Clarification: the directory /george/ exists when I am given the task, and I do know how to read the attribute in cli showing that it is a directory. The balance of the path requested does not exist, so the only way I can tell if 'poppet' is a file or directory is by convention. This question comes from an issue I had with interpretation of this line, so am interested in knowing if there is a convention. This occurred using Ubuntu, but would be relevant in any distro, methinks.
Related
Is there a way to open a directory and create it if doesn't exist atomically ?
My use case is simple, I use a directory to watch every public key that are allowed to connect to my server, but if the directory doesn't exist I want to create it, unfortunately for now I only see a two step solution, create the path with create_dir_all() and then open it with read_dir() but this create a possible situation where the directory is delete between the two calls (very unlucky in my docker container... but anyway !)
I didn't find any solution in linux to do that and I'm quite surprise cause that a very common operation for file.
I found a related question Create a directory and return a dirfd with `open` but it's focus on file descriptor and so is more specific. The answer seem to say this doesn't prevent race condition, I don't really understand the context, my only concern is to avoid to create the directory and than try to open it and it's fail. It's more for convenience and robustness of code.
TLDR
I need to get paths to system directories like "Screenshots":
On an English system. I can just use this one:
C:/Users/User/Pictures/Screenshots
How do I get the path to "Screenshots" directory on a non-English system?
C:/Users/User/Pictures/[NAME]
Description
I have a file manager app, it displays system directories and loads them on click.
The app can run system commands via Powershell and use Node.js (preferred)
Problem
The problem is, it only works if the system has English system language.
Currently, to resolve the "Screenshots" directory path, the app simply joins the User directory with the word "Screenshots"
const pictures = electronRemote.app.getPath('pictures')
const screenshots = PATH.join(pictures, 'Screenshots')
link to the line in code
Expectedly, the C:/Users/User/Screenshots path only exists on English systems.
One way to solve this is to use short names, at least on Windows, I know that system directories have short names like SCREEN~1 and WALLPA~1 for Screenshots and Wallpapers directories, but if I use these names the paths will look like this:
C:/Users/User/SCREEN~1 instead of C:/Users/User/Screenshots throughout the app.
And even if I were to run these paths through a function to convert it to readable name, how would I know which word to replace it with? I need to get the name in the system's language.
Are these translations stored somewhere on the system? Can I just retrieve the translated directory name and use that in the code above?
Question
How do I make it to get / resolve the actual path of system directories like Screenshots and Wallpapers, independently of system locale?
If you know how to do it, could you please suggest the solution for all platforms (Win, Mac, Linux)?
Should I just use the short names like SCREEN~1 and then automatically replace all the occurrences in UI and also filter all paths through a function that replaces this short name with the actual path throughout the whole app? Seems like a lot of work, this approach
I've been developing an nw.js project and use node.js file system functions in it as normal. In my application there is a file manager and I list folders and files according to user navigation. In Windows, for example, if I scan drive C: I get the Turkish named folder "Kullanıcılar" as "Users". I know it's real name in operating system is "Users" and just seen on the screen according to Languages. I can replace names of such folders when dispaying in my file manager but I'm searching for better solution if exists. Thanks in advance.
There's an SO answer here that reads the localized name of a folder in C# using the SHGetFileInfo function which might help you along.
Now I know you didn't ask, but in case you want to know where the information is stored... It's within the directory, in the Desktop.ini file.
For instance, my Windows 10 installation has this in it for "Users":
[.ShellClassInfo]
LocalizedResourceName=#%SystemRoot%\system32\shell32.dll,-21813
And this for the Images folder within my user folder (bringing this up to show you the additional keys):
[.ShellClassInfo]
LocalizedResourceName=#%SystemRoot%\system32\shell32.dll,-21779
InfoTip=#%SystemRoot%\system32\shell32.dll,-12688
IconResource=%SystemRoot%\system32\imageres.dll,-113
IconFile=%SystemRoot%\system32\shell32.dll
IconIndex=-236
The #%SystemRoot%\system32\shell32.dll,-21813 points to having to read the MUI (multilingual user interface) resources, key 21813 for the given file (presumably the # means that it's in this file, not this literal value, but don't quote me on that). %SystemRoot% is an environment variable that points to the Windows directory.
The actual MUI files and their locations are handled by Windows (see the MSDN link above), but we'll just happen to handily know that the MUI file for the US English localization of shell32.dll is system32\en-US\shell32.dll.mui.
Opening up that file with Resource Hacker, we can search for 21813 -- and voila! We can find STRINGTABLE resource #1364 that contains:
[...snip...]
21812, "Extras and Upgrades"
21813, "Users"
21814, "Saved Games"
[...snip...]
I unfortunately don't have tr-TR/shell32.dll.mui available, so you'll just have to trust me that you'd find the Kullanıcılar string there.
I know puppet modules always have a files directory and I know where it's supposed to be and I have used the source => syntax effectively from my own, handwritten modules but now I need to learn how to deploy files using Hiera.
I'm starting with the saz-sudo module and I've read the docs but I can't see anything about where to put the sudoers file; the one I want to distribute.
I'm not sure whether I need to set up a site-wide files dir in /etc/puppetlabs/puppet and then make subdirs in there for every module or what. And does Hiera know to look in /etc/puppetlabs/puppet/files/sudo if I say, source => "puppet:///files/etc/sudoers" ? Do I need to add a pathname in /etc/hiera.yaml? Add a line - files ?
Thanks for any clues.
My cursory view of the puppet module, given their example of using hiera:
sudo::configs:
'web':
'source' : 'puppet:///files/etc/sudoers.d/web'
'admins':
'content' : "%admins ALL=(ALL) NOPASSWD: ALL"
'priority' : 10
'joe':
'priority' : 60
'source' : 'puppet:///files/etc/sudoers.d/users/joe'
Suggest it assumes you have a "files" puppet module. So under you puppet modules section:
mkdir -p files/files/etc/sudoers.d/
Drop your files in there.
Explanation:
The url 'puppet:///files/etc/sudoers.d/users/joe' is broken down thus:
puppet: protocol
///: Three slashes indicate the source of the file is in a module.
files: name of the module
etc/sudoers.d/users/joe: full path to the file within the module's "files" directory.
You don't.
The idea of a module (Hiera backed or not) is to lift the need to manage the whole sudoers file from you. Instead, you can manage each single entry in the sudoers file.
I recommend reviewing the documentation carefully. You should definitely not have a file { "/etc/sudoers": } resource in your manifest.
Hiera doesn't have to do anything with Files.
Hiera is like a Variables Database, and servers you based on the hierarchy you have.
the files inside puppet, are usually accessed in methods like source => but also these files are using some basic structure.
In most cases when you call an file or template.
A template can serve your needs to automatically build an sudoers based on that.
There are also modules that supports modifying sudoers too.
It is up to you what to do.
In this case, saz stores the location of the file in hiera, but the real location can be a file inside your puppet (like a module file or something similar).
Which is completely unrelated.
Read about puppet file server
If you have questions, just ask.
V
I am currently facing a problem in locating the syscall_table.S file in my arch/x86/kernel/ directory. In the online tutorail that i am following, it is gievn that i will find the file in this location. I am using linux-3.11.10. Please tell me how to locate this file. However, I have found this file in some other folders. If i were to modify one of these,which one should I modify ?
The following folders have syscall_table.S :
arch/microblaze/kernel
arch/m32r/kernel
arch/avr32/kernel
arch/parisc/kernel
Your question isn't very specific about what exactly you are trying to do.
sys_call_table is defined in arch/x86/kernel/syscall_64.c
The syscall entry is located in arch/x86/kernel/entry_64.S
routines are associated with their syscall number in include/uapi/asm-generic/unistd.h and arch/x86/syscalls/syscall_64.tbl
You might also want to look at include/linux/syscalls.h.