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

   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.

其他回答

在静态目录中,在该目录中创建模板目录,添加所有的html文件,为css和javascript创建单独的目录,因为flask将处理或识别模板目录中的所有html文件。

static -
       |_ templates
       |_ css
       |_javascript
       |_images

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)。

我遇到的问题与使用static_url_path和static_folder时没有为目录提供index.html文件有关。

以下是我的解决方案:

import os
from flask import Flask, send_from_directory
from flask.helpers import safe_join

app = Flask(__name__)
static = safe_join(os.path.dirname(__file__), 'static')

@app.route('/')
def _home():
  return send_from_directory(static, 'index.html')

@app.route('/<path:path>')
def _static(path):
  if os.path.isdir(safe_join(static, path)):
    path = os.path.join(path, 'index.html')
  return send_from_directory(static, path)

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:

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。