I'm working on a Flask extension from which I want to create a directory in the project's root path on the file system.
Suppose we have this directory structure
/project
/app
/tests
/my_folder
manage.py
my_folder should be created dynamically by the extension, which is a test utility and wraps the application under test in the /tests directory. However, I'm struggling to determine the project's root path within my extension.
For now, I am trying to guess the path from the run file:
def root_path(self):
# Infer the root path from the run file in the project root (e.g. manage.py)
fn = getattr(sys.modules['__main__'], '__file__')
root_path = os.path.abspath(os.path.dirname(fn))
return root_path
This obviously breaks as soon as the tests are run from within the IDE instead of the manage.py. I could simply infer the project's root relative to the app or tests directory, but I don't want to make any assumptions regarding the name or structure of these directories (since multiple apps might be hosted as subpackages in a single package).
I was wondering if there is a best practice for this type of problem or an undocumented method which the Flask object provides (such as get_root_path).
app.root_path contains the root path for the application. This is determined based on the name passed to Flask. Typically, you should use the instance path (app.instance_path) not the root path, as the instance path will not be within the package code.
filename = os.path.join(app.instance_path, 'my_folder', 'my_file.txt')
app.root_path is the absolute path to the root directory containing your app code.
app.instance_path is the absolute path to the instance folder. os.path.dirname(app.instance_path) is the directory above the instance folder. During development, this is next to or the same as the root path, depending on your project layout.
Related
How do would I serve static files from my base directory? Would it just be a / or would I have to include the name of the base directory, which in this case would be Scanning
app.use(express.static(join(__dirname, '/')));
Pretty close, just a few adjustments!
app.use(express.static(`${__dirname}/Scanning`))
You need to use or construct the actual full path to the base directory. You don't show your actual directory structure and where the desired directory is relative to the directory that your code is running from.
If you wanted express.static() to serve from the Scanning directory which is a sub-directory of the directory you code is located in, you would do this:
app.use(express.static(path.join(__dirname, 'Scanning')));
Or, if Scanning is a sibling of __dirname, then it would be this:
app.use(express.static(path.join(__dirname, '../Scanning')));
You should never be serving files with express.static() directly from __dirname because that would allow your server to serve up your actual source files (and sometimes things like credentials).
Question
When building Singularity images using definition files, is there a way to specify the path to a file on the host system relative to the definition file (i.e. independent of where the build command is called)?
Example to Illustrate the Problem
I have the following files in the same directory (e.g. a git repository):
foobar.def
some_file.txt
foobar.def looks as follows:
Bootstrap: library
From: ubuntu:20.04
Stage: build
%files
# Add some_file.txt at the root of the image
some_file.txt /some_file.txt
This works fine when I build with the following command in the directory which contains the files:
singularity build --fakeroot foobar.sif foobar.def
However, it fails if I call the build command from anywhere else (e.g. from a dedicated "build" directory) because it searches some_file.txt relative to the current working directory of the build command, not relative to the definition file.
Is there a way to implement the definition file such that the build works independently of where the command is called? I know that I could use absolute paths but this is not a viable solution in my case.
To make it even more complicated: My actual definition file is bootstrapping from another local image, which is located in the build directory. So ideally I would need a solution where some files are found relative the working directory while others are found relative to the location of the definition file.
Short answer: Not really
Longer answer: Not really, but there's a reason why and it shouldn't really matter for most use cases. While Docker went the route of letting you specify what your directory context is, Singularity decided to base all of its commands off the current directory where it is being executed. This also follows with $PWD being auto-mounted into the container, so it makes sense for it to be consistent.
That said, is there a reason you can't run singularity build --fakeroot $build_dir/foobar.sif foobar.def from the repo directory? There isn't any other output written besides the final image and it makes more sense for the directory with the data being used to be the context to work from.
I have this module which needs a specific file to work. You can pass the path of the file you want to use or not. If you don't, then a default file is taken. That default file is located at the resources folder, so I typed the path as: "resources/data/type-graph.txt". The problem is that does not work because it takes my CWD as root directory.
Do you know how to make the path relative to the module dir?
Any suggestion is appreciated :).
You should take a look at the Modules documentation page.
There this example is given to access a file placed in the resources folder:
my $template-text = %?RESOURCES<templates/default-template.mustache>.slurp;
You also need to list the file in META6.json so the file will be accessible once the module is installed.
{
...,
"resources": [ "templates/default-template.mustache"]
}
As guifa noted in a comment %?RESOURCES works with individual files, not directory structures. It makes no guarantees of how the files are actually stored. So %?RESOURCES<templates>.dir will not work.
I'm writing a project documenter and I write out the full file path of each compiled file. This is for the VB.NET language so .proj files are written in xml.
Any file that is linked to the project exists on the same drive so at least one of the directory levels are the same for all files. I currently have it set up to put the project directory path on files which exist inside the project since it only shows the name of the file and the residing directory it lives in if its in a directory inside the project. For files outside (linked in) to the project I initially saw their files paths were "..\..\..\dir\filename". So I set it up to take off all the "..\" and put the necessary directories in front of it and all that worked fine. Now for this one .proj file some of the linked in files have their full file path with no "..\".
How can I properly distinguish these three possible inputs?
System.IO.Path.IsRooted will tell you whether a path is rooted, i.e. is a full path, or not. If the path is not rooted it is a relative path. You can use Path.Combine to resolve the full path from a relative path.
I have deployed my web application in GlassFish server 3.0. When I execute it, I get an error: The system cannot find the file specified (MobileOntologyRev1.owl) , which is a file from which I read in my code (I haven't specified the absolute path for this file in my code, and simply refer it using the file name without any addtional path) . Where should this file be kept in order to access it? I have presently tried keeping it inside the WEB-INF/Classes folder and in the root dir of the application inside glassfish/domains/domain1/
Where should I place this file??
You may consider taking advantage of the FaceContext as mentioned below.
You can create a folder (repors) inside your WEB-INF for example.
String pathToFile=
FacesContext.getCurrentInstance().getExternalContext().getRealPath("/WEB-INF/reports/MobileOntologyRev1.owl");