为什么flask-socketio 给我一个错误的请求错误？

问题描述

我正在创建一个出价网站，我目前想要实现一项功能，在用户被出价高出时通知他们。我看到了flask-socketIO并决定使用它。我目前正在尝试建立测试连接，但无法正常工作。

main.py

library(ggplot2)
library(ggnewscale)

sum_ua <- data.frame(
  ID = c(1L,3L,5L,11L,13L,18L,20L,24L,33L,34L,36L),sum_hrs = c(12,8,68,44,16,12,36,4,20,4),avg_workload = c(263.1615,275.312,269.462444444444,268.867666666667,276.686,257.3605,267.8695,268.3355,239.409,260.230333333333,330.061),avg_achv = c(92.5,98,94.2222222222222,94.8333333333333,92,94.25,93.5,95.3333333333333,100),mchild = c(1L,0L,1L,2L,4L,1L),drinker = as.factor(c("No","Yes","No","Yes"))
)

ggplot(data=sum_ua,aes(x=avg_achv,y=sum_hrs,size=mchild,color=drinker,alpha=sum_hrs)) +
  geom_point() +
  geom_text(data=sum_ua,label=as.character(mchild)),color='#A9A9A9') +
  scale_size(range=c(4,20)) +
  scale_alpha_binned(range = c(0.01,1),guide = 'none') +   
  scale_discrete_identity(guide = "legend",aesthetics = "label",name = "mchild") +
  new_scale_color()

index.html

from flask import Flask,render_template
from flask_socketio import SocketIO

app = Flask(__name__)
app.config['SECRET_KEY'] = 'secret!'
socketio = SocketIO(app)

@app.route('/')
def home():
    return render_template('index.html')

@socketio.on('connect')
def handle_connection():
    print('User has connected!')

if __name__ == '__main__':
    socketio.run(app,debug=True)

解决方法

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flask flask-socketio socket.io