Question

 class MySync extends AsyncTask{ ProgressDialog mProgressDialog;

    protected void onPreExecute(){
       mProgressDialog = ProgressDialog.show(MainActivity.this, "Loading...", "Data is Loading...");
    }
    @Override
    protected Integer doInBackground(String... params)  {
        int result = 0;
       //String url="http://192.168.0.108:8080/sbi/login?"+"key1="+params[0]+"&key2="+params[1]";
        int code;
        try {
            URL hp=new URL("http://192.168.0.108:8080/sbi/login?"+"key1="+params[0]+"&key2="+params[1]);
            HttpURLConnection urlConnection=(HttpURLConnection)hp.openConnection();
            urlConnection.connect();
            Log.i("A", "connect");
            code=urlConnection.getResponseCode();
            Log.i("A","code");
            boolean a=check(code);

           if(a){
                //urlConnection.setDoInput(true);
                Log.i("A", "input");
               // urlConnection.setDoOutput(true);
                Log.i("A", "output");
                urlConnection.setRequestMethod("GET");
                Log.i("A", "get");
                byte [] buf=("key1=" + params[0] + "&key2=" + params[1]).getBytes();
                urlConnection.getOutputStream().write(buf);
                Log.i("A", "sent");
            }
            else{
                Log.i("A","error");
                result=3;
            }
        }
        catch (MalformedURLException e) {
            Log.i("e", "Error");
        }
        catch (IOException e){
            e.printStackTrace();
        }
    protected boolean check(int c){
            if(c==200) return true;
            else return false;
        }
   }

这段代码给出的错误方法不支持请求体：get？如果我插入setdooutput（true），那么它会给出已经连接的错误。我是android的新手，我正在制作我的大学项目

Answer 1

如果您确实要将键值对发送到请求正文中的服务器，请更改

urlConnection.setRequestMethod("GET");

到

urlConnection.setRequestMethod("POST");

或者，如果服务器不支持POST，但要求您执行GET，请删除这些行

byte [] buf=("key1=" + params[0] + "&key2=" + params[1]).getBytes();
urlConnection.getOutputStream().write(buf);

正如我从这一行所见

URL hp=new URL("http://192.168.0.108:8080/sbi/login?"+"key1="+params[0]+"&key2="+params[1]);

您已经为GET Http请求正确构建了Url，但是您正在向不支持请求正文的HTTP请求方法添加请求正文（在这种情况下为GET http方法）。

请查看this wikipidea页面，了解有关使用HTTP / S的REST和REST的更多详细信息，以便更详细地了解此架构

Answer 2

准备URL连接如下

HttpURLConnection conn = null;
    try {
        conn = (HttpURLConnection) (url.openConnection());
    } catch (IOException e) {
        e.printStackTrace();
    }

致电

String dataToSend = "(\"key1=\" + params[0] + \"&key2=\" + params[1])";
conn.setRequestMethod("POST");// do not use "GET" in your case

conn.setRequestProperty("Content-Type", "application/json");//whatever you want

conn.setRequestProperty("Content-Length", "" + dataToSend.getBytes().length);

conn.setUseCaches(false);//set true to enable Cache for the req
conn.setDoOutput(true);//enable to write data to output stream
OutputStream os = conn.getOutputStream();
os.write(dataToSend.getBytes());
os.flush();
os.close();

dot忘记在所有

之后调用以下方法

conn.connect();

方法不支持请求体：GET

2 个答案: