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

如果你只是想打开一个文件，你可以使用app.open_resource()。读取文件看起来就像这样

with app.open_resource('/static/path/yourfile'):
      #code to read the file and do something

2017-08-31 02:26:59

其他回答

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。

2015-02-03 23:05:58

如果您只想移动静态文件的位置，那么最简单的方法是在构造函数中声明路径。在下面的示例中，我已经将模板和静态文件移动到名为web的子文件夹中。

app = Flask(__name__,
            static_url_path='', 
            static_folder='web/static',
            template_folder='web/templates')

static_url_path= "从URL中删除任何前面的路径(即。默认/static)。 Static_folder ='web/static'提供在文件夹中找到的任何文件 Web /static作为静态文件。 Template_folder ='web/templates'类似地，这将改变模板文件夹。

使用这个方法，下面的URL将返回一个CSS文件:

<link rel="stylesheet" type="text/css" href="/css/bootstrap.min.css">

最后，这里是文件夹结构的快照，其中flask_server.py是Flask实例:

2017-03-14 16:44:27

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

这对我来说很管用:

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


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

@app.route('/', methods=['GET'])
def main(request):
    path = request.path
    if (path == '/'):
        return send_from_directory(root, 'index.html')
    else:
        return send_from_directory(root, path[1:])

2021-09-26 04:45:47

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

如何在Flask中提供静态文件

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