如何在Flask中提供静态文件

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

最简单的方法是在主项目文件夹中创建一个静态文件夹。包含。css文件的静态文件夹。

主文件夹

/Main Folder
/Main Folder/templates/foo.html
/Main Folder/static/foo.css
/Main Folder/application.py(flask script)

图像的主文件夹包含静态和模板文件夹和烧瓶脚本

瓶

from flask import Flask, render_template

app = Flask(__name__)

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

html(布局)

<!DOCTYPE html>
<html>
    <head>
        <title>Project(1)</title>
        <link rel="stylesheet" href="/static/styles.css">
     </head>
    <body>
        <header>
            <div class="container">
                <nav>
                    <a class="title" href="">Kamook</a>
                    <a class="text" href="">Sign Up</a>
                    <a class="text" href="">Log In</a>
                </nav>
            </div>
        </header>  
        {% block body %}
        {% endblock %}
    </body>
</html>

html

{% extends "layout.html" %}

{% block body %}
    <div class="col">
        <input type="text" name="username" placeholder="Username" required>
        <input type="password" name="password" placeholder="Password" required>
        <input type="submit" value="Login">
    </div>
{% endblock %}

2019-10-18 00:53:52

其他回答

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])```

2019-08-13 18:30:26

例如，要返回我使用过的Adsense文件:

@app.route('/ads.txt')
def send_adstxt():
    return send_from_directory(app.static_folder, 'ads.txt')

2022-02-17 19:44:02

基于其他答案的一个最简单的工作示例如下:

from flask import Flask, request
app = Flask(__name__, static_url_path='')

@app.route('/index/')
def root():
    return app.send_static_file('index.html')

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

使用名为index.html的HTML:

<!DOCTYPE html>
<html>
<head>
    <title>Hello World!</title>
</head>
<body>
    <div>
         <p>
            This is a test.
         </p>
    </div>
</body>
</html>

重要提示:index.html在一个名为static的文件夹中，这意味着<projectpath>有.py文件，<projectpath>\static有html文件。

如果你想让服务器在网络上可见，使用app.run(debug=True, host='0.0.0.0')

EDIT:如果需要显示文件夹中的所有文件，请使用此选项

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

这基本上就是黑曼巴的答案，所以给他们点赞。

2014-08-26 08:56:59

在我的情况下，我需要一个静态文件夹中的所有文件都可以由用户访问，以及我需要为我的一些html文件使用模板，这样公共html代码可以放在模板中，代码得到重用。以下是我如何同时实现这两个目标的:

from flask import Flask, request, render_template
from flask.json import JSONEncoder

app = Flask(__name__, template_folder='static')


@app.route('/<path:path>')
def serve_static_file(path):
    # In my case, only html files are having the template code inside them, like include.
    if path.endswith('.html'):
        return render_template(path)
    # Serve all other files from the static folder directly.
    return app.send_static_file(path)

我所有的文件都保存在静态文件夹下，与主flask文件平行。

2021-08-29 14:43:10