Maybe there's a better way to do this, but I'm trying to organize my Flask application so that both static Javascript and HTML templates sit in the same folder (similar to components)
This is just for organization purposes, I won't be rendering the statics in jinja2
So, instead of this :
├── templates/
│ ├── login.html
├── static/
│ ├── login.js
I'm looking for this (just an idea):
├── components/
│ ├── login/
│ │ ├── login.html
│ │ └── login.js
Or maybe this, so I can make a wrapper that renders a component just by its folder name, ex: return render_component('login')
├── components/
│ ├── login/
│ │ ├── template.html
│ │ └── login.js
The concept is that inside login.html template it adds a script tag with url_for to login.js
I'm also open to different ideas/ways to achieve this component structure
I'm sure there are other ways to achieve this but one way is to use blueprint structure and contain all static files for that specific blueprint inside the blueprint folder.
app
|- collections
| - blueprint_one
| - configs
| - static
| - css
| - js
| js_file_one.js
| - templates
| - blueprint_one
| html_file_one.html
| blue_print_one_view_file_one.py
| __init__.py
| - blueprint_two
| - blueprint_three
|- static
| - js
| - css
|- templates
__init__.py
Above folder structure allows you to separate out not only the static files but also the Flask view files.
Below are examples of activating/importing files:
1) app/init.py
app = Flask(__name__, static_folder='static')
app.config.from_pyfile('app_config.py')
# Do all your db set up and etcs..
# Import blueprints below set ups to prevent circular import
app.collections.blueprint_one import blueprint_one
app.collections.blueprint_two import blueprint_two
# Now register blueprints that you've imported
app.register_blueprint(blueprint_one, url_prefix='/blueprint_one')
app.register_blueprint(blueprint_two, url_prefix='/blueprint_two')
2) blueprint_one/init.py
# Create a blueprint
from flask import Blueprint
blueprint_one = Blueprint('blueprint_one_name', __name__, template_folder='templates', static_folder='static')
# Import required views to use
from . import blue_print_one_view_file_one.py
3) blueprint_one/blue_print_one_view_file_one.py
# Import blueprint
from app.collections.blueprint_one import blueprint_one
# Any view functions using above blueprint
#blueprint_one.route('/login/, methods=['GET'])
def login():
return render_template('blueprint_one/html_file_one.html')
4) blueprint_one/templates/blueprint_one/html_file_one.html
// Note: instead of .static, you can use blueprint_one.static.
<script type="text/javascript" src="{{ url_for('.static', filename = 'js/js_file_one.js') }}"></script>
5) app/static and app/templates still can be used as you are using it currently.
This solution doesn't solve the problem of putting js and html into same folder but at least you can divide each feature into component like blueprint for more modularized project structure.
There isn't a best way to do this so I like to suggest that you review the way flask projects are supposed to be set up.
http://flask.pocoo.org/docs/1.0/tutorial/layout/
Also I think this question may have an answer that solves your problem here.
Flask url_for URLs in Javascript
Now here's a few thoughts.
To me it sounds like you want to use Templating in files that are in your static folder. This is not what the static folder is intended for. Static means unchanging, so the files in your static folder should not change. Templates are intended to be dynamic which usually means that they change. If you plan to render templates from files in your static directory you will encounter numerous issues and possibly even not be able to do it. One of the most prominent issues I can see you possibly encountering is that browsers will not want to grab the new javascript files if it thinks they're in your static directory. Browsers may assume they haven't changed. Also Flask does some magic behind the scenes. When a url for a file in your static directory is processed from a client who has already requested the file flask returns HTTP CODE 304 which essentially means "File hasn't changed".
I could go on and on about why you shouldn't do this. Instead, if none of information above solves your question, I would like to suggest that you put your javascript templates in a sub-directory of your template directory. Then use the following code to return them.
#app.route('/js/<file_name>',methods=['GET'])
def get_js_templates(file_name):
return render_template('js/'+file_name)
I am by no means an expert in any capacity but I have done full stack development using the flask web framework. If you have more questions leave a comment on this post.
Related
I have build an Nextjs application which display data from database through an API. The data should be displayed as folder structures. If a row in database is marked as folder then it should be displayed as folder otherwise it should be displayed as file. If I click on folder then an API call is made which displays the files and folders inside it. And Likewise there can be multiple folders or folder within the folders.
The application is hosted as Azure static web app. The issue which I am facing is when I am inside a folder and when I refresh the page, I get 404 error. The issue seems to be related the route configuration. I tried by adding staticwebapp.config.json, fallback route but in this case it redirects to home page.
Ideally it should stay on same page with the folder on which I refreshed the page.
Any guidance on how to setup dynamic route on Nextjs/static web app will be helpfull.
Thanks,
Rajesh
As its mentioned in the document that "You can control which requests return the fallback file by defining a filter.
In the following example, requests for certain routes in the /images folder and all files in the /css folder are excluded from returning the fallback file.
{
"navigationFallback": {
"rewrite": "/index.html",
"exclude": ["/images/*.{png,jpg,gif}", "/css/*"]
}
}
The example file structure below, the following outcomes are possible with this rule.
├── images
│ ├── logo.png
│ ├── headshot.jpg
│ └── screenshot.gif
│
├── css
│ └── global.css
│
└── index.html
Also refer to this similar discussion here.
I prefer to not use index.js as a file name for the main file of a component in React and in node.js, but instead use the name of the parent directory (for example: ...somePath.../ComponentName/ComponentName.js would just be ...somePath.../ComponentName).
Is there any way to have
require('..somePath../ComponentName')
and
import ComponentName from '..somePath../ComponentName'
work without having to put in the file name too, similar to how you can drop the index.js from the import path.
I am guessing there would be two different solutions for node.js and React, but hopefully there is just one solution.
Maybe this is because of my IDE (I use VScode), but the extension for any file is not needed. This example works in React. Note: Does not work with the styles sheet.
Components
├── Header
│ ├── index.js
├── Footer
│ ├── index.js
import Header from '../Header';
import Footer from '../Footer';
I have my own Yeoman generator.
I created a sub-generator to create a new view folder.
Basically, the usage is:
open a new terminal
cd into the parent folder
run the yeoman command yo my-generator:view
follow the instructions
This view sub-generator prompt a folder name.
For example:
If I want to create the view authentication on the default views directory.
cd views
yo my-generator:view
The result should be:
views //- Already created by the main generator
├── authentication
│ ├── authentication.controller.js
│ ├── authentication.template.html
Now, if I want to create a sub-view login for the authentication view.
cd views/authentication
yo my-generator:view
The result should be:
views //- Already created by the main generator
├── authentication
│ ├── authentication.controller.js
│ ├── authentication.template.html
│ ├── login
│ │ ├── login.controller.js
│ │ ├── login.template.html
Instead, the current (wrong) result is:
views //- Already created by the main generator
├── authentication
│ ├── authentication.controller.js
│ ├── authentication.template.html
├── login
│ ├── login.controller.js
│ ├── login.template.html
My struggle here is that I don't know how to get the current path when I run the command.
Actually, I just create the new folder with a default prefix path which is app/views/.
This is why Authentication example works.
However when my current path is deeper in the views folder, it will add the new folder at the root of the views folder.
If I could get the current path (of the cmd), I should be able to add this path as the prefix instead of setting a default and not static one.
This is why Login example doesn't works.
Some code example:
$that is the current generator object
$that.viewNameCamel is the name of the folder set by the user
I use a .txt file as template and then create the controller.js file.
const filePrefix = 'app/views/' + $that.viewNameCamel + '/' + $that.viewNameCamel + '.';
const exampleData = {
controllerAlias: 'vm',
otherVar: 'example'
};
$that.fs.copyTpl(
$that.templatePath('controller.txt'),
filePrefix + 'controller.js',
exampleData
);
Tried:
$that.env.cwd
process.cwd()
__dirname
path.js library
Similar:
Issue 1037
Question 28481715
So guys, do you have a clue on how do I get the current folder path ?
Is there an alternative solution here ?
Thanks !
EDIT 1:
The problem here is the .yo-rc.json present on the root directory of the projet.
The file rewrite the path so I should delete it to fix the problem.
However if I delete this file, the user configuration will no longer be saved.
And I need it for later sub-generator usage.
Is there another way to save the user configuration ?
Or once again, is there another way to get the current real path ?
I know this is kinda old but for any one who comes later,
from the documentation https://yeoman.io/authoring/file-system.html
If you want to know from where the user is running yo, then you can get the path with this.contextRoot. This is the raw path where yo was invoked from; before we determine the project root with .yo-rc.json.
from my experience removing the .yo-rc.json would mean that you need to manually check if the path provided is ok, having a fixed root point is something helpful.
Just remove the .yo-rc.json file from your root directory. This is the file that is responsible for locating your root irrespective of where you are in the file system.
I am not sure about the repercussions of removing it but nothing seem to have happened to the generators I built.
Now you can use process.cwd() and it will get the correct working directory.
For your use case to have a prefix app/views, you probably need to write some Javascript to append or not append the prefix based on where you are, which should be trivial.
I'm newbie developing MEAN apps, i'm coming from a Laravel structure to develop web applications, what i don't like about Laravel is that is not so modular, everything is separated, especially Models, Views, and Controllers, to navigate through the folders you have to scroll a lot of times when building large apps...
My question is all about Express under the MEAN stack, what i'm looking for is a structure where i create modules for every entity of an app, for example:
I have three modules: users, questions, and answers; each module will contain the routes, the model, and the controller for that specific module, for example:
├── node_modules
├── src
│ ├── client
│ │ └── ... //Frontend things managed by angular (like views, etc...)
│ └── server
│ ├── modules //By module i mean an entity
| | ├── users
| | | ├── users.model.js
| | | ├── users.controller.js
| | | ├── users.routes.js
| | | └── index.js
| | ├── questions
| | └── answers
| ├── config
| └── etc... //Suggestions please...
|
├── package.json
└── server.js
How could i set up that structure?, so far i've found this tutorial about modularizing an Express app, but i would like to extend it, so that i keep it DRY, and make use of the LIFT principle described by John Papa in his Angular Style Guide but under the backend.
Why am i asking this? Simple, i don't like to scroll through large folders to get open a file and then scroll again to open another file that are related, i want to work with a structure easy to maintain, easy to understand, etc...
If possible could someone explain an example of how to setup an application using this structure and upload it to github?
Place one node_modules at the top level.
Give each module a package.json
Now write a script in your bootstrap that finds all unique node_modules required from the package.json.
There, node_modules solved. Make sure it identifies the variances of versions too.
Build a universal run script file that reads a universal configuration file that points to all of the other modules.
Angular 2..4..5 whatever is a great modularized file structure to model after.
I found that Express has an application generator, however the documentation does not explain the purpose of each directory and file. If someone could just give me a short explanation of which files I should be putting where, that would be much appreciated. Here's the generated app structure:
├── app.js
├── bin
│ └── www
├── package.json
├── public
│ ├── images
│ ├── javascripts
│ └── stylesheets
│ └── style.css
├── routes
│ ├── index.js
│ └── users.js
└── views
├── error.jade
├── index.jade
└── layout.jade
7 directories, 9 files
The app.js file is the entry-point of your application.
The package.json file contains all of your dependencies and various details regarding your project.
The bin folder should contain the various configuration startup scripts for your application.
For example, instead of applying all the Express middleware in the app.js file, you module.exports = {} them from their own configuration file and require them in app.js. [additional info LINK]
The views folder contains all of your server-side views.
The public folder contains all of your front-end code.
The routes folder contains all the routes that you have created for your application.
As stated in the official documentation, be aware that this is just one way to organize your code.
You should test it out and see if it fits your project.
This thread gives a deeper answer about the www file specifically: What does "./bin/www" do in Express 4.x?
Basically, running your app from the www file (which calls app.js) allows you to start your app with different configurations. You might have a "www" file representing the way the app should be run when on the web, a "dev" file that you, as the developer, would run, a "test" file you would run when running tests, etc. Read the thread linked above for more detail!
This structure is a standard organization for web-app
public contains all client static files (css, client javascript (ex. jQuery), images, fonts...)
routes contains the main back-end code (server side), which compute data before calling a template engine (see below) or respond to the client (via json of xml).
views contains each page template, see jade template. These files are used by scripts in "route"
app.js contains the express core, such as uri parser, modules, database...
package.json is the project descriptor file (used by npm for dependencies and sharing)
If the application generator provide a full example, don't hesitate to open each file (starting from app.js) to understand the project's organization.