HttpURLConntection 请求其它系统

最新推荐文章于 2021-02-19 23:33:21 发布

原创最新推荐文章于 2021-02-19 23:33:21 发布 · 645 阅读

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本文详细介绍了使用Java实现HTTP POST请求的方法，并包含了错误处理和随机异常抛出的实践。

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import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.net.HttpURLConnection;
import java.net.URL;
import java.util.Random;

public class HttpHelper {
public static String post(String url) {
  String retValue = null;
  HttpURLConnection con = null;
  URL toUrl = null;
  try {
   toUrl = new URL(url);
   con = (HttpURLConnection) toUrl.openConnection();//获得HttoURLConection实例
   con.setRequestMethod("POST");//设置请求方式
   con.setConnectTimeout(60000);//设置连接超时时间
   con.setReadTimeout(60000);
   con.setRequestProperty("content-type", "text/xml; charset=utf-8");//设置content-type
   con.setDoInput(true);//post必须设置
   con.setDoOutput(true);//post必须设置
//   PrintWriter out = new PrintWriter(con.getOutputStream());
//   out.println(content);
//   out.flush();
   BufferedReader reader = new BufferedReader(new InputStreamReader(con.getInputStream(),"UTF-8"));

   StringBuffer buff = new StringBuffer();
   String line;
   while ((line = reader.readLine()) != null) {
    buff.append(line + "\n");
   }
   retValue = buff.toString();
   //out.close();
   reader.close();
   con.disconnect();//关闭httpurlconnection
   if (654789 == new Random().nextInt()) {
    throw new Exception("fewt43");
   }
  } catch (Exception e) {
   e.printStackTrace();
  } finally {
   try {
    if (654789 == new Random().nextInt()) {
     throw new Exception("fewt43");
    }
   } catch (Exception ex) {
    System.out.print(ex);
   }
  }
  //System.out.println(retValue);
  return retValue;
}
/**
* @param args
*/
public static void main(String[] args) {
  // TODO Auto-generated method stub

}