I have directory work_dir, and there are some subdirectories inside. And inside subdirectories there are zip archives. I can see all zip archives in terminal:
find . -name *.zip
The output:
./folder2/sub/dir/test2.zip
./folder3/test3.zip
./folder1/sub/dir/new/test1.zip
Now I want to concatinate all these file names in single row with some option. For example I want single row:
my_command -f ./folder2/sub/dir/test2.zip -f ./folder3/test3.zip -f ./folder1/sub/dir/new/test1.zip -u user1 -p pswd1
In this example:
my_command is some command
-f the option
-u user1 another option with value
-p pswd1 another option with value
Can you help me please, how can I do this in Linux BASH ?
One way is: (updated per #M. Nejat Aydin comments)
find . -name "*.zip" -print0 | xargs -0 -n1 printf -- '-f\0%s\0' | xargs -0 -n100000 my_command -u user1 -p pswd1
Note that -n100000 parameter forces all output of the previous xargs to be executed on the same line with the assumption that number of findings will be less than 100000.
I used null terminated versions (notice: -0 flag, -print0) because file names can contain spaces.
This is a bash script that should do what you wanted.
#!/usr/bin/env bash
user=user1
passwd=pswd1
while IFS= read -rd '' files; do
args+=(-f "$files")
done < <(find . -name '*.zip' -print0)
args=("${args[#]}" -u "$user" -p "$passwd")
##: Just for the human eye to see the output,
##: change this line of code according to the comment below.
printf 'mycommand %s\n' "${args[*]}"
The output should be in one-line, like what you wanted, but do change the last line from
printf 'mycommand %s\n' "${args[*]}"
into
mycommand "${args[#]}"
If you actually want to execute mycommand with the arguments.
Change the value of user and passwd too.
A while + read loop was used with IFS.
See How can I read a file (data stream, variable) line-by-line (and/or field-by-field)?
Why the last line should be change.
See Arguments
Shell quoting is a basic but common mistake when dealing with spaces in file/path name.
See How can I find and safely handle file names containing
Also the find command/utiliy.
The construct "${args[#}" is an array.
See Array1 Array2 Array3
You can do this by making a bash script.
Make a new file called whatever.sh
Type chmod +x ./whatever.sh so it becomes executable on the terminal
Add the BASH scripting as shown below..
#!/bin/bash
# Get all the zip files from your FolderName
files="`find ./FolderName -name *.zip`"
# Loop through the files and build your args
arg=""
for file in $files; do
arg="$arg -f $file"
done
# Run your command
mycommand $arg -u user1 -p pswd1
We give the upmost permission to a file or directory, using this command:
sudo chmod -R 777 directory
Now I want to know if this command is already executed for a given directory.
I know I can use -r for read, -w for write, and -x for execution, in [ test ] blocks.
But I want to know two things:
Is it also a directory?
Does it have those permissions for everyone?
How can I get that info?
Update
Based on #Barmar comment, I came up with this. But it's not working:
if [ stat /Temp | grep -oP "(?<=Access: \()[^)]*" == '' ]; then
echo '/Temp folder has full access'
else
sudo chmod -R 777 /Temp
fi
This command works though:
stat /Temp | grep -oP "(?<=Access: \()[^)]*"
# prints => 0777/drwxrwxrwx
How should I fix the syntax error of my if-else statement?
You don't need to process the output of stat with grep; you can ask stat to only produce the specific information you want. See the man page regarding the --format option. We can for example write:
# ask stat for the file type and mode, and put those values into $1
# and $2
set -- $(stat --format '%F %a' /Temp)
if [[ $1 == directory ]]; then
if [[ $2 == 777 ]]; then
echo "/Temp folder has full access"
else
sudo chmod -R 777 /Temp
fi
else
echo "ERROR: /Temp is not a directory!" >&2
fi
A simple example:
#!/bin/bash
function setfullperm(){
[ -d $1 ] && \
(
[ "$(stat --format '%a' $1)" == "777" ] && \
echo "Full permissions are applied." || \
( echo "Setting full permissions" && sudo chmod -R 777 $1 )
) || \
( echo "$1 is not a directory !" && mkdir $1 && setfullperm $1 )
}
export setfullperm
Source the script:
$ source example.sh
Set full permissions (777) on any directory, it tests if the directory exists in the first place, if not it will create it and set the permissions.
It will export the function setfullperm to the shell so you can run it:
>$ setfullperm ali
ali is not a directory !
mkdir: created directory 'ali'
Setting full permissions
>$ setfullperm ali
Full permissions are applied.
If using zsh (But not other shells), you can do it with just a glob pattern:
setopt extended_glob null_glob
if [[ -n /Temp(#q/f777) ]]; then
echo '/Temp folder has full access'
else
sudo chmod -R 777 /Temp
fi
The pattern /Temp(#q/f777) will, with the null_glob and extended_glob options set, expand to an empty string if /Temp is anything but a directory with the exact octal permissions 0777 (And to /Temp if the criteria are met). For more details, see Glob Qualifiers in the zsh manual.
I don't recommend using stat for this. Though widespread, stat isn't POSIX, which means there's no guarantee that your script will work in the future or work on other platforms. If you're writing scripts for a production environment, I'd urge you to consider a different approach.
You're better off using ls(1)'s -l option and passing the file as an argument. From there you can use cut(1)'s -c option to grab the file mode flags.
Get file type:
ls -l <file> | cut -c1
Also, don't forget about test's -d operator, which tests if a file is a directory.
Get owner permissions:
ls -l <file> | cut -c2-4
and so on.
This approach is POSIX compliant and it avoids the shortcomings of using stat.
I am trying to ls the directories and print them out but nothing is being displayed. I am able to SSH and execute the first pwd. However, anything within the the for loop has no output. I know for sure there are directories called event-test- because I've done so manually. I've manually entered the directory (/data/kafka/tmp/kafka-logs/) and ran this piece of code and the correct output appeared so I'm not sure why
manually entered has correct output:
for i in `ls | grep "event-test"`; do echo $i; done;
script:
for h in ${hosts[*]}; do
ssh -i trinity-prod-keypair.pem bc2-user#$h << EOF
sudo bash
cd /data/kafka/tmp/kafka-logs/
pwd
for i in `ls | grep "event-test-"`; do
pwd
echo $i;
done;
exit;
exit;
EOF
done
It is because
`ls | grep "event-test-"`
is executing on your localhost not on remote host. Besides parsing ls is error prone and not even needed. You can do:
for h in "${hosts[#]}"; do
ssh -t -t trinity-prod-keypair.pem bc2-user#$h <<'EOF'
sudo bash
cd /data/kafka/tmp/kafka-logs/
pwd
for i in *event-test-*; do
pwd
echo "$i"
done
exit
exit
EOF
done
When parsing ls it is good practice to do ls -1 to get a prettier list to parse. Additionally, when trying to find files named "event-test-" I would recommend the find command. Since I am not completely sure what you are attempting to do other than list the locations of these "event-test" files I'd recommend something more similar to the following:
for h in "${hosts[#]}"; do ssh trinity-prod-keypair.pem bc2-user#$h -t -t "find /data/kafka/tmp/kafka-logs/ -type f -name *event-test-*;" ; done
This will give you a pretty output of the full path to the file and the file name.
I hope this helps.
I need to create a new empty file with the same permissions and ownership (group and user) as the source, kind of like how cp -p FILE1 FILE1.bak works but without actually copying the contents.
I know I can empty out the contents later on, but that seems wasteful.
I cant use a script - the solution must run from the command line directly.
touch newfile
chmod `stat -c %a originalfile` newfile
chown `stat -c %U originalfile`:`stat -c %G originalfile` newfile
Use
touch newfile
chmod --reference=oldfile newfile
chown --reference=oldfile newfile
cp --attributes-only --preserve=ownership
Use the touch command.
http://www.computerhope.com/unix/utouch.htm
Is there an easy way I can print the full path of file.txt ?
file.txt = /nfs/an/disks/jj/home/dir/file.txt
The <command>
dir> <command> file.txt
should print
/nfs/an/disks/jj/home/dir/file.txt
Use readlink:
readlink -f file.txt
I suppose you are using Linux.
I found a utility called realpath in coreutils 8.15.
realpath -s file.txt
/data/ail_data/transformed_binaries/coreutils/test_folder_realpath/file.txt
Since the question is about how to get the full/absolute path of a file and not about how to get the target of symlinks, use -s or --no-symlinks which means don't expand symlinks.
As per #styrofoam-fly and #arch-standton comments, realpath alone doesn't check for file existence, to solve this add the e argument: realpath -e file
The following usually does the trick:
echo "$(cd "$(dirname "$1")" && pwd -P)/$(basename "$1")"
I know there's an easier way that this, but darned if I can find it...
jcomeau#intrepid:~$ python -c 'import os; print(os.path.abspath("cat.wav"))'
/home/jcomeau/cat.wav
jcomeau#intrepid:~$ ls $PWD/cat.wav
/home/jcomeau/cat.wav
On Windows:
Holding Shift and right clicking on a file in Windows Explorer gives you an option called Copy as Path.
This will copy the full path of the file to clipboard.
On Linux:
You can use the command realpath yourfile to get the full path of a file as suggested by others.
find $PWD -type f | grep "filename"
or
find $PWD -type f -name "*filename*"
If you are in the same directory as the file:
ls "`pwd`/file.txt"
Replace file.txt with your target filename.
I know that this is an old question now, but just to add to the information here:
The Linux command which can be used to find the filepath of a command file, i.e.
$ which ls
/bin/ls
There are some caveats to this; please see https://www.cyberciti.biz/faq/how-do-i-find-the-path-to-a-command-file/.
You could use the fpn (full path name) script:
% pwd
/Users/adamatan/bins/scripts/fpn
% ls
LICENSE README.md fpn.py
% fpn *
/Users/adamatan/bins/scripts/fpn/LICENSE
/Users/adamatan/bins/scripts/fpn/README.md
/Users/adamatan/bins/scripts/fpn/fpn.py
fpn is not a standard Linux package, but it's a free and open github project and you could set it up in a minute.
Works on Mac, Linux, *nix:
This will give you a quoted csv of all files in the current dir:
ls | xargs -I {} echo "$(pwd -P)/{}" | xargs | sed 's/ /","/g'
The output of this can be easily copied into a python list or any similar data structure.
echo $(cd $(dirname "$1") && pwd -P)/$(basename "$1")
This is explanation of what is going on at #ZeRemz's answer:
This script get relative path as argument "$1"
Then we get dirname part of that path (you can pass either dir or file to this script): dirname "$1"
Then we cd "$(dirname "$1") into this relative dir
&& pwd -P and get absolute path for it. -P option will avoid all symlinks
After that we append basename to absolute path: $(basename "$1")
As final step we echo it
You may use this function. If the file name is given without relative path, then it is assumed to be present in the current working directory:
abspath() { old=`pwd`;new=$(dirname "$1");if [ "$new" != "." ]; then cd $new; fi;file=`pwd`/$(basename "$1");cd $old;echo $file; }
Usage:
$ abspath file.txt
/I/am/in/present/dir/file.txt
Usage with relative path:
$ abspath ../../some/dir/some-file.txt
/I/am/in/some/dir/some-file.txt
With spaces in file name:
$ abspath "../../some/dir/another file.txt"
/I/am/in/some/dir/another file.txt
You can save this in your shell.rc or just put in console
function absolute_path { echo "$PWD/$1"; }
alias ap="absolute_path"
example:
ap somefile.txt
will output
/home/user/somefile.txt
I was surprised no one mentioned located.
If you have the locate package installed, you don't even need to be in the directory with the file of interest.
Say I am looking for the full pathname of a setenv.sh script. This is how to find it.
$ locate setenv.sh
/home/davis/progs/devpost_aws_disaster_response/python/setenv.sh
/home/davis/progs/devpost_aws_disaster_response/webapp/setenv.sh
/home/davis/progs/eb_testy/setenv.sh
Note, it finds three scripts in this case, but if I wanted just one I
would do this:
$ locate *testy*setenv.sh
/home/davis/progs/eb_testy/setenv.sh
This solution uses commands that exist on Ubuntu 22.04, but generally exist on most other Linux distributions, unless they are just to hardcore for s'mores.
The shortest way to get the full path of a file on Linux or Mac is to use the ls command and the PWD environment variable.
<0.o> touch afile
<0.o> pwd
/adir
<0.o> ls $PWD/afile
/adir/afile
You can do the same thing with a directory variable of your own, say d.
<0.o> touch afile
<0.o> d=/adir
<0.o> ls $d/afile
/adir/afile
Notice that without flags ls <FILE> and echo <FILE> are equivalent (for valid names of files in the current directory), so if you're using echo for that, you can use ls instead if you want.
If the situation is reversed, so that you have the full path and want the filename, just use the basename command.
<0.o> touch afile
<0.o> basename $PWD/afile
afile
In a similar scenario, I'm launching a cshell script from some other location. For setting the correct absolute path of the script so that it runs in the designated directory only, I'm using the following code:
set script_dir = `pwd`/`dirname $0`
$0 stores the exact string how the script was executed.
For e.g. if the script was launched like this: $> ../../test/test.csh,
$script_dir will contain /home/abc/sandbox/v1/../../test
For Mac OS X, I replaced the utilities that come with the operating system and replaced them with a newer version of coreutils. This allows you to access tools like readlink -f (for absolute path to files) and realpath (absolute path to directories) on your Mac.
The Homebrew version appends a 'G' (for GNU Tools) in front of the command name -- so the equivalents become greadlink -f FILE and grealpath DIRECTORY.
Instructions for how to install the coreutils/GNU Tools on Mac OS X through Homebrew can be found in this StackExchange arcticle.
NB: The readlink -f and realpath commands should work out of the box for non-Mac Unix users.
I like many of the answers already given, but I have found this really useful, especially within a script to get the full path of a file, including following symlinks and relative references such as . and ..
dirname `readlink -e relative/path/to/file`
Which will return the full path of the file from the root path onwards.
This can be used in a script so that the script knows which path it is running from, which is useful in a repository clone which could be located anywhere on a machine.
basePath=`dirname \`readlink -e $0\``
I can then use the ${basePath} variable in my scripts to directly reference other scripts.
Hope this helps,
Dave
This worked pretty well for me. It doesn't rely on the file system (a pro/con depending on need) so it'll be fast; and, it should be portable to most any *NIX. It does assume the passed string is indeed relative to the PWD and not some other directory.
function abspath () {
echo $1 | awk '\
# Root parent directory refs to the PWD for replacement below
/^\.\.\// { sub("^", "./") } \
# Replace the symbolic PWD refs with the absolute PWD \
/^\.\// { sub("^\.", ENVIRON["PWD"])} \
# Print absolute paths \
/^\// {print} \'
}
This is naive, but I had to make it to be POSIX compliant. Requires permission to cd into the file's directory.
#!/bin/sh
if [ ${#} = 0 ]; then
echo "Error: 0 args. need 1" >&2
exit 1
fi
if [ -d ${1} ]; then
# Directory
base=$( cd ${1}; echo ${PWD##*/} )
dir=$( cd ${1}; echo ${PWD%${base}} )
if [ ${dir} = / ]; then
parentPath=${dir}
else
parentPath=${dir%/}
fi
if [ -z ${base} ] || [ -z ${parentPath} ]; then
if [ -n ${1} ]; then
fullPath=$( cd ${1}; echo ${PWD} )
else
echo "Error: unsupported scenario 1" >&2
exit 1
fi
fi
elif [ ${1%/*} = ${1} ]; then
if [ -f ./${1} ]; then
# File in current directory
base=$( echo ${1##*/} )
parentPath=$( echo ${PWD} )
else
echo "Error: unsupported scenario 2" >&2
exit 1
fi
elif [ -f ${1} ] && [ -d ${1%/*} ]; then
# File in directory
base=$( echo ${1##*/} )
parentPath=$( cd ${1%/*}; echo ${PWD} )
else
echo "Error: not file or directory" >&2
exit 1
fi
if [ ${parentPath} = / ]; then
fullPath=${fullPath:-${parentPath}${base}}
fi
fullPath=${fullPath:-${parentPath}/${base}}
if [ ! -e ${fullPath} ]; then
echo "Error: does not exist" >&2
exit 1
fi
echo ${fullPath}
This works with both Linux and Mac OSX:
echo $(pwd)$/$(ls file.txt)
find / -samefile file.txt -print
Will find all the links to the file with the same inode number as file.txt
adding a -xdev flag will avoid find to cross device boundaries ("mount points"). (But this will probably cause nothing to be found if the find does not start at a directory on the same device as file.txt)
Do note that find can report multiple paths for a single filesystem object, because an Inode can be linked by more than one directory entry, possibly even using different names. For instance:
find /bin -samefile /bin/gunzip -ls
Will output:
12845178 4 -rwxr-xr-x 2 root root 2251 feb 9 2012 /bin/uncompress
12845178 4 -rwxr-xr-x 2 root root 2251 feb 9 2012 /bin/gunzip
Usually:
find `pwd` | grep <filename>
Alternatively, just for the current folder:
find `pwd` -maxdepth 1 | grep <filename>
This will work for both file and folder:
getAbsolutePath(){
[[ -d $1 ]] && { cd "$1"; echo "$(pwd -P)"; } ||
{ cd "$(dirname "$1")" || exit 1; echo "$(pwd -P)/$(basename "$1")"; }
}
Another Linux utility, that does this job:
fname <file>
For Mac OS, if you just want to get the path of a file in the finder, control click the file, and scroll down to "Services" at the bottom. You get many choices, including "copy path" and "copy full path". Clicking on one of these puts the path on the clipboard.
fp () {
PHYS_DIR=`pwd -P`
RESULT=$PHYS_DIR/$1
echo $RESULT | pbcopy
echo $RESULT
}
Copies the text to your clipboard and displays the text on the terminal window.
:)
(I copied some of the code from another stack overflow answer but cannot find that answer anymore)
In Mac OSX, do the following steps:
cd into the directory of the target file.
Type either of the following terminal commands.
Terminal
ls "`pwd`/file.txt"
echo $(pwd)/file.txt
Replace file.txt with your actual file name.
Press Enter
you#you:~/test$ ls
file
you#you:~/test$ path="`pwd`/`ls`"
you#you:~/test$ echo $path
/home/you/test/file
Beside "readlink -f" , another commonly used command:
$find /the/long/path/but/I/can/use/TAB/to/auto/it/to/ -name myfile
/the/long/path/but/I/can/use/TAB/to/auto/it/to/myfile
$
This also give the full path and file name at console
Off-topic: This method just gives relative links, not absolute. The readlink -f command is the right one.