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。
最简单的方法是在主项目文件夹中创建一个静态文件夹。包含。css文件的静态文件夹。
主文件夹
/Main Folder
/Main Folder/templates/foo.html
/Main Folder/static/foo.css
/Main Folder/application.py(flask script)
图像的主文件夹包含静态和模板文件夹和烧瓶脚本
瓶
from flask import Flask, render_template
app = Flask(__name__)
@app.route("/")
def login():
return render_template("login.html")
html(布局)
<!DOCTYPE html>
<html>
<head>
<title>Project(1)</title>
<link rel="stylesheet" href="/static/styles.css">
</head>
<body>
<header>
<div class="container">
<nav>
<a class="title" href="">Kamook</a>
<a class="text" href="">Sign Up</a>
<a class="text" href="">Log In</a>
</nav>
</div>
</header>
{% block body %}
{% endblock %}
</body>
</html>
html
{% extends "layout.html" %}
{% block body %}
<div class="col">
<input type="text" name="username" placeholder="Username" required>
<input type="password" name="password" placeholder="Password" required>
<input type="submit" value="Login">
</div>
{% endblock %}
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
基于其他答案的一个最简单的工作示例如下:
from flask import Flask, request
app = Flask(__name__, static_url_path='')
@app.route('/index/')
def root():
return app.send_static_file('index.html')
if __name__ == '__main__':
app.run(debug=True)
使用名为index.html的HTML:
<!DOCTYPE html>
<html>
<head>
<title>Hello World!</title>
</head>
<body>
<div>
<p>
This is a test.
</p>
</div>
</body>
</html>
重要提示:index.html在一个名为static的文件夹中，这意味着<projectpath>有.py文件，<projectpath>\static有html文件。
如果你想让服务器在网络上可见，使用app.run(debug=True, host='0.0.0.0')
EDIT:如果需要显示文件夹中的所有文件，请使用此选项
@app.route('/<path:path>')
def static_file(path):
return app.send_static_file(path)
这基本上就是黑曼巴的答案，所以给他们点赞。
