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.
Related
I was working on a project which requires me to add a user.cmake file in the root directory. Can anyone help me out hot to create the .cmake file...
Link to Project Directory
According to the link you provided user.cmake just needs to point where so-called eego sdk is located in your file system:
set(EEGO_SDK_ZIP /path/to/download/eego-sdk-1.3.19.40453.zip)
There is nothing fancy actually here, just make a plain text file, put this one line (don't forget to replace the EEGO_SDK_ZIP variable content) and save it with the name user.cmake
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.
I have tried to install PhpWord to yii. I have downloaded zip file and extracted it into extentions folder:
extenstions
--PHPWord
--PHPWord.php
However, I cannot make it to run. I got following error:
include(PHPWord.php): failed to open stream: No such file or directory
How can i solve it?
After extracting the file in extension folder, you have to import that file in controller.
Yii::import('ext.phpword.PHPWord');
First of all, you didn't say if it's Yii 1 or 2. They have different autoloading methods.
Second, you have extracted it into extension folder, and I assume your file where you want to include it is in a completely different folder.
You should do it like this
include('/full/path/to/PHPWord.php');
You need either absolute or a relative path to the file (I suggest using abosulte path (the one I used as an example).
Relative path means the path to the file you want to include compared to where your file, in which you are including it, is.
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 a dir structure for Intellij 12:
...
...test
- java
- com.mycompany.myproject
- package1 (contains code, etc,.)
- resourcePackage (want to contain .json files but can't mark as a resource)
- myOtherJunk.json
- o o o
- resources
- aResource.json
The thing is if I right click on my package name (com.mycompany.myproject) I can only add packages and not directories (like that of an existing resource folder).
However, I don't want to use that existing resource folder for the .json files that I'm going to read into per my test class.
So, I need something to support:
// this already works for the resources directory per the .json file but doesn't for the
// myOtherJunk.json per the resourcePackage.
InputStream is = MyClassTest.class.getResourceAsStream("aResource.json");
This can be solved in several ways. An example of a good approach would be the following folder structure:
src
main
java
resources
test
java
resources
When this is done, you put all you java classes under src/main/java/com.mycompany package and any resources under /src/main/resources/com/mycompany folder.
To link them together, go to the project properties, and find the Path tab. Mark the src/main/java and src/main/resources as source folders. (see the screen-shot attached)
If you link them together, you'll be able to use getResourceAsStream() method.
If you wonder why you should use the following folder structure - this is standard maven way of keeping things clean and tidy.
Directories Creation
Intellij creates directories when you ask her to create package. It is not an error.
If you create package "com", it will create the dir "com", and if you create a source file there, it will think that the file is in the package "com".
If you create package "com.next.pack", it will create three nested dirs "com", then "next", then "pack", and if you create a source file there, it will think that the file is in the package "com.next.pack".
Directories Structures
Only the path under the source root is taken as a package. Any directory(ies) can be set as a source root(s).
Resources roots
Make practically any structure of directories. Somewhere in it there is the root dir of resources. Right-click it and Mark Directory As ... Resources Root.
Notice, the same method can be used for the directories structures for tests, test classes, and test resources. Look here.
Please use #ContextConfiguration annotation to load the resource files. Please see below example.
#ContextConfiguration( { "/app-config.xml", "/test-data-access-config.xml",application-test.yml })