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

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])```

其他回答

你可以使用这个函数:

send_static_file(文件名) 内部用于发送静态数据的函数 文件从静态文件夹转移到浏览器。

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

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。

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

   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.

我使用的是一个“模板”目录和一个“静态”目录。我把所有的.html文件/Flask模板放在模板目录中,静态包含CSS/JS。据我所知,render_template适用于通用html文件,不管你在多大程度上使用Flask的模板语法。下面是views.py文件中的一个示例调用。

@app.route('/projects')
def projects():
    return render_template("projects.html", title = 'Projects')

只要确保当您想要引用单独静态目录中的某个静态文件时使用url_for()即可。你可能会在CSS/JS文件链接的html中这样做。例如……

<script src="{{ url_for('static', filename='styles/dist/js/bootstrap.js') }}"></script>

这里有一个链接到“规范的”非正式的Flask教程——这里有很多很棒的提示,可以帮助你快速上手。

http://blog.miguelgrinberg.com/post/the-flask-mega-tutorial-part-i-hello-world