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"
Related
I want to exclude some directory in my script (directory name >1000) for deletion and here is my directory look like:
/home/tester/100
/home/tester/1000
/home/tester/1020 # delete all files inside
/home/tester/2000 # delete all files inside
My bash script:
cd /home/tester
for dir in */ ; do
echo -n $dir": ";
find "$dir" -type f | wc -l;
if [ $dir -gt 1000 ]; then
cd $dir;
rm *;
cd ..;
fi
done
I got error on the if line and have no idea how to fix it ... Is it possible to do with bash script ?
Thank you for your help
for dir in */ ; do will set dir to things like "1000/" -- and the "/" makes it not a valid number. You can trim off the trailing "/" with ${dir%/}. I'd also recommend double-quoting it to prevent possible weird parsing:
if [ "${dir%/}" -gt 1000 ]; then
Note that if the directory name isn't a number (even after the "/" is removed), you'll get an error from the comparison, and the then clause won't run (which is probably what you want). If you want to handle other (non-numeric) directory names more gracefully, you should add some appropriate is-this-a-number test first.
Also, using cd in scripts tends to be problematic, because if a cd fails for any reason, the rest of the script will continue running, but in the wrong place. This can cause all sorts of chaos. Consider what'd happen if one of the cd $dir commands fails: it'd run rm * in the /home/tester directory, deleting all the non-subdirectory files there, then it'd cd .., leaving it in /home. The next iteration would try to cd down to something like 2000, which doesn't exist under /home, so that cd would fail too, and then it'd delete all files in /home. This repeats indefinitely, potentially all the way up to running rm * in /, the root directory. Not good at all.
I recommend either putting error checks on cd commands, or just avoiding them entirely in favor of using explicit paths to files.
#!/bin/bash
cd /home/tester || {
echo "Couldn't cd to /home/tester, quitting here..." >&2
exit 1
}
for dir in */ ; do
echo -n "$dir: "
find "$dir" -type f | wc -l
if [ "${dir%/}" -gt 1000 ]; then
rm "$dir"/* # Explicit path -- the / is redundant, but won't hurt
fi
done
I've also added an explicit shebang line, double-quoted all the variable references (good general scripting hygiene), and removed the semicolons from the ends of lines (not needed in shell syntax).
Another recommendation: run your scripts through shellcheck.net -- it'll point out a lot of common mistakes like unquoted variable references and unchecked cds.
The value of $dir is not numeric. Add set -x at the to of your script to debug.
Use "$(basename "$dir")" to get the numeric value.
When I did not have my first cup of coffee, I would do
for dir in */ ; do
echo -n $dir": ";
find "$dir" -type f | wc -l;
done
mv /home/tester/1000 /home/tester/some_unique_name
rm /home/tester/[1-9][0-9][0-9][0-9]/*
mv /home/tester/some_unique_name /home/tester/1000
This will not work when you have directories > 9999.
Perhaps rm /home/tester/[1-9][0-9][0-9][0-9]*/* will work, when you don't have directories like 1000backup or 2000my_unique_name.
A better solution is
find . -regextype sed -regex '/home/tester/[0-9]\{4,\}' ! -name 1000 |
xargs -L1 -I{} echo rm {}/*
I have some files in Pictures\ with extension *.png and directories like 12-21-20, 12-20-20. These directories was created with dir=mkdir $(date +'%m'-'%d'-'%Y')
At the end of the day I want to run a script which will create a folder $dir and copy all png files I've made for today into that folder. How can I do that? Any information you could give me would be greatly appreciated.
date +'%m-%d-%Y' is the date command outputting, e,g. 12-22-2020. $(..) is called command substitution that captures the result of the date command allowing it to be assigned to the variable dir.
To create a directory with the contents of $dir (e.g. 12-22-2020) you would use the mkdir command, providing the -p option to suppress the error if that directory already exists (and also create parent directories as necessary -- not relevant here). You want to ensure it succeeds before you attempt to copy files to the new directory, so you would use:
mkdir -p "$dir" || exit 1
Which simply exits if the command fails.
At this point, you can simply use cp (or preferably mv to move the files) from whatever source directory they currently reside in. That you can do with:
mv /path/to/source/dir/*.png "$dir"
Or the copy command,
cp -a /path/to/source/dir/*.png "$dir"
Both cp -a and mv will preserve the original file attributes (time, date, permissions, etc...).
From a script standpoint, you will either want to change to the directory above the new "$dir" or use the full path, e.g.
mv /path/to/source/dir/*.png "/path/to/$dir"
Short Example
If you want to provide the directory containing the .png files to move to "$dir" created with today's date, you could write a short script like the following. You provide the directory containing the .png files you would like to copy or move as the first argument to the script on the command-line, e.g. usage would be bash pngscript.sh /path/to/source/dir.
#!/bin/bash
[ -z "$1" ] && { ## validate one argument given for source directory
printf "usage: ./%s /path/to/images" "${0##*/}" >&2
exit 1
}
[ "$(ls -1 "$1"/*.png | wc -l)" -gt 0 ] || { ## validate png files in source dir
printf "error: no .png files in '%s'\n" "$1" >&2
exit 1
}
dir=$(date +'%m-%d-%Y') ## get current date
mkdir -p "$dir" || exit 1 ## create directory, exit on failure
mv "$1"/*.png "$dir" ## move .png files from source to "$dir"
(note: it will create the "$dir" directory below the current working directory and then move files from the path provided as the first argument (positional parameter) to the newly created directory. Change mv to cp -a if you want to leave a copy of the files in the original directory)
You can make the script executable with chmod +x pngscript.sh and then you can simply run it from the current directory as:
./pngscript.sh /path/to/source/dir
Let me know if you have further questions.
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 ($2).
The -R flag means it does this recursively, so the cp command will go through all the files and folders from your first variable, and copy them to the directory of your second variable.
Your problem might be caused by an alias for cp command created in your system by default (you can see al your aliases by typing "alias").
For example, my system has the following alis by default: alias cp='cp -i', where -i overrides -f option, i.e. cp will always prompt for overwriting confirmation.
What you need in such case (that'll actually work even if you don't have an alias) is to feed "yes" to that confirmation. To do that simply modify your cp command to look like this:
yes | cp /source/file /destination
/bin/cp -rf src dst
or
/usr/bin/env cp -rf
I have several hundred rar files. I would like to create a directory for each rar file then move the file into the newly created directory.
This is the code I am using to create the rar's
#!bin/bash
for f in *; do
rar a -s -m5 "${f%.*}.rar" "$f";
done
This is the code I am using to move the files.
#!/bin/bash
for i in *.rar; do
dir=$(echo "$i" | \
sed 's/\(.\)\([^ ]\+\) \([^ ]\+\) - \(.*\)\.pdf/\1\/\1\2 \3/')
dir="DestinationDirectory/$dir"
mkdir -p -- "$dir" && mv -uv "$i" "$dir/$i"
done
The problem is that it creates the directory with the extension name.
ie: file irclog3_26_198.rar is moved into folder /DestinationDirectory/irclog3_26_1988.rar/irclog3_26_1988.rar
I would like the folder to be created ignoring the .rar and just use the name of the file.
How about:
dir="${dir%.rar}"
mkdir -p -- "$dir" ...
Read more about it at the abs.
dir=$(echo ${i[#]::-4})
${name[#]:pos:len}) gets the substring/subarray of the string/array, for string, [#] can be avoid.
dir=$(echo ${i::-4})
You can use the normal bash shell parameter expension https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html
FILE="TEST.rar"
echo "${FILE%%.*}"
--> TEST
zip -r 1.zip /home/username/the_folder
At here, when i unzip 1.zip, it will create /home/username/the_folder, from whichever folder i am unzipping from.
How do I zip, stating the full absolute paths, but make the zip only contain the folder structure starting at, in this case for instance, /home/username?
That way I could be at whatever path i wanted, unzip and it would just create the_folder, and not /home/username/the_folder.
Use this command:
cd path_under_folder_to_zip && \
zip -r 1.zip folder_to_zip >/dev/null && \
mv 1.zip my_current_path
Use relative path when specifying the file to zip.
cd /home/username
zip -r 1.zip ./the_folder
Then when you unzip, it'll be a relative path starting at whichever folder you're in when unzipping.
Just use the -j option, works on OSX, I don't know about linux.
zip -j -r 1.zip /home/username/the_folder
List item
How about this:
function zipExtraFolder {
if [ $# -lt 2 ]; then
echo "provide at least two arguments"
return
fi
folder=$2
mkdir del
echo cp -r `dirname $folder` del
cd del
echo zip -r ../$1 .
cd -
rm -rf del
}
Define above as a shell function in your .bashrc and you should be able to use it whenever you want.
The usage will be like below.
zipExtraFolder 1.zip /home/username/the_folder