is it possible to put comments in a directory about a dir or file locally on my own machine, appearing in my terminal. I don't have any kind of GUI, so this would be helpful.
Maybe something like, when I do ls -l I will see:
file.txt #this is that file I made on tuesday
files #this is the directory I made with all those other files
If so, what is the tool to do it?
Is it available for Arch Linux?
Yes, there is a tool for that. It is called GIT or SVN.
Related
Microsoft just introduced a Linux subsystem in its Windows 10 Anniversary Edition. The installation is pretty straight forward, but I could not locate bash files on Windows.
How does it work? What does ~ refer to in Windows? Where to find .bashrc?
Since the Windows 10 Fall Creators Update, the location changed to:
C:\Users\USERNAME\AppData\Local\Packages\{DIST}\LocalState\rootfs\home\{LINUXUSER}\
Where:
{DIST} is equal to CanonicalGroupLimited.UbuntuonWindows_79rhkp1fndgsc
{LINUXUSER} is the user for which you are looking for the .bashrc file
Just for anyone wondering that came here from Google.
Sorry for the misunderstanding, I check on google and it will be at C:\Users\USERNAME\AppData\Local\Lxss\home\USERNAME .
I tried and it works, in the cmd just type cd\ && dir *bashrc* /s it will locate the file, and in my case i see the line C:\Users\USERNAME\AppData\Local\Lxss\home\USERNAME but when I want to navigate it with the window browser it doesn't work, but if you copy paste it, it works :-)
I found it here.
Considering that you need to know where a file is located you can use the find command.
The syntax of the command is find {search-path} {file-names-to-search} {action-to-take}by default the action to take is printing the file name.
So if you are finding .bashrc file you can use find / -name .bashrc the bash will return you /home/yourusername/.bashrc
Also, if you want to access to your home directory you can use cd ~
Hope my answer will be helpful :-)
just type
vi ~/.bashrc
and that should put you into the file where ever it is.
You can navigate there simply by doing cd ~
List all files with ls -a and you should be able to see it.
~ means that is user home folder, way like /home/%username%/
you can list files like ls -al and see .bashrc file.
Right now on WSL 2 you can find it under /home/{user_name} and the file is hidden.
You can access it from Ubuntu console by {text_editor} .bashrc
If you want to edit that in Windows just type in ubuntu console explorer.exe . and it opens the current folder and shows all hidden files.
It's weird but works fine.
Other answers doesn't work for me using WSL 2.
The LocalState folder contains a virtual disk so rootfs does not exist,
and AppData\Local folder does not have the Lxss folder.
The solution for me is surprisingly simple:
wsl -u root
This will allow you to get into wsl as root.
From here, you have access to the whole linux.
Fix the .bashrc or anything you want.
Don't screw up the root user. :)
I find my .bashrc file in:
/home/your_user_name
you can run cd /home/your_user_name or cd ~ should work as well
If you previously installed git bash for window, you may also find .bashrc file in your window user profile folder. In Linux subsystem, you may local the file under /mnt/c/Users/your_window_user_name/.bashrc However, modifying that file only works for git bash in window but not for the shell terminal of the Linux subsystem.
Note: my installation of the Ubuntu is 20.04 LTS straight from window store.
At first I had
touch $NAME_OF_FILE$DATE.$FILE_EXT
then I changed it to
PATH="Logs/"
touch $PATH$NAME_OF_FILE$DATE.$FILE_EXT
The file is created correctly in the folder, however only echos are being printed in there because says commands are not found like grep, awk, and others.
EDIT: The folder is already created on my desktop
Thanks
Alan
PATH is an environment variable that specifies where executables are located and is used by your shell to look for commands executables (grep, awk, ...). You should not override it in your script.
Try:
MYPATH="Logs/"
touch $MYPATH$NAME_OF_FILE$DATE.$FILE_EXT
To understand what PATH is open a shell and type echo $PATH. You will see it contains the directories where your commands executables are.
I used command zip in linux (RedHat), this is my command:
zip -r /home/username/folder/compress/zip.zip /home/username/folder/compressed/*
Then, i open file zip.zip, i see architecture as path folder compress.
I want to in folder zip only consist list file *.txt
Because i used this command in script crontab hence i can't use command cd to path folder before run command zip
Please help me
I skimmed the zip man page and this is what I have found. There is not an option archive files relative to a different directory. The closest I have found is zip -j which removes the entire path and stores the files directly in the zip rather than sub directories. I do not know what happens in the case of file name conflicts such as if /home/username/folder/compressed/a.txt and /home/username/folder/compressed/subdir/a.txt both exist. If this is not a problem for you, you can use this option, but I am concerned because you did specify the -r option indicating that you expect zip to traverse sub folders.
I also thought of the possibility that your script could somehow call zip with a different working directory, but I took a look at this unix stack exchange page and it looks like their options use cd.
I have to admit I do not understand why you cannot use cd and I am very curious about it. You said something about using crontab, but I have never heard of anything wrong with changing directories in a crontab script.
I used option -j in command zip
zip -jr /home/username/folder/compress/zip.zip /home/username/folder/compressed/*
and i was yet settled this problem, thanks
I have lots of directories with the same structure. A specific folder in a directory can contain a file, that I want to rename.
mv /home/*?*/domains/*?*/public/old.text /home/*?*/domains/*?*/public/new.txt
The ? can be different each time.
I tried the above, but didn't work of course.
It's on a Linux CentOS machine.
Does this do what you want?
for i in /home/*/domains/*/public/; do
echo mv "$i/old.txt" "$i/new.txt"
done
In my NSIS script, I use this line:
File "..\help\*.*"
My problem is that I have the help directory in my subversion repository (its constantly updated as we add new functionality). This means that the help directory contains a .svn directory.
I wish to view the contents of the setup.exe that NSIS created to verify that it does not have the .svn directory.
P.s. I experimented to see if NSIS recursively adds files when wildcards are used. It doesn't. But I want to verify this, hence the question.
These things are typically compressed files.
You could check with 7z/7-zip to open the EXE archive.
As a record, after the comments below,
I'd like to point to my recent notes on the merits of 7-zip at Superuser,
Compressing with RAR vs ZIP
Rather than look at what's in your NSIS exe itself, just exclude the .svn directories so you know they'll never be in there.
Something like this will do the trick:
File /r /x .svn "..\help\*.*"
The /x .svn bit tells NSIS to exclude those directories.
Coincidentally, if you're not using the /r switch, then you're not adding files and folders recursively, so it wouldn't add the .svn subdirectories anyway.
Instead of unzipping, my suggestion is to look at the NSIS compilation log. It will tell you everything about files included. When doing changes in my NSIS scripts I always check the logs to make sure that everything is going according to plan. Streaming the log from the command line to a text file, then read it from your favorite editor.
I use 7zip File Manager and the "Open Inside" command.