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

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)

2021-10-11 17:32:46

其他回答

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

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

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

默认文件夹名为“static”，包含所有静态文件下面是一个代码示例:

<link href="{{ url_for('static', filename='vendor/bootstrap/css/bootstrap.min.css') }}" rel="stylesheet">

2020-08-21 15:54:32

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

2013-12-18 00:52:38

如何在Flask中提供静态文件

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