Question

我正在尝试使用org.apache.http api编写带有SOAP操作的硬编码HTTP Post请求。我的问题是我没有找到添加请求体的方法（在我的情况下 - SOAP动作）。我很乐意提供一些指导。

import java.net.URI;
import org.apache.http.HttpResponse;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.impl.client.RequestWrapper;
import org.apache.http.protocol.HTTP;

public class HTTPRequest
{
    @SuppressWarnings("unused")
    public HTTPRequest()
    {
        try {
            HttpClient httpclient = new DefaultHttpClient();
            String body="DataDataData";
            String bodyLength=new Integer(body.length()).toString();
            System.out.println(bodyLength);
//          StringEntity stringEntity=new StringEntity(body);

            URI uri=new URI("SOMEURL?Param1=1234&Param2=abcd");
            HttpPost httpPost = new HttpPost(uri);
            httpPost.addHeader("Test", "Test_Value");

//          httpPost.setEntity(stringEntity);

            StringEntity entity = new StringEntity(body, "text/xml",HTTP.DEFAULT_CONTENT_CHARSET);
            httpPost.setEntity(entity);

            RequestWrapper requestWrapper=new RequestWrapper(httpPost);
            requestWrapper.setMethod("POST");
            requestWrapper.setHeader("LuckyNumber", "77");
            requestWrapper.removeHeaders("Host");
            requestWrapper.setHeader("Host", "GOD_IS_A_DJ");
//          requestWrapper.setHeader("Content-Length",bodyLength);          
            HttpResponse response = httpclient.execute(requestWrapper);
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

Answer 1

这是一个完整的工作示例：

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpPost; 
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.util.EntityUtils;

public void callWebService(String soapAction, String soapEnvBody)  throws IOException {
    // Create a StringEntity for the SOAP XML.
    String body ="<?xml version=\"1.0\" encoding=\"UTF-8\"?><SOAP-ENV:Envelope xmlns:SOAP-ENV=\"http://schemas.xmlsoap.org/soap/envelope/\" xmlns:ns1=\"http://example.com/v1.0/Records\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:SOAP-ENC=\"http://schemas.xmlsoap.org/soap/encoding/\" SOAP-ENV:encodingStyle=\"http://schemas.xmlsoap.org/soap/encoding/\"><SOAP-ENV:Body>"+soapEnvBody+"</SOAP-ENV:Body></SOAP-ENV:Envelope>";
    StringEntity stringEntity = new StringEntity(body, "UTF-8");
    stringEntity.setChunked(true);

    // Request parameters and other properties.
    HttpPost httpPost = new HttpPost("http://example.com?soapservice");
    httpPost.setEntity(stringEntity);
    httpPost.addHeader("Accept", "text/xml");
    httpPost.addHeader("SOAPAction", soapAction);

    // Execute and get the response.
    HttpClient httpClient = new DefaultHttpClient();
    HttpResponse response = httpClient.execute(httpPost);
    HttpEntity entity = response.getEntity();

    String strResponse = null;
    if (entity != null) {
        strResponse = EntityUtils.toString(entity);
    }
}

Answer 2

soapAction必须作为http-header参数传递 - 使用时，它不是http-body / payload的一部分。

在这里查看apache httpclient：http://svn.apache.org/repos/asf/httpcomponents/oac.hc3x/trunk/src/examples/PostSOAP.java

的示例

Answer 3

... using org.apache.http api. ...

您需要在请求中包含SOAPAction作为标头。由于您有httpPost和requestWrapper句柄，因此添加标题有三种方法。

 1. httpPost.addHeader( "SOAPAction", strReferenceToSoapActionValue );
 2. httpPost.setHeader( "SOAPAction", strReferenceToSoapActionValue );
 3. requestWrapper.setHeader( "SOAPAction", strReferenceToSoapActionValue );

唯一的区别是addHeader允许具有相同标题名称的多个值，而setHeader仅允许唯一标题名称。 setHeader(... over写出具有相同名称的第一个标头。

您可以根据自己的要求选择其中任何一项。

Answer 4

通过Java客户端调用WCF服务时，确定需要在soap操作上设置的内容的最简单方法是加载wsdl，转到与服务匹配的操作名称。从那里拿起动作URI并在soap动作头中设置它。你完成了。

例如：来自wsdl

<wsdl:operation name="MyOperation">
  <wsdl:input wsaw:Action="http://tempuri.org/IMyService/MyOperation" message="tns:IMyService_MyOperation_InputMessage" />
  <wsdl:output wsaw:Action="http://tempuri.org/IMyService/MyServiceResponse" message="tns:IMyService_MyOperation_OutputMessage" />

现在在java代码中我们应该将soap操作设置为Action URI。

//The rest of the httpPost object properties have not been shown for brevity
string actionURI='http://tempuri.org/IMyService/MyOperation';
httpPost.setHeader( "SOAPAction", actionURI);

Answer 5

它将 415 Http响应代码视为错误，

所以我添加了

httppost.addHeader("Content-Type", "text/xml; charset=utf-8");

现在一切都好了，Http：200

Answer 6

这是我尝试过的示例，它对我有用：

创建XML文件SoapRequestFile.xml

stack

这里是Java中的代码：

<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:tem="http://tempuri.org/">
       <soapenv:Header/>
       <soapenv:Body>
          <tem:GetConversionRate>
             <!--Optional:-->
             <tem:CurrencyFrom>USD</tem:CurrencyFrom>
             <!--Optional:-->
             <tem:CurrencyTo>INR</tem:CurrencyTo>
             <tem:RateDate>2018-12-07</tem:RateDate>
          </tem:GetConversionRate>
       </soapenv:Body>
    </soapenv:Envelope>

使用org.apache.http使用SOAP操作发送HTTP Post请求

6 个答案: