我试图将从client.py创建的python字典发送到我的webservice,让webservice对数据执行某些操作,并将一个布尔值返回给client.py.这是我到目前为止服务器和客户端的代码:
服务器端(在webservice.py内):
from flask import Flask
from flask import request
import json
app = Flask(__name__)
@app.route('/determine_escalation',methods = ['POST'])
def determine_escalation():
jsondata = request.form['jsondata']
data = json.loads(jsondata)
#stuff happens here that involves data to obtain a result
result = {'escalate':'True'}
return json.dumps(result)
if __name__ == '__main__':
app.run(debug=True)
客户端(在client.py内):
import sys
import json
import requests
conv = [{'input': 'hi', 'topic': 'Greeting'}]
s = json.dumps(conv)
res = requests.post("http://127.0.0.1:5000/determine_escalation",data=s)
print res.text
但是当我打印出res.text时,我得到了这个:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<title>400 Bad Request</title>
<h1>Bad Request</h1> …我是新来的Haskell和实践我一直在执行一组函数(map,length使用等)foldr和foldl.现在我想搬到树上!
我的树数据结构:
data Tree a = Node a [Tree a] deriving (Show)
我编写了一个Haskell函数来折叠树:
treeFold :: (b->a->b) -> b -> Tree a -> b
treeFold f s (Node a []) = f s a
treeFold f s (Node a xs) = foldl f (f s a) (dFSList xs)
其中dFSList是树中所有节点的列表.所以做以下事情:
treeFold (+) 0 (Node 1 [Node 2 [], Node 3 []])
返回6.很酷.
现在我想写一个Haskell函数来映射树,但我想用我的treeFold函数来做它.这是我到目前为止:
treeMap f (Node a []) = (Node …