发送肥皂请求到wsdl网站

Question

如何从网站获取请求？ 我发现了这一点： http : //www.yodaspeak.co.uk/webservice/yodatalk.php?wsdl

但是我可以弄清楚如何向它发送请求并获得响应。

到目前为止，我已经尝试过：

import requests


    yoda_params = {"inputText": 'Is this working?'}
    yoda_url = 'http://www.yodaspeak.co.uk/webservice/yodatalk.php?'
    yoda_re = requests.get(yoda_url, params=yoda_params)
    yoda_text = yoda_re.json()
    print(yoda_text)

但这没有用。

Name: yodaTalk
Binding: http://www.yodaspeak.co.uk/webservice/yodatalkBinding
Endpoint: http://www.yodaspeak.co.uk/webservice/yodatalk.php
SoapAction: uri:http://www.yodaspeak.co.uk/webservice/yodatalk#yodaTalk
Style: rpc
Input:
  use: literal
  namespace: uri:http://www.yodaspeak.co.uk/webservice/yodatalk
  encodingStyle:
  message: yodaTalkRequest
  parts:
    inputText: xsd:string
Output:
  use: literal
  namespace: uri:http://www.yodaspeak.co.uk/webservice/yodatalk
  encodingStyle:
  message: yodaTalkResponse
  parts:
    return: xsd:string
Namespace: uri:http://www.yodaspeak.co.uk/webservice/yodatalk
Transport: http://schemas.xmlsoap.org/soap/http
Documentation: Pass any string and it will be returned as Yoda-Speak.

我还在尝试将InputText = Something放入url时错误地发现了这一点

更新：

我尝试使用zeep，但是当我运行python -mzeep 'http://www.yodaspeak.co.uk/webservice/yodatalk.php?wsdl我得到了：

No namespace defined for 'http' ('http://www.yodaspeak.co.uk/webservice/yodatalkPortType')

Answer 1

尝试使用任何肥皂库（例如zeep ）。 http://www.yodaspeak.co.uk/webservice/yodatalk.php?wsdl有wsdl ，所以我想这与soap使用有关。