I have a desktop shortcut e.g /home/user/Desktop/myfolder/link.desktop which invokes a bash script located somewhere else, e.g. /tmp/myscript.sh
Within my script, how can I find the path of the shortcut which invoked my script? Is it possible at all?
What I actually want to achieve is that there is a subfolder where the shortcut link is, e.g. /home/user/Desktop/myfolder/subfolder. And in my script I would like to be able to access the subfolder.
I have tried readlink -f but that will always return /home/user no matter where the shortcut icon lies.
I cannot set the work path as the shortcut link is generated and dynamically placed in different locations.
Well at least you can do something like this in your script:
find / -iname '*desktop' -exec fgrep -l $0 \{\} \; 2>/dev/null
That will travell your filesystem and searches every .desktop file for your script in it... But note, this can be misleading, as Someone can put comments in a .desktop... so you might create a searchstring first like ^Exec=/PATH/TO/$0 and use egrep instead of fgrep.
Or you can do a copy function which edits the .desktop files when it copies it to it's location and adds it's new location as a parameter to the Exec line, e.g.:
mycp() {
sed "s/^Exec=.*/& $2/" $1 > $2
}
Or (and I'd go with it) use the %k param in your Exec line, according to the spec.
Related
I'm using the newest version of macOS Monterey.
Sometimes I have to merge/combine all files that are in a specific folder into one txt file.
I currently do that by typing this in Terminal:
cd /Users/my_name/Desktop/test_folder ; cat * >merged.txt
This will merge/combine all files in folder test_folder into one file called merged.txt. The file merged.txt will get saved into the folder test_folder.
Every time I need this I have to open Terminal copy/paste the command and replace test_folder with the right folder name, since it's not always the same.
I want to make this easier by just make a right click on a folder, go to Quick Actions and select e.g. Merge all files to merge/combine all files inside the folder I just clicked on.
But I stuck at getting the folder name. How can I dynamically get the folder name and path I clicked on to start this Quick Action instead of the hard coded /Users/my_name/Desktop/test_folder?
Or, is there another and easier solution?
This is what I have so far:
I wouldn’t do this with AppleScript, especially if all it’s ultimately doing is calling out to a shell script.
Stick with the Run Shell Script action except change the option for passing the input as arguments rather than to stdin.
The folders selected in Finder will then be available to your script via $#, so you can do something like:
for d in "$#"; do
cat "$d"/* > "$d/merged.txt"
open -R "$d/merged.txt"
done 2>/dev/null
This loops through the selected directories and concatenates the files to merged.txt in the respective directory. The open -R line reveals the merged.txt file in Finder.
Errors are written to /dev/null, i.e. discarded, as cat will throw an error if any of the directories, themselves, contain directories.
Instead of adding a Run Shell Script to your workflow, try adding a Run AppleScript command instead. Copy this following AppleScript code to the Run AppleScript command.
on run {input, parameters}
try
do shell script "cd " & quoted form of POSIX path of input & " && cat *.txt > merged.txt"
on error
try
do shell script "cd " & quoted form of POSIX path of input & " && rm merged.txt"
end try
end try
end run
I have a c5 project with a directory C5PROJECTDIR it has a bin directory. Is there a way (without adding C5PROJECTDIR/bin to $PATH) to access commands from any other subdirectory of C5PROJECTDIR?
For example:
There is an x command located here: C5PROJECTDIR/bin/x.
I want to run this command from a bash script from C5PROJECTDIR/packages/some-other-dir.
UPDATE: The sub directory can be more or less deeper in the tree...
How can it be possible?
There are a few ways:
Use relative paths:
Essentially call the binary directly with its path, but relative to the current script. If your script is located in /home/user/foo/C5PROJECTDIR/packages/somedir/script, then you call the script relatively:
#!/usr/bin/env bash
../../bin/x
The problem with this method is maintainability. Imagine you want to move your script to /home/user/foo/C5PROJECTDIR/packages/scripts, then you would have to update all your scripts.
Use a shell variable:
Instead of relative paths, you can define a variable PROJECTHOME which contains the base value.
#!/usr/bin/env bash
PROJECTHOME=/home/user/foo/C5PROJECTDIR
$PROJECTHOME/bin/x
This generally solves most problems unless you move project location from /home/user/foo/C5PROJECTDIR to /home/user/random-dir. But this can be solved with a simple sed command that searches for the line /home/user/foo/C5PROJECTDIR and replaces it.
PATH variable only make things easier.
You able to run any commands with absolute path like /bin/echo or /home/jon/myscript. Relative path works as well ~/../../bin/ls.
Script have its own PATH variable, so you can add the desired location to PATH in your script if you like.
Apart from this you have to deal with permissions.
The user that have script executed by, must have execute permission on the x script.
To find and execute command:
find C5PROJECTDIR -type f -executable -name "x" -exec {} \;
Find's must be strict to match only one command else it will execute all that it fund. You cannot have script with the same name under C5PROJECTDIR in this case.
From security point of view. I do not recommend this, as anyone can put an executable file under C5PROJECTDIR with a name used in the script. things can go nasty.
I just during the weekend decided to try out zsh and have a bit of fun with it. Unfortunately I'm an incredible newbie to shell scripting in general.
I have this folder with a file, which filename is a hash (4667e85581f80b6936f8811f0a7493c70eae4ee7) without a file-extension.
What I would like to do is copy this file to another folder and rename it to "screensaver.png".
I've tried with the following code:
#!/usr/bin/zsh
KUVVA_CACHE="$HOME/Library/Containers/com.kuvva.Kuvva-Wallpapers/Data/Library/Application Support/Kuvva"
DEST_FOLDER="/Library/Desktop Pictures/Kuvva/$USERNAME/screensaver.png"
for wallpaper in ${KUVVA_CACHE}; do
cp -f ${wallpaper} ${DEST_FOLDER}
done
This returns the following error:
cp: /Users/Morten/Library/Containers/com.kuvva.Kuvva-Wallpapers/Data/Library/Application Support/Kuvva is a directory (not copied).
And when I try to echo the $wallpaper variable instead of doing "cp" then it just echo's the folder path.
The name of the file changes every 6 hour, which is why I'm doing the for-loop. So I never know what the name of the file will be, but I know that there's always only ONE file in the folder.
Any ideas how I can manage to do this? :)
Thanks a lot!
Morten
It should work with regular filename expansion (globbing).
KUVVA_CACHE="$HOME/Library/Containers/com.kuvva.Kuvva-Wallpapers/Data/Library/Application Support/Kuvva/"
And then copy
cp -f ${KUVVA_CACHE}/* ${DEST_FOLDER}
You can add the script to your crontab so it will be run at a certain interval. Edit it using 'crontab -e' and add
30 */3 * * * /location/of/your/script
This will run it every third hour. First digit is minutes. Star indicates any. Exit the editor by pressing the escape-key, then shift+: and type wq and press enter. These vi-commands.
Don't forget to 'chmod 0755 file-name' the script so it becomes executable.
Here is the script.
#!/bin/zsh
KUVVA_CACHE="$HOME/Library/Containers/com.kuvva.Kuvva-Wallpapers/Data/Library/Application Support/Kuvva"
DEST_FOLDER="/Library/Desktop Pictures/Kuvva/$USERNAME/screensaver.png"
cp "${KUVVA_CACHE}/"* "${DEST_FOLDER}"
I have a lot of files named the same, with a directory structure (simplified) like this:
../foo1/bar1/dir/file_1.ps
../foo1/bar2/dir/file_1.ps
../foo2/bar1/dir/file_1.ps
.... and many more
As it is extremely inefficient to view all of those ps files by going to the
respective directory, I'd like to copy all of them into another directory, but include
the name of the first two directories (which are those relevant to my purpose) in the
file name.
I have previously tried like this, but I cannot get which file is from where, as they
are all named consecutively:
#!/bin/bash -xv
cp -v --backup=numbered {} */*/dir/file* ../plots/;
Where ../plots is the folder where I copy them. However, they are now of the form file.ps.~x~ (x is a number) so I get rid of the ".ps.~*~" and leave only the ps extension with:
rename 's/\.ps.~*~//g' *;
rename 's/\~/.ps/g' *;
Then, as the ps files have hundreds of points sometimes and take a long time to open, I just transform them into jpg.
for file in * ; do convert -density 150 -quality 70 "$file" "${file/.ps/}".jpg; done;
This is not really a working bash script as I have to change the directory manually.
I guess the best way to do it is to copy the files form the beginning with the names
of the first two directories incorporated in the copied filename.
How can I do this last thing?
If you just have two levels of directories, you can use
for file in */*/*.ps
do
ln "$file" "${file//\//_}"
done
This goes over each ps file, and hard links them to the current directory with the /s replaced by _. Use cp instead of ln if you intend to edit the files but don't want to update the originals.
For arbitrary directory levels, you can use the bash specific
shopt -s globstar
for file in **/*.ps
do
ln "$file" "${file//\//_}"
done
But are you sure you need to copy them all to one directory? You might be able to open them all with yourreader */*/*.ps, which depending on your reader may let browse through them one by one while still seeing the full path.
You should run a find command and print the names first like
find . -name "file_1.ps" -print
Then iterate over each of them and do a string replacement of / to '-' or any other character like
${filename/\//-}
The general syntax is ${string/substring/replacement}. Then you can copy it to the required directory. The complete script can be written as follows. Haven't tested it (not on linux at the moment), so you might need to tweak the code if you get any syntax error ;)
for filename in `find . -name "file_1.ps" -print`
do
newFileName=${filename/\//-}
cp $filename YourNewDirectory/$newFileName
done
You will need to place the script in the same root directory or change the find command to look for the particular directory if you are placing the above script in some other directory.
References
string manipulation in bash
find man page
To create a playlist for all of the music in a folder, I am using the following command in bash:
ls > list.txt
I would like to use the result of the pwd command for the name of the playlist.
Something like:
ls > ${pwd}.txt
That doesn't work though - can anyone tell me what syntax I need to use to do something like this?
Edit: As mentioned in the comments pwd will end up giving an absolute path, so my playlist will end up being named .txt in some directory - d'oh! So I'll have to trim the path. Thanks for spotting that - I would probably have spent ages wondering where my files went!
The best way to do this is with "$(command substitution)" (thanks, Landon):
ls > "$(pwd).txt"
You will sometimes also see people use the older backtick notation, but this has several drawbacks in terms of nesting and escaping:
ls > "`pwd`.txt"
Note that the unprocessed substitution of pwd is an absolute path, so the above command creates a file with the same name in the same directory as the working directory, but with a .txt extension. Thomas Kammeyer pointed out that the basename command strips the leading directory, so this would create a text file in the current directory with the name of that directory:
ls > "$(basename "$(pwd)").txt"
Also thanks to erichui for bringing up the problem of spaces in the path.
This is equivalent to the backtick solution:
ls > $(pwd).txt
To do literally what you said, you could try:
ls > `pwd`.txt
which will use the full pathname, which should be fine.
Note that if you do this in your home directory, which might
be in /home/hoboben, you will be trying the create /home/hoboben.txt,
a text file in the directory above.
Is this what you wanted?
If you wanted the directory to contain a file named after it, you would get
the basename of the current directory and append that with .txt to the pwd.
Now, rather than use the pwd command... why not use the PWD environment variable?
For example:
ls > $PWD.txt
or
ls > ${PWD}.txt
is probably what you were trying to remember with your second example.
If you're in /home/hoboben and you want to create /home/hoboben/hoboben.txt, try:
ls > ${PWD}/${PWD##*/}.txt
If you do this, the file will contain its own name, so most often, you would remedy this in one of a few ways. You could redirect to somewhere else and move the file or name the file beginning with a dot to hide it from the ls command as long as you don't use the -a flag (and then optionally rename the resulting file).
I write my own scripts to manage a directory hierarchy of music files and I use subdirectories named ".info", for example, to contain track data in some spare files (basically, I "hide" metadata this way). It works out okay because my needs are simple and my collection small.
I suspect the problem may be that there are spaces in one of the directory names. For example, if your working directory is "/home/user/music/artist name". Bash will be confused thinking that you are trying to redirect to /home/user/music/artist and name.txt. You can fix this with double quotes
ls > "$(pwd).txt"
Also, you may not want to redirect to $(pwd).txt. In the example above, you would be redirecting the output to the file "/home/user/music/artist name.txt"
The syntax is:
ls > `pwd`.txt
That is the '`' character up underneath the '~', not the regular single quote.
Using the above method will create the files one level above your current directory. If you want the play lists to all go to one directory you'd need to do something like:
#!/bin/sh
MYVAR=`pwd | sed "s|/|_|g"`
ls > /playlistdir/$MYVAR-list.txt
to strip all but the directory name
ls >/playlistdir/${PWD##/*}.txt
this is probably not what you want because then you don't know where the files are (unless you change the ls command)
to replace "/" with "_"
ls >/playlistdir/${PWD//\//_}.txt
but then the playlist would look ugly and maybe not even fit in the selection window
So this will give you both a short readable name and usable paths inside the file
ext=.mp3 #leave blank for all files
for FILE in "$PWD/*$ext"; do echo "$FILE";done >/playlistdir/${PWD##/*}.txt