使用 org.apache.http 发送带有 SOAP 操作的 HTTP Post 请求

post http java httpwebrequest apache

2022-09-02 10:14:43

我正在尝试使用org.apache.http api编写一个带有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正文的方法（在我的情况下 - SOAP操作）。我会很高兴得到一些指导。

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();
        }
    }
}

答案 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);
    }
}

答案 2

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

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