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

我相信你会在那里找到你需要的东西:http://flask.pocoo.org/docs/quickstart/#static-files

基本上你只需要在你的包的根目录下有一个“静态”文件夹,然后你就可以使用url_for('static', filename='foo.bar')或者直接用http://example.com/static/foo.bar链接到你的文件。

编辑:正如评论中所建议的那样,你可以直接使用'/static/foo. xml '。但是url_for()开销(性能明智)是相当低的,使用它意味着你将能够轻松地自定义行为之后(改变文件夹,改变URL路径,移动你的静态文件到S3等)。

其他回答

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:])

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
app = Flask(__name__, static_folder="your path to static")

如果你的根目录中有模板,放置app=Flask(name)将工作,如果文件包含这个也在相同的位置,如果这个文件在另一个位置,你必须指定模板的位置,以使Flask指向该位置

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])```

你也可以,这是我最喜欢的,将一个文件夹设置为静态路径,这样每个人都可以访问其中的文件。

app = Flask(__name__, static_url_path='/static')

有了这个设置,你可以使用标准的HTML:

<link rel="stylesheet" type="text/css" href="/static/style.css">