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

想到分享....这个例子。

from flask import Flask
app = Flask(__name__)

@app.route('/loading/')
def hello_world():
    data = open('sample.html').read()    
    return data

if __name__ == '__main__':
    app.run(host='0.0.0.0')

这样工作起来更好，也更简单。

2018-10-03 14:01:09

其他回答

使用重定向和url_for

from flask import redirect, url_for

@app.route('/', methods=['GET'])
def metrics():
    return redirect(url_for('static', filename='jenkins_analytics.html'))

这个服务器在你的html中引用的所有文件(css & js…)

2015-12-28 22:21:59

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

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

2019-05-30 14:04:38

By default, flask use a "templates" folder to contain all your template files(any plain-text file, but usually .html or some kind of template language such as jinja2 ) & a "static" folder to contain all your static files(i.e. .js .css and your images). In your routes, u can use render_template() to render a template file (as I say above, by default it is placed in the templates folder) as the response for your request. And in the template file (it's usually a .html-like file), u may use some .js and/or `.css' files, so I guess your question is how u link these static files to the current template file.

2013-12-18 13:49:13

在生产环境中，在应用程序前面配置HTTP服务器(Nginx, Apache等)，从静态文件夹向/static提供请求。专用的web服务器非常擅长有效地提供静态文件，尽管与低容量的Flask相比，您可能不会注意到差异。

Flask自动创建一个/static/<path:filename>路由，它将在定义你的Flask应用程序的Python模块旁边的静态文件夹下提供任何文件名。使用url_for链接到静态文件:url_for('static'， filename='js/ analysis .js')

您还可以使用send_from_directory在自己的路由中提供来自某个目录的文件。这接受一个基本目录和一个路径，并确保该路径包含在目录中，从而可以安全地接受用户提供的路径。如果您希望在提供文件之前检查某些内容，例如登录用户是否具有权限，那么这可能非常有用。

from flask import send_from_directory

@app.route('/reports/<path:path>')
def send_report(path):
    return send_from_directory('reports', path)

不要对用户提供的路径使用send_file或send_static_file。这将使您暴露于目录遍历攻击。Send_from_directory的设计目的是安全地处理已知目录下用户提供的路径，如果该路径试图逃离该目录，则会引发错误。

如果在内存中生成一个文件而不将其写入文件系统，则可以将BytesIO对象传递给send_file，使其像文件一样提供服务。在本例中，您需要将其他参数传递给send_file，因为它不能推断文件名或内容类型等内容。

2013-12-18 01:25:40

如何在Flask中提供静态文件

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