Dav*_*arx 5 python multithreading network-programming stream flask
这基于/sf/answers/937224081/发布的答案
我想监视数据流并将其推送到类似于上述答案的前端，但是一旦应用启动，该流就开始生成/监视数据，并且客户端始终会看到数据流的当前状态。数据流（无论它们是否正在从服务器请求数据，它都会一直运行）。
我很确定我需要通过线程将数据流与前端分开，但是我对线程/异步编程不是很熟练，并且认为我做错了。也许不是threading我需要使用多重处理？这大致就是我想做的事情（根据上面链接的答案进行了修改）：
app.py
#!/usr/bin/env python
from __future__ import division
import itertools
import time
from flask import Flask, Response, redirect, request, url_for
from random import gauss
import threading
app = Flask(__name__)
# Generate streaming data and calculate statistics from it
class MyStreamMonitor(object):
def __init__(self):
self.sum = 0
self.count = 0
@property
def mu(self):
try:
outv = self.sum/self.count
except:
outv = 0
return outv
def generate_values(self):
while True:
time.sleep(.1) # an artificial delay
yield gauss(0,1)
def monitor(self, report_interval=1):
print "Starting data stream..."
for x in self.generate_values():
self.sum += x
self.count += 1
stream = MyStreamMonitor()
@app.route('/')
def index():
if request.headers.get('accept') == 'text/event-stream':
def events():
while True:
yield "data: %s %d\n\n" % (stream.count, stream.mu)
time.sleep(.01) # artificial delay. would rather push whenever values are updated.
return Response(events(), content_type='text/event-stream')
return redirect(url_for('static', filename='index.html'))
if __name__ == "__main__":
# Data monitor should start as soon as the app is started.
t = threading.Thread(target=stream.monitor() )
t.start()
print "Starting webapp..." # we never get to this point.
app.run(host='localhost', port=23423)
static/index.html
<!doctype html>
<title>Server Send Events Demo</title>
<style>
#data {
text-align: center;
}
</style>
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script>
if (!!window.EventSource) {
var source = new EventSource('/');
source.onmessage = function(e) {
$("#data").text(e.data);
}
}
</script>
<div id="data">nothing received yet</div>
此代码无效。“正在启动webapp ...”消息从不打印，也没有正常的烧瓶消息，并且访问所提供的URL确认该应用程序未在运行。
如何使数据监视器在后台运行，以便Flask可以访问其看到的值并将当前状态推送到客户端（甚至更好：只要客户端正在侦听，在以下情况下推送当前状态：相关值会发生变化）？
我只需要改变这一行
t = threading.Thread(target=stream.monitor())
对此：
t = threading.Thread(target=stream.monitor)
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