Question

如何使用HTTP POST将XML请求发布到URL并检索响应？

更新抱歉，我的问题不明确。我想知道如何使用HttpClient或URLConnection将XML请求发布到网址，并将响应作为POST参数获取，并将其显示在网页中。

Answer 1

以下是如何使用java.net.URLConnection：

执行此操作的示例

String url = "http://example.com";
String charset = "UTF-8";
String param1 = URLEncoder.encode("param1", charset);
String param2 = URLEncoder.encode("param2", charset);
String query = String.format("param1=%s&param2=%s", param1, param2);

URLConnection urlConnection = new URL(url).openConnection();
urlConnection.setUseCaches(false);
urlConnection.setDoOutput(true); // Triggers POST.
urlConnection.setRequestProperty("accept-charset", charset);
urlConnection.setRequestProperty("content-type", "application/x-www-form-urlencoded");

OutputStreamWriter writer = null;
try {
    writer = new OutputStreamWriter(urlConnection.getOutputStream(), charset);
    writer.write(query); // Write POST query string (if any needed).
} finally {
    if (writer != null) try { writer.close(); } catch (IOException logOrIgnore) {}
}

InputStream result = urlConnection.getInputStream();
// Now do your thing with the result.
// Write it into a String and put as request attribute
// or maybe to OutputStream of response as being a Servlet behind `jsp:include`.

Answer 2

此示例发布一个xml文件，它依赖于Jakarta HttpClient API（jakarta.apache.org）

import java.io.File;
import java.io.FileInputStream;

import org.apache.commons.httpclient.HttpClient;
import org.apache.commons.httpclient.methods.InputStreamRequestEntity;
import org.apache.commons.httpclient.methods.PostMethod;

/**
 * This is a sample application that demonstrates
 * how to use the Jakarta HttpClient API.
 *
 * This application sends an XML document
 * to a remote web server using HTTP POST
 *
 * @author Sean C. Sullivan
 * @author Ortwin Glück
 * @author Oleg Kalnichevski
 */
public class PostXML {

    /**
     *
     * Usage:
     * java PostXML http://mywebserver:80/ c:\foo.xml
     *
     * @param args command line arguments
     * Argument 0 is a URL to a web server
     * Argument 1 is a local filename
     *
     */
    public static void main(String[] args) throws Exception {

        if (args.length != 2) {
            System.out.println(
                "Usage: java -classpath <classpath> [-Dorg.apache.commons."+
                "logging.simplelog.defaultlog=<loglevel>]" +
                " PostXML <url> <filename>]");

            System.out.println("<classpath> - must contain the "+
                "commons-httpclient.jar and commons-logging.jar");

            System.out.println("<loglevel> - one of error, "+
                    "warn, info, debug, trace");

            System.out.println("<url> - the URL to post the file to");
            System.out.println("<filename> - file to post to the URL");
            System.out.println();
            System.exit(1);
        }

        // Get target URL
        String strURL = args[0];

        // Get file to be posted
        String strXMLFilename = args[1];
        File input = new File(strXMLFilename);

        // Prepare HTTP post
        PostMethod post = new PostMethod(strURL);

        // Request content will be retrieved directly
        // from the input stream
        // Per default, the request content needs to be buffered
        // in order to determine its length.
        // Request body buffering can be avoided when
        // content length is explicitly specified
        post.setRequestEntity(new InputStreamRequestEntity(
                new FileInputStream(input), input.length()));

        // Specify content type and encoding
        // If content encoding is not explicitly specified
        // ISO-8859-1 is assumed
        post.setRequestHeader(
                "Content-type", "text/xml; charset=ISO-8859-1");

        // Get HTTP client
        HttpClient httpclient = new HttpClient();

        // Execute request
        try {

            int result = httpclient.executeMethod(post);

            // Display status code
            System.out.println("Response status code: " + result);

            // Display response
            System.out.println("Response body: ");
            System.out.println(post.getResponseBodyAsString());

        } finally {
            // Release current connection to the connection pool 
            // once you are done
            post.releaseConnection();
        }
    }
}

Answer 3

使用InputStreamEntity。我使用 httpclient 4.2.1 。

例如：

HttpPost httppost = new HttpPost(url);
InputStream inputStream=new ByteArrayInputStream(xmlString.getBytes());//init your own inputstream
InputStreamEntity inputStreamEntity=new InputStreamEntity(inputStream,xmlString.getBytes());
httppost.setEntity(inputStreamEntity);

Answer 4

警告此代码已超过5年。我为这篇文章做过一些修改，从未测试过。希望它有所帮助。

将XML（数据）发布到服务器并下载resp：

 public int uploadToServer(String data) throws Exception {
        OutputStream os;


      URL url = new URL("someUrl");


           HttpURLConnection httpConn= (HttpURLConnection) url.openConnection();
            os = httpConn.getOutputStream();



        BufferedWriter osw = new BufferedWriter(new OutputStreamWriter(os));

        osw.write(data);
        osw.flush();
        osw.close();

        return httpConn.getResponseCode();

    }



 public String downloadFromServer()
            throws MalformedURLException, IOException {

        String returnString = null;
        StringBuffer sb = null;
        BufferedInputStream in;



//set up httpConn code not included same as previous

            in = new BufferedInputStream(httpConn.getInputStream());


        int x = 0;

        sb = new StringBuffer();

        while ((x = in.read()) != -1) {
            sb.append((char) x);
        }

        in.close();
        in = null;


        if (httpConn != null) {
            httpConn.disconnect();
        }


        returnString = sb.toString();



        return returnString;

    }

其他地方......

int respCode = uploadToServer(someXmlData);



if (respCode  == 200) {

    String respData = downloadFromServer();

}

在java中发布XML请求

4 个答案: