我有一个应用程序,该应用程序已集成在一起,Flask现在它只提供一个静态HTML页面,其中包含指向CSS和JS的链接。而且我找不到文档中Flask描述返回静态文件的位置。是的,我可以使用,render_template但是我知道数据没有模板化。我还以为send_file或者url_for是正确的事情,但我不能让这些工作。同时,我正在打开文件,阅读内容,并装配Response具有适当mimetype的:
Flask
HTML
render_template
send_file
url_for
Response
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)
有人要为此提供代码示例或网址吗?我知道这将变得简单。
首选方法是使用nginx或其他Web服务器提供静态文件。他们将比Flask更有效率。
但是,你可以用来send_from_directory从目录发送文件,这在某些情况下非常方便:
from flask import Flask, request, send_from_directory # set the project root directory as the static folder, you can set others. app = Flask(__name__, static_url_path='') @app.route('/js/<path:path>') def send_js(path): return send_from_directory('js', path) if __name__ == "__main__": app.run() 千万不能使用send_file或send_static_file与用户提供的路径。 send_static_file 例: from flask import Flask, request # set the project root directory as the static folder, you can set others. app = Flask(__name__, static_url_path='') @app.route('/') def root(): return app.send_static_file('index.html')