向wsdl-websi发送soap请求

2024-10-01 09:16:04 发布

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如何从网站获取请求?我发现这个: http://www.yodaspeak.co.uk/webservice/yodatalk.php?wsdl

但我可以想出如何向它发送请求并得到响应。你知道吗

到目前为止,我试过:

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)

但没用。你知道吗

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.

我在尝试将InputText=Something放入url时也发现了这个错误

更新:

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

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

Tags: webhttpurlstringwwwserviceuriparams
1条回答
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1楼 · 发布于 2024-10-01 09:16:04

尝试使用任何soap库(例如,zeep)。 在http://www.yodaspeak.co.uk/webservice/yodatalk.php?wsdl中有wsdl,所以我想是关于soap使用的。你知道吗

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