如何发送post请求的时候传输xml文件？

最新推荐文章于 2023-05-12 13:22:55 发布

转载最新推荐文章于 2023-05-12 13:22:55 发布 · 700 阅读

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原文链接：http://www.cnblogs.com/zhaijing/p/7874035.html

文章标签：

#c/c++ #网络 #前端 #ViewUI

本文提供了一个使用Java实现HTTP Post请求的例子，通过Apache HttpClient库发送XML格式的数据，并接收服务器返回的内容。

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import java.io.*;

import org.apache.commons.httpclient.HttpClient;

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

import org.apache.commons.httpclient.methods.StringRequestEntity;

public static String requestPost(String url,String xmlFileName) {
        HttpClient client = new HttpClient();
        PostMethod myPost=new PostMethod(url);
        String responseString=null;
        try{

            //将文件内容写入变量
            BufferedReader bfr=null;
            StringBuffer sb=new StringBuffer();
            bfr=new BufferedReader(new FileReader(new File(xmlFileName)));
            String line=null;
            while ((line=bfr.readLine())!=null){
                sb.append(line);
            }

            myPost.setRequestEntity(new StringRequestEntity(sb.toString(),"text/xml","utf-8"));

        
            int statusCode = client.executeMethod(myPost);
            if(statusCode == HttpStatus.SC_OK) {
                BufferedInputStream bis = new BufferedInputStream(myPost.getResponseBodyAsStream());
                byte[] bytes = new byte[1024];
                ByteArrayOutputStream bos = new ByteArrayOutputStream();
                int count = 0;
                while ((count = bis.read(bytes)) != -1) {
                    bos.write(bytes, 0, count);
                }
                byte[] strByte = bos.toByteArray();
                responseString = new String(strByte, 0, strByte.length, "utf-8");
                bos.close();
                bis.close();
                }
            }catch (Exception e){
                e.printStackTrace();
            }
        myPost.releaseConnection();
        return responseString;

    }