So this is embarrassing. I've got an application that I threw together in Flask and for now it is just serving up a single static HTML page with some links to CSS and JS. And I can't find where in the documentation Flask describes returning static files. Yes, I could use render_template but I know the data is not templatized. I'd have thought send_file or url_for was the right thing, but I could not get those to work. In the meantime, I am opening the files, reading content, and rigging up a Response with appropriate mimetype:

import os.path

from flask import Flask, Response


app = Flask(__name__)
app.config.from_object(__name__)


def root_dir():  # pragma: no cover
    return os.path.abspath(os.path.dirname(__file__))


def get_file(filename):  # pragma: no cover
    try:
        src = os.path.join(root_dir(), filename)
        # Figure out how flask returns static files
        # Tried:
        # - render_template
        # - send_file
        # This should not be so non-obvious
        return open(src).read()
    except IOError as exc:
        return str(exc)


@app.route('/', methods=['GET'])
def metrics():  # pragma: no cover
    content = get_file('jenkins_analytics.html')
    return Response(content, mimetype="text/html")


@app.route('/', defaults={'path': ''})
@app.route('/<path:path>')
def get_resource(path):  # pragma: no cover
    mimetypes = {
        ".css": "text/css",
        ".html": "text/html",
        ".js": "application/javascript",
    }
    complete_path = os.path.join(root_dir(), path)
    ext = os.path.splitext(path)[1]
    mimetype = mimetypes.get(ext, "text/html")
    content = get_file(complete_path)
    return Response(content, mimetype=mimetype)


if __name__ == '__main__':  # pragma: no cover
    app.run(port=80)

有人愿意给出一个代码示例或url吗?我知道这非常简单。


当前回答

这是最简单的方法之一。干杯!

demo.py

from flask import Flask, render_template
app = Flask(__name__)

@app.route("/")
def index():
   return render_template("index.html")

if __name__ == '__main__':
   app.run(debug = True)

现在创建名为templates的文件夹。 将index.html文件添加到templates文件夹中

index . html

<!DOCTYPE html>
<html>
<head>
    <title>Python Web Application</title>
</head>
<body>
    <div>
         <p>
            Welcomes You!!
         </p>
    </div>
</body>
</html>

项目结构

-demo.py
-templates/index.html

其他回答

你可以使用这个函数:

send_static_file(文件名) 内部用于发送静态数据的函数 文件从静态文件夹转移到浏览器。

app = Flask(__name__)
@app.route('/<path:path>')
def static_file(path):
    return app.send_static_file(path)

所以我把事情搞定了(基于@user1671599的答案),想和你们分享。

(我希望我做得对,因为这是我在Python中的第一个应用程序)

我做了这个

项目结构:

server.py:

from server.AppStarter import AppStarter
import os

static_folder_root = os.path.join(os.path.dirname(os.path.abspath(__file__)), "client")

app = AppStarter()
app.register_routes_to_resources(static_folder_root)
app.run(__name__)

AppStarter.py:

from flask import Flask, send_from_directory
from flask_restful import Api, Resource
from server.ApiResources.TodoList import TodoList
from server.ApiResources.Todo import Todo


class AppStarter(Resource):
    def __init__(self):
        self._static_files_root_folder_path = ''  # Default is current folder
        self._app = Flask(__name__)  # , static_folder='client', static_url_path='')
        self._api = Api(self._app)

    def _register_static_server(self, static_files_root_folder_path):
        self._static_files_root_folder_path = static_files_root_folder_path
        self._app.add_url_rule('/<path:file_relative_path_to_root>', 'serve_page', self._serve_page, methods=['GET'])
        self._app.add_url_rule('/', 'index', self._goto_index, methods=['GET'])

    def register_routes_to_resources(self, static_files_root_folder_path):

        self._register_static_server(static_files_root_folder_path)
        self._api.add_resource(TodoList, '/todos')
        self._api.add_resource(Todo, '/todos/<todo_id>')

    def _goto_index(self):
        return self._serve_page("index.html")

    def _serve_page(self, file_relative_path_to_root):
        return send_from_directory(self._static_files_root_folder_path, file_relative_path_to_root)

    def run(self, module_name):
        if module_name == '__main__':
            self._app.run(debug=True)

对于angular+样板流,它创建下一个文件夹树:

backend/
|
|------ui/
|      |------------------build/          <--'static' folder, constructed by Grunt
|      |--<proj           |----vendors/   <-- angular.js and others here
|      |--     folders>   |----src/       <-- your js
|                         |----index.html <-- your SPA entrypoint 
|------<proj
|------     folders>
|
|------view.py  <-- Flask app here

我使用以下解决方案:

...
root = os.path.join(os.path.dirname(os.path.abspath(__file__)), "ui", "build")

@app.route('/<path:path>', methods=['GET'])
def static_proxy(path):
    return send_from_directory(root, path)


@app.route('/', methods=['GET'])
def redirect_to_index():
    return send_from_directory(root, 'index.html')
...

它有助于重新定义“静态”文件夹自定义。

所有的答案都很好,但对我来说工作得很好,只是使用Flask的简单函数send_file。当host:port/ApiName将在浏览器中显示文件的输出时,当你只需要发送一个html文件作为响应时,这种方法很有效


@app.route('/ApiName')
def ApiFunc():
    try:
        return send_file('some-other-directory-than-root/your-file.extension')
    except Exception as e:
        logging.info(e.args[0])```

默认文件夹名为“static”,包含所有静态文件 下面是一个代码示例:

<link href="{{ url_for('static', filename='vendor/bootstrap/css/bootstrap.min.css') }}" rel="stylesheet">