java实现URL带参数请求(get/post)

最新推荐文章于 2024-07-07 03:34:31 发布

原创最新推荐文章于 2024-07-07 03:34:31 发布 · 2.9k 阅读

2 ·

CC 4.0 BY-SA版权

文章标签：

#url #java #string #exception #null

javabase 专栏收录该内容

16 篇文章

订阅专栏

本文介绍如何使用Java通过GET和POST方式发送带有参数的URL请求。提供了两种方法的具体代码实现，可用于网络请求的数据交互。

java实现URL带参数请求(get/post)

2007年06月27日星期三 09:23

param 为aaa=111&bbb=222.....格式

-------------------------------------------------
public static String sendGet(String url,String param)
{
   String result = "";
   try{
    String urlName = url + "?"+param;//

    URL U = new URL(urlName);
    URLConnection connection = U.openConnection();
    connection.connect();

    BufferedReader in = new BufferedReader(new InputStreamReader(connection.getInputStream()));
    String line;
    while ((line = in.readLine())!= null)
    {
     result += line;
    }
    in.close();
   }catch(Exception e){
    System.out.println("没有结果！"+e);
   }
   return result;
}

public static String sendPost(String url,String param)
{
   String result="";
   try{
    URL httpurl = new URL(url);
    HttpURLConnection httpConn = (HttpURLConnection)httpurl.openConnection();
    httpConn.setDoOutput(true);
    httpConn.setDoInput(true);
    PrintWriter out = new PrintWriter(httpConn.getOutputStream());
    out.print(param);
    out.flush();
    out.close();
    BufferedReader in = new BufferedReader(new InputStreamReader(httpConn.getInputStream()));
    String line;
    while ((line = in.readLine())!= null)
    {
     result += line;
    }
    in.close();
   }catch(Exception e){
    System.out.println("没有结果！"+e);
   }
   return result;
}