如何将 JSON 响应转换为字符串并显示在屏幕上?

发布于 2024-11-08 05:37:27 字数 1676 浏览 5 评论 0原文

我已经编写了 android 类,它将调用 RESTful Web 服务。如果请求成功, 响应将是 JSON 对象。我这样写 android 类:

public class android extends Activity {

    public void onCreate(Bundle savedInstanceState)
    {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    TextView txt = (TextView) findViewById(R.id.textView1);
    txt.setText(getInputStreamFromUrl("http://localhost:8080/kyaw"));
    }

    public static String getInputStreamFromUrl(String url) {
          InputStream content = null;
          try {
            HttpClient httpclient = new DefaultHttpClient();
            HttpResponse response = httpclient.execute(new HttpGet(url));
            content = response.getEntity().getContent();
          } catch (Exception e) {
            Log.e("[GET REQUEST]", "Network exception");
          }
            String result=convert(content);
            return result;
        }

    private static String convert(InputStream in)
    {
        BufferedReader reader=new BufferedReader(new InputStreamReader(in));
        StringBuilder sb=new StringBuilder();
        String line=null;
        try{
            while((line=reader.readLine())!=null){
                sb.append(line+"\n");
            }
        }catch(Exception e)
        {
            e.printStackTrace();
        }finally{
            try{
                in.close();
            }catch(IOException e){
                e.printStackTrace();
            }
        }
        return sb.toString();
    }

}

我运行后遇到问题。我会得到异常,因为我作为字符串返回。但响应是 JSON。我应该如何将 JSON 转换为字符串或其他方式,然后如何在android屏幕上显示结果?

谢谢在此处输入图像描述

I have written android class which will call RESTful web service. If request is successful,
response will be JSON object. I write android class like this:

public class android extends Activity {

    public void onCreate(Bundle savedInstanceState)
    {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    TextView txt = (TextView) findViewById(R.id.textView1);
    txt.setText(getInputStreamFromUrl("http://localhost:8080/kyaw"));
    }

    public static String getInputStreamFromUrl(String url) {
          InputStream content = null;
          try {
            HttpClient httpclient = new DefaultHttpClient();
            HttpResponse response = httpclient.execute(new HttpGet(url));
            content = response.getEntity().getContent();
          } catch (Exception e) {
            Log.e("[GET REQUEST]", "Network exception");
          }
            String result=convert(content);
            return result;
        }

    private static String convert(InputStream in)
    {
        BufferedReader reader=new BufferedReader(new InputStreamReader(in));
        StringBuilder sb=new StringBuilder();
        String line=null;
        try{
            while((line=reader.readLine())!=null){
                sb.append(line+"\n");
            }
        }catch(Exception e)
        {
            e.printStackTrace();
        }finally{
            try{
                in.close();
            }catch(IOException e){
                e.printStackTrace();
            }
        }
        return sb.toString();
    }

}

I have a problem after i run.I will get exception coz i return as a string.But response is JSON .How should I convert JSON to String or other way and then how to show result in android screen?

thanksenter image description here

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评论(2

薯片软お妹 2024-11-15 05:37:27

那么“http://localhost:8080/kyaw”是一个问题...通过这个你指向模拟器而不是模拟器的主机...并且你收到网络错误

尝试“http://ip.of.your.host:8080 /kyaw”

编辑:

        content = response.getEntity().getContent();// here comes an error
      } catch (Exception e) {
        Log.e("[GET REQUEST]", "Network exception");//we catch it here
      }
      //here you got content == null
      //so you getting null point exception 

well "http://localhost:8080/kyaw" is a problem ... by this you poining to emulator not emulator's host ... and you getting network error

try "http://ip.of.your.host:8080/kyaw"

EDIT:

        content = response.getEntity().getContent();// here comes an error
      } catch (Exception e) {
        Log.e("[GET REQUEST]", "Network exception");//we catch it here
      }
      //here you got content == null
      //so you getting null point exception 
×纯※雪 2024-11-15 05:37:27

我只是这样做:

HttpClient client = new DefaultHttpClient();
HttpPost poster = new HttpPost("http://www.example.com/json.php");
poster.addHeader("Content-Type", "text/json");
poster.setEntity(new StringEntity(data.toString()));
//data being a json object created and filled earlier
HttpResponse response = client.execute(poster);                                 
if(response.getStatusLine().getStatusCode() == HttpStatus.SC_OK)
{
    DataInputStream input = new DataInputStream(response.getEntity().getContent());
    JSONObject json = new JSONObject(input.readLine());
    json.toString(2);
}

I just do this:

HttpClient client = new DefaultHttpClient();
HttpPost poster = new HttpPost("http://www.example.com/json.php");
poster.addHeader("Content-Type", "text/json");
poster.setEntity(new StringEntity(data.toString()));
//data being a json object created and filled earlier
HttpResponse response = client.execute(poster);                                 
if(response.getStatusLine().getStatusCode() == HttpStatus.SC_OK)
{
    DataInputStream input = new DataInputStream(response.getEntity().getContent());
    JSONObject json = new JSONObject(input.readLine());
    json.toString(2);
}
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