如何使用node-soap服务器返回XML？

Question

我正在用NodeJS制作一个包含API REST和SOAP方法(Im使用https://www.npmjs.com/package/soap)的Express服务。使用API REST我没有任何问题，但是对于SOAP我有一个不方便的地方，当我尝试使用来自测试C#应用程序的SOAP方法时，我可以看到参数运行良好，但是在响应中，我在C#中有下一个错误(响应不是正确的XML代码)

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当我通过node-soap从NodeJS客户端使用该方法时，响应也可以正常工作。

我的NodeJS代码的一部分：

const express = require('express');
const bodyParser = require('body-parser');
const soap = require('soap');
const fs = require('fs');

const xml = fs.readFileSync('src/templates/ws_soap.wsdl', 'utf8');

const app = express();

app.use(bodyParser.urlencoded({
  extended: false
}));

app.use(bodyParser.json());

const soap_service = {
  integrations: {
    pull: {
      getSnapshotGIGA: function(args) {
        return {
          res: "HOLA"
        };
      },
    }
  }
};

app.listen(port, ip, function() {

  soap.listen(app, '/integrations_service', soap_service, xml, function() {
    console.log('SOAP web service started on ' + ip + ':' + port);
  });

  console.log('API REST started on ' + ip + ':' + port);
});

我的WSDL文件是next (作为响应，我有一个string类型，因为我想看看它是如何运行的，但我需要返回一个对象XML)：

<?xml version="1.0" encoding="UTF-8"?>
<wsdl:definitions name="integrations_service" targetNamespace="http://localhost:4205/integrations_service" xmlns="http://localhost:4205/integrations_service" xmlns:http="http://schemas.xmlsoap.org/wsdl/http/" xmlns:soap="http://schemas.xmlsoap.org/wsdl/soap/" xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/" xmlns:xs="http://www.w3.org/2001/XMLSchema">
  <wsdl:message name="getSnapshotGIGARequest">
    <wsdl:part name="User" type="xs:string"/>
    <wsdl:part name="Password" type="xs:string"/>
  </wsdl:message>
  <wsdl:message name="getSnapshotGIGAResponse">
    <wsdl:part name="res" type="xs:string"/>
  </wsdl:message>
  <wsdl:portType name="pull_integrations">
    <wsdl:operation name="getSnapshotGIGA">
      <wsdl:input message="getSnapshotGIGARequest"/>
      <wsdl:output message="getSnapshotGIGAResponse"/>
    </wsdl:operation>
  </wsdl:portType>
  <wsdl:binding name="pull_integrations_binding" type="pull_integrations">
    <soap:binding style="rpc" transport="http://schemas.xmlsoap.org/soap/http"/>
    <wsdl:operation name="getSnapshotGIGA">
      <soap:operation soapAction="getSnapshotGIGA"/>
      <wsdl:input>
        <soap:body encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" use="literal"/>
      </wsdl:input>
      <wsdl:output>
        <soap:body encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" use="literal"/>
      </wsdl:output>
    </wsdl:operation>
  </wsdl:binding>
  <wsdl:service name="integrations">
    <wsdl:port binding="pull_integrations_binding" name="pull">
      <soap:address location="http://localhost:4205/integrations_service"/>
    </wsdl:port>
  </wsdl:service>
</wsdl:definitions>

在C#中，我有一个控制台应用程序，并且我已经将SOAP服务注册为web引用。

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我使用SOAP方法的方式是(当我使用C#制作SOAP服务时，我也是通过这种方式测试方法的，因为这是客户端的工作方式)：

Console.WriteLine("Consume NodeJS SOAP service");
Thread.Sleep(500);
integrations_service.integrations integrations = new integrations_service.integrations();
integrations.Url = "http://localhost:4205/integrations_service?wsdl";
var some_response = integrations.getSnapshotGIGA("myuser", "123456");
Console.WriteLine("Press enter to out...");

我想要在XmlNode中获得响应，如本例所示：

Console.WriteLine("Consume C# SOAP service");
Thread.Sleep(500);
serviceSOAP sSOAP = new serviceSOAP ();
sSOAP.Url = "http://my.domain.com.mx/";
XmlNode xmlNode = sSOAP .anyMethodSoap("yomero", "123456");
Console.WriteLine(XElement.Parse(xmlNode.OuterXml).ToString());
Thread.Sleep(500);

如果您知道如何从NodeJS返回XML，并在C#或任何想法中正确获取它，我将不胜感激。雷加兹。

Answer 1

回答我的问题时，问题出在WSDL配置上。通过这个配置，我让它与C# web reference一起工作。主要的问题是样式，我将其从"rpc“改为"document”，并为响应正确配置了元素。

<?xml version="1.0" encoding="UTF-8"?>
<wsdl:definitions
  name="devices_service"
  targetNamespace="http://localhost:4205/devices_service"
  xmlns:http="http://schemas.xmlsoap.org/wsdl/http/"
  xmlns:s="http://schemas.xmlsoap.org/wsdl/soap/"
  xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/"
  xmlns:xs="http://www.w3.org/2001/XMLSchema"
  xmlns="http://localhost:4205/devices_service">
  <wsdl:types>
    <xs:schema targetNamespace="http://localhost:4205/devices_service" xmlns="http://localhost:4205/devices_service" attributeFormDefault="qualified" elementFormDefault="qualified">
      <xs:element name="GetDevicesRequest">
        <xs:complexType>
          <xs:sequence>
            <xs:element minOccurs="0" maxOccurs="1" name="User" type="xs:string"/>
            <xs:element minOccurs="0" maxOccurs="1" name="Password" type="xs:string"/>
          </xs:sequence>
        </xs:complexType>
      </xs:element>
      <xs:element name="GetDevicesResponse">
        <xs:complexType>
          <xs:sequence>
            <xs:any/>
          </xs:sequence>
        </xs:complexType>
      </xs:element>
    </xs:schema>
  </wsdl:types>
  <wsdl:message name="GetDevicesSoapIn">
    <wsdl:part name="parameters" element="GetDevicesRequest"/>
  </wsdl:message>
  <wsdl:message name="GetDevicesSoapOut">
    <wsdl:part name="parameters" element="GetDevicesResponse"/>
  </wsdl:message>
  <wsdl:portType name="user_devices">
    <wsdl:operation name="GetDevices">
      <wsdl:input message="GetDevicesSoapIn"/>
      <wsdl:output message="GetDevicesSoapOut"/>
    </wsdl:operation>
  </wsdl:portType>
  <wsdl:binding name="user_devices_binding" type="user_devices">
    <s:binding transport="http://schemas.xmlsoap.org/soap/http" style="rpc"/>
    <wsdl:operation name="GetDevices">
      <s:operation soapAction="http://localhost:4205/devices_services/GetDevices"/>
      <wsdl:input>
        <s:body use="literal" encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"/>
      </wsdl:input>
      <wsdl:output>
        <s:body use="literal" encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"/>
      </wsdl:output>
    </wsdl:operation>
  </wsdl:binding>
  <wsdl:service name="devices">
    <wsdl:port binding="user_devices_binding" name="user">
      <s:address location="http://localhost:4205/devices_service"/>
    </wsdl:port>
  </wsdl:service>
</wsdl:definitions>**

Answer 2

您可能想要使用以下选项之一：

node-soap强soap(重写node-soap) easysoap