Python 向wsdl网站发送soap请求
如何从网站获取请求?我发现: 但是我可以想出如何向它发送请求并得到响应 到目前为止,我已经尝试过:Python 向wsdl网站发送soap请求,python,soap,wsdl,python-requests,Python,Soap,Wsdl,Python Requests,如何从网站获取请求?我发现: 但是我可以想出如何向它发送请求并得到响应 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)
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)
但它不起作用
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.
在尝试将InputText=Something放入url时,我也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时http://www.yodaspeak.co.uk/webservice/yodatalk.php?wsdl
我得到:
No namespace defined for 'http' ('http://www.yodaspeak.co.uk/webservice/yodatalkPortType')
尝试使用任何soap库(例如,)。
在
http://www.yodaspeak.co.uk/webservice/yodatalk.php?wsdl
,我想这是关于soap
使用的。我尝试过,但是当我运行'python3-mzeep'时,我没有为'http'定义名称空间。谢谢。。。使用soap库帮助很大。不幸的是,soap根本不起作用,所以我不得不搜索另一个soap库——一个由另一位作者制作的更新版本的suds,但它在Python3.x上工作