Python 获取web服务的完整路径'；从wsdl文件中删除方法

我正在使用suds库python2.7，我想从wsdl文件中获得所有方法的完整路径。 我使用以下代码从wsdl文件获取方法列表：

from suds.client import Client

url = "http://www.webservicex.net/country.asmx?WSDL"
client = Client(url)
list_of_methods = [method for method in client.wsdl.services[0].ports[0].methods]

我得到了这样的结果：

[GetCurrencyCodeByCurrencyName, GetCurrencyByCountry, GetCurrencyCode, GetCountries, GetCurrencies, GetISOCountryCodeByCountyName, GetISD, GetCountryByCurrencyCode, GetGMTbyCountry, GetCountryByCountryCode]

[http://www.webserviceX.NET/GetCountryByCountryCode, ...]

但我想得到这样的东西：

[GetCurrencyCodeByCurrencyName, GetCurrencyByCountry, GetCurrencyCode, GetCountries, GetCurrencies, GetISOCountryCodeByCountyName, GetISD, GetCountryByCurrencyCode, GetGMTbyCountry, GetCountryByCountryCode]

[http://www.webserviceX.NET/GetCountryByCountryCode, ...]

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那么，我该怎么做呢？

使用端口的

qname

属性怎么样

from suds.client import Client
url = "http://www.webservicex.net/country.asmx?WSDL"

client = Client(url)
port = client.wsdl.services[0].ports[0]
list_of_methods = [port.qname[1] + '/' + method for method in port.methods]
print(list_of_methods)

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使用端口的

qname

属性怎么样

from suds.client import Client
url = "http://www.webservicex.net/country.asmx?WSDL"

client = Client(url)
port = client.wsdl.services[0].ports[0]
list_of_methods = [port.qname[1] + '/' + method for method in port.methods]
print(list_of_methods)

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谢谢你的回答谢谢你的回答