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 %}
在我的情况下，我需要一个静态文件夹中的所有文件都可以由用户访问，以及我需要为我的一些html文件使用模板，这样公共html代码可以放在模板中，代码得到重用。以下是我如何同时实现这两个目标的:
from flask import Flask, request, render_template
from flask.json import JSONEncoder
app = Flask(__name__, template_folder='static')
@app.route('/<path:path>')
def serve_static_file(path):
# In my case, only html files are having the template code inside them, like include.
if path.endswith('.html'):
return render_template(path)
# Serve all other files from the static folder directly.
return app.send_static_file(path)
我所有的文件都保存在静态文件夹下，与主flask文件平行。
对于angular+样板流，它创建下一个文件夹树:
backend/
|
|------ui/
| |------------------build/ <--'static' folder, constructed by Grunt
| |--<proj |----vendors/ <-- angular.js and others here
| |-- folders> |----src/ <-- your js
| |----index.html <-- your SPA entrypoint
|------<proj
|------ folders>
|
|------view.py <-- Flask app here
我使用以下解决方案:
...
root = os.path.join(os.path.dirname(os.path.abspath(__file__)), "ui", "build")
@app.route('/<path:path>', methods=['GET'])
def static_proxy(path):
return send_from_directory(root, path)
@app.route('/', methods=['GET'])
def redirect_to_index():
return send_from_directory(root, 'index.html')
...
它有助于重新定义“静态”文件夹自定义。
