所以这很尴尬.我有一个应用程序,我把它放在一起,Flask现在它只是提供一个静态HTML页面,其中包含一些CSS和JS的链接.我无法在文档Flask中找到返回静态文件的位置.是的,我可以使用,render_template但我知道数据没有模板化.我曾经想过send_file或者url_for是对的,但是我无法让它们发挥作用.与此同时,我正在打开文件,阅读内容,并Response使用适当的mimetype来装配:
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: …Run Code Online (Sandbox Code Playgroud) 我有一个通过端口5000运行的Flask服务器,它很好.我可以访问http://example.com:5000
但是可以在http://example.com上简单地访问它吗?我假设这意味着我必须将端口从5000更改为80.但是当我在Flask上尝试时,我在运行它时收到此错误消息.
Traceback (most recent call last):
File "xxxxxx.py", line 31, in <module>
app.run(host="0.0.0.0", port=int("80"), debug=True)
File "/usr/local/lib/python2.6/dist-packages/flask/app.py", line 772, in run
run_simple(host, port, self, **options)
File "/usr/local/lib/python2.6/dist-packages/werkzeug/serving.py", line 706, in run_simple
test_socket.bind((hostname, port))
File "<string>", line 1, in bind
socket.error: [Errno 98] Address already in use
Run Code Online (Sandbox Code Playgroud)
运行lsof -i :80回报
COMMAND PID USER FD TYPE DEVICE SIZE/OFF NODE NAME
apache2 467 root 3u IPv4 92108840 0t0 TCP *:www (LISTEN)
apache2 4413 www-data 3u IPv4 92108840 …Run Code Online (Sandbox Code Playgroud)