is of course notwithstanding the fact that you can assign anything you want to the PWD variable after performing a cd and it takes that value:
$ cd /tmp/fakedir
$ PWD=/i/love/cake
$ echo $PWD
/i/love/cake
but that's not really what you were asking.
It is not necessarily the case even when symbolic links are not used and PWD is not set by the user:
vinc17#xvii:~$ mkdir my_dir
vinc17#xvii:~$ cd my_dir
vinc17#xvii:~/my_dir$ rmdir ../my_dir
vinc17#xvii:~/my_dir$ echo $PWD
/home/vinc17/my_dir
vinc17#xvii:~/my_dir$ realpath .
.: No such file or directory
Note that under zsh, ${${:-.}:A} still gives the same answer as $PWD (the zshexpn(1) man page says about the A modifier: "Note that the transformation takes place even if the file or any intervening directories do not exist.").
Note that however, $PWD contains obsolete information. Using it may be a bad idea if some other process can remove the directory. Consider the following script:
rm -rf my_dir
mkdir my_dir
cd my_dir
echo 1 > file
cat $PWD/file
rm -r ../my_dir
mkdir ../my_dir
echo 2 > ../my_dir/file
cat ./file
cat $PWD/file
rm -r ../my_dir
It will output:
1
cat: ./file: No such file or directory
2
i.e. $PWD/file has changed.
I am writing a bash script that creates a folder, and copies files to that folder. It works from the command line, but not from my script. What is wrong here?
#! /bin/sh
DIR_NAME=files
ROOT=..
FOOD_DIR=food
FRUITS_DIR=fruits
rm -rf $DIR_NAME
mkdir $DIR_NAME
chmod 755 $DIR_NAME
cp $ROOT/$FOOD_DIR/"*" $DIR_NAME/
I get:
cp: cannot stat `../food/fruits/*': No such file or directory
You got that exactly backwards -- everything except the * character should be double-quoted:
#!/bin/sh
dir_name=files
root=..
food_dir=food
fruits_dir=fruits
rm -rf "$dir_name"
mkdir "$dir_name"
chmod 755 "$dir_name"
cp "$root/$food_dir/"* "$dir_name/"
Also, as a matter of best-practice / convention, non-environment variable names should be lower case to avoid name conflicts with environment variables and builtins.
I want to copy a certain file to a location, irrespective of that file already exists in the destination or not. I'm trying to copy through shell script.But the file is not getting copied. I'm using the following command
/bin/cp -rf /source/file /destination
but that doesn't work.
Use
cp -fr /source/file /destination
this should probably solve the problem.
This question has been already discussed, however you can write a little script like this:
#!/bin/bash
if [ ! -d "$2" ]; then
mkdir -p "$2"
fi
cp -R "$1" "$2"
Explaining this script a little bit
#!/bin/bash: tells your computer to use the bash interpreter.
if [ ! -d "$2" ]; then: If the second variable you supplied does not already exist...
mkdir -p "$2": make that directory, including any parent directories supplied in the path.
Running mkdir -p one/two/three will make:
$ mkdir -p one/two/three
$ tree one
one/
└── two
└── three
If you don't supply the -p tag then you'll get an error if directories one and two don't exist:
$ mkdir one/two/three
mkdir: cannot create directory ‘one/two/three’: No such file or directory
fi: Closes the if statement.
cp -R "$1" "$2": copies files from the first variable you supplied to the directory of the second variable you supplied.
So if you ran script.sh mars pluto, mars would be the first variable ($1) and pluto would be the second variable ($2).
The -R flag means it does this recursively, so the cp command will go through all the files and folders from your first variable, and copy them to the directory of your second variable.
Your problem might be caused by an alias for cp command created in your system by default (you can see al your aliases by typing "alias").
For example, my system has the following alis by default: alias cp='cp -i', where -i overrides -f option, i.e. cp will always prompt for overwriting confirmation.
What you need in such case (that'll actually work even if you don't have an alias) is to feed "yes" to that confirmation. To do that simply modify your cp command to look like this:
yes | cp /source/file /destination
/bin/cp -rf src dst
or
/usr/bin/env cp -rf
So, if I'm in my home directory and I want to move foo.c to ~/bar/baz/foo.c , but those directories don't exist, is there some way to have those directories automatically created, so that you would only have to type
mv foo.c ~/bar/baz/
and everything would work out? It seems like you could alias mv to a simple bash script that would check if those directories existed and if not would call mkdir and then mv, but I thought I'd check to see if anyone had a better idea.
How about this one-liner (in bash):
mkdir --parents ./some/path/; mv yourfile.txt $_
Breaking that down:
mkdir --parents ./some/path
# if it doesn't work; try
mkdir -p ./some/path
creates the directory (including all intermediate directories), after which:
mv yourfile.txt $_
moves the file to that directory ($_ expands to the last argument passed to the previous shell command, ie: the newly created directory).
I am not sure how far this will work in other shells, but it might give you some ideas about what to look for.
Here is an example using this technique:
$ > ls
$ > touch yourfile.txt
$ > ls
yourfile.txt
$ > mkdir --parents ./some/path/; mv yourfile.txt $_
$ > ls -F
some/
$ > ls some/path/
yourfile.txt
mkdir -p `dirname /destination/moved_file_name.txt`
mv /full/path/the/file.txt /destination/moved_file_name.txt
Save as a script named mv.sh
#!/bin/bash
# mv.sh
dir="$2" # Include a / at the end to indicate directory (not filename)
tmp="$2"; tmp="${tmp: -1}"
[ "$tmp" != "/" ] && dir="$(dirname "$2")"
[ -a "$dir" ] ||
mkdir -p "$dir" &&
mv "$#"
Or put at the end of your ~/.bashrc file as a function that replaces the default mv on every new terminal. Using a function allows bash keep it memory, instead of having to read a script file every time.
function mvp ()
{
dir="$2" # Include a / at the end to indicate directory (not filename)
tmp="$2"; tmp="${tmp: -1}"
[ "$tmp" != "/" ] && dir="$(dirname "$2")"
[ -a "$dir" ] ||
mkdir -p "$dir" &&
mv "$#"
}
Example usage:
mv.sh file ~/Download/some/new/path/ # <-End with slash
These based on the submission of Chris Lutz.
You can use mkdir:
mkdir -p ~/bar/baz/ && \
mv foo.c ~/bar/baz/
A simple script to do it automatically (untested):
#!/bin/sh
# Grab the last argument (argument number $#)
eval LAST_ARG=\$$#
# Strip the filename (if it exists) from the destination, getting the directory
DIR_NAME=`echo $2 | sed -e 's_/[^/]*$__'`
# Move to the directory, making the directory if necessary
mkdir -p "$DIR_NAME" || exit
mv "$#"
It sounds like the answer is no :). I don't really want to create an alias or func just to do this, often because it's one-off and I'm already in the middle of typing the mv command, but I found something that works well for that:
mv *.sh shell_files/also_with_subdir/ || mkdir -p $_
If mv fails (dir does not exist), it will make the directory (which is the last argument to the previous command, so $_ has it). So just run this command, then up to re-run it, and this time mv should succeed.
The simpliest way to do that is:
mkdir [directory name] && mv [filename] $_
Let's suppose I downloaded pdf files located in my download directory (~/download) and I want to move all of them into a directory that doesn't exist (let's say my_PDF).
I'll type the following command (making sure my current working directory is ~/download):
mkdir my_PDF && mv *.pdf $_
You can add -p option to mkdir if you want to create subdirectories just like this: (supposed I want to create a subdirectory named python):
mkdir -p my_PDF/python && mv *.pdf $_
Making use of the tricks in "Getting the last argument passed to a shell script" we can make a simple shell function that should work no matter how many files you want to move:
# Bash only
mvdir() { mkdir -p "${#: -1}" && mv "$#"; }
# Other shells may need to search for the last argument
mvdir() { for last; do true; done; mkdir -p "$last" && mv "$#"; }
Use the command like this:
mvdir foo.c foo.h ~/some/new/folder/
rsync command can do the trick only if the last directory in the destination path doesn't exist, e.g. for the destination path of ~/bar/baz/ if bar exists but baz doesn't, then the following command can be used:
rsync -av --remove-source-files foo.c ~/bar/baz/
-a, --archive archive mode; equals -rlptgoD (no -H,-A,-X)
-v, --verbose increase verbosity
--remove-source-files sender removes synchronized files (non-dir)
In this case baz directory will be created if it doesn't exist. But if both bar and baz don't exist rsync will fail:
sending incremental file list
rsync: mkdir "/root/bar/baz" failed: No such file or directory (2)
rsync error: error in file IO (code 11) at main.c(657) [Receiver=3.1.2]
So basically it should be safe to use rsync -av --remove-source-files as an alias for mv.
The following shell script, perhaps?
#!/bin/sh
if [[ -e $1 ]]
then
if [[ ! -d $2 ]]
then
mkdir --parents $2
fi
fi
mv $1 $2
That's the basic part. You might want to add in a bit to check for arguments, and you may want the behavior to change if the destination exists, or the source directory exists, or doesn't exist (i.e. don't overwrite something that doesn't exist).
Sillier, but working way:
mkdir -p $2
rmdir $2
mv $1 $2
Make the directory with mkdir -p including a temporary directory that is shares the destination file name, then remove that file name directory with a simple rmdir, then move your file to its new destination.
I think answer using dirname is probably the best though.
This will move foo.c to the new directory baz with the parent directory bar.
mv foo.c `mkdir -p ~/bar/baz/ && echo $_`
The -p option to mkdir will create intermediate directories as required.
Without -p all directories in the path prefix must already exist.
Everything inside backticks `` is executed and the output is returned in-line as part of your command.
Since mkdir doesn't return anything, only the output of echo $_ will be added to the command.
$_ references the last argument to the previously executed command.
In this case, it will return the path to your new directory (~/bar/baz/) passed to the mkdir command.
I unzipped an archive without giving a destination and wanted to move all the files except demo-app.zip from my current directory to a new directory called demo-app. The following line does the trick:
mv `ls -A | grep -v demo-app.zip` `mkdir -p demo-app && echo $_`
ls -A returns all file names including hidden files (except for the implicit . and ..).
The pipe symbol | is used to pipe the output of the ls command to grep (a command-line, plain-text search utility).
The -v flag directs grep to find and return all file names excluding demo-app.zip.
That list of files is added to our command-line as source arguments to the move command mv. The target argument is the path to the new directory passed to mkdir referenced using $_ and output using echo.
Based on a comment in another answer, here's my shell function.
# mvp = move + create parents
function mvp () {
source="$1"
target="$2"
target_dir="$(dirname "$target")"
mkdir --parents $target_dir; mv $source $target
}
Include this in .bashrc or similar so you can use it everywhere.
Code:
if [[ -e $1 && ! -e $2 ]]; then
mkdir --parents --verbose -- "$(dirname -- "$2")"
fi
mv --verbose -- "$1" "$2"
Example:
arguments: "d1" "d2/sub"
mkdir: created directory 'd2'
renamed 'd1' -> 'd2/sub'
((cd src-path && tar --remove-files -cf - files-to-move) | ( cd dst-path && tar -xf -))
I frequently stumble upon this issue while bulk moving files to new subdirectories. Ideally, I want to do this:
mv * newdir/
Most of the answers in this thread propose to mkdir and then mv, but this results in:
mkdir newdir && mv * newdir
mv: cannot move 'newdir/' to a subdirectory of itself
The problem I face is slightly different in that I want to blanket move everything, and, if I create the new directory before moving then it also tries to move the new directory to itself. So, I work around this by using the parent directory:
mkdir ../newdir && mv * ../newdir && mv ../newdir .
Caveats: Does not work in the root folder (/).
My one string solution:
test -d "/home/newdir/" || mkdir -p "/home/newdir/" && mv /home/test.txt /home/newdir/
i accomplished this with the install command on linux:
root#logstash:# myfile=bash_history.log.2021-02-04.gz ; install -v -p -D $myfile /tmp/a/b/$myfile
bash_history.log.2021-02-04.gz -> /tmp/a/b/bash_history.log.2021-02-04.gz
the only downside being the file permissions are changed:
root#logstash:# ls -lh /tmp/a/b/
-rwxr-xr-x 1 root root 914 Fev 4 09:11 bash_history.log.2021-02-04.gz
if you dont mind resetting the permission, you can use:
-g, --group=GROUP set group ownership, instead of process' current group
-m, --mode=MODE set permission mode (as in chmod), instead of rwxr-xr-x
-o, --owner=OWNER set ownership (super-user only)
There's a lot of conflicting solutions around for this, here's what worked for us:
## ss_mv ##
function ss_mv {
mkdir -p $(dirname "$2") && mv -f "$#"
}
This assumes commands in the following syntax:
ss_mv /var/www/myfile /var/www/newdir/myfile
In this way the directory path /var/www/newdir is extracted from the 2nd part of the command, and that new directory is then created (it's critical that you use the dirname tag to avoid myfile being added to the new directory being created).
Then we go ahead and mv on the entire string again by using the "$#" tag.
You can even use brace extensions:
mkdir -p directory{1..3}/subdirectory{1..3}/subsubdirectory{1..2}
which creates 3 directories (directory1, directory2, directory3),
and in each one of them two subdirectories (subdirectory1, subdirectory2),
and in each of them two subsubdirectories (subsubdirectory1 and subsubdirectory2).
You have to use bash 3.0 or newer.
$what=/path/to/file;
$dest=/dest/path;
mkdir -p "$(dirname "$dest")";
mv "$what" "$dest"