Python使用urllib或requests调用WSDL接口

import urllib.request
import html

url = "http://xx.xx.xx/xxx"
target_namespace = "http://xx.xx.xx/"
target_method = "xxxxxxxxxx"

# 下面是接口的两个参数，这个按接口的要求来。
# 需要注意的是参数在拼接成xml请求body时需要经过html编码，防止参数中的恶意字符导致xml注入
param_aaa = "xxxxxxxxxxxx"
param_bbb = "xxxxxxxxxx"

body = '<?xml version="1.0"?>'
body += f'<SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" ' \
        f'xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" ' \
        f'xmlns:ns0="{target_namespace}" ' \
        f'xmlns:ns1="http://schemas.xmlsoap.org/soap/envelope/">'
body += '<SOAP-ENV:Header/>'
body += '   <ns1:Body>'
body += f'      <ns0:{target_method}>'
body += f'         <ns0:sTaskType>{html.escape(param_aaa)}</ns0:sTaskType>'  # 需要经过html编码
body += f'         <ns0:sImport>{html.escape(param_bbb)}</ns0:sImport>'  # 需要经过html编码
body += f'      </ns0:{target_method}>'
body += '   </ns1:Body>'
body += '</SOAP-ENV:Envelope>'

# headers也可能是{'content-type': 'application/soap+xml'}，需按实际的进行配置
headers = {'content-type': 'text/xml'}
req = urllib.request.Request(url, data=body.encode(), headers=headers)
response = urllib.request.urlopen(req)
if response.status != 200:
    raise Exception("query WSDL failed.")
print(response.read().decode())

# 如果使用requests库，则会更简便一些，速度也更快：
import requests
response = requests.post(url, data=body, headers=headers)
if response.status_code != 200:
    raise Exception("query WSDL failed.")
print(response.text)