67

我是 Java 和 Android 开发的新手，我尝试创建一个简单的应用程序，它应该联系 Web 服务器并使用 http get 将一些数据添加到数据库中。

当我使用计算机中的网络浏览器拨打电话时，它工作得很好。但是，当我在 Android 模拟器中运行应用程序时，不会添加任何数据。

我已将 Internet 权限添加到应用程序的清单中。Logcat 不报告任何问题。

谁能帮我找出问题所在？

这是源代码：

package com.example.httptest;

import java.io.IOException;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
import android.app.Activity;
import android.os.Bundle;
import android.util.Log;
import android.widget.TextView;

public class HttpTestActivity extends Activity {
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);

        TextView tv = new TextView(this);
        setContentView(tv);

        try {
            URL url = new URL("http://www.mysite.se/index.asp?data=99");
            HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
            urlConnection.disconnect();
            tv.setText("Hello!");
        }
        catch (MalformedURLException ex) {
            Log.e("httptest",Log.getStackTraceString(ex)); 
        }
        catch (IOException ex) {
            Log.e("httptest",Log.getStackTraceString(ex));
        }   
    }        
}
4

7 回答 7

76

尝试从中获取输入流，然后可以这样获取文本数据：-

    URL url;
    HttpURLConnection urlConnection = null;
    try {
        url = new URL("http://www.mysite.se/index.asp?data=99");

        urlConnection = (HttpURLConnection) url
                .openConnection();

        InputStream in = urlConnection.getInputStream();

        InputStreamReader isw = new InputStreamReader(in);

        int data = isw.read();
        while (data != -1) {
            char current = (char) data;
            data = isw.read();
            System.out.print(current);
        }
    } catch (Exception e) {
        e.printStackTrace();
    } finally {
        if (urlConnection != null) {
            urlConnection.disconnect();
        }    
    }

您也可以使用其他输入流阅读器，例如缓冲阅读器。

问题是当您打开连接时 - 它不会“拉”任何数据。

于 2011-12-28T11:03:21.523 回答
43

这是一个完整的AsyncTask类

public class GetMethodDemo extends AsyncTask<String , Void ,String> {
    String server_response;

    @Override
    protected String doInBackground(String... strings) {

        URL url;
        HttpURLConnection urlConnection = null;

        try {
            url = new URL(strings[0]);
            urlConnection = (HttpURLConnection) url.openConnection();

            int responseCode = urlConnection.getResponseCode();

            if(responseCode == HttpURLConnection.HTTP_OK){
                server_response = readStream(urlConnection.getInputStream());
                Log.v("CatalogClient", server_response);
            }

        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

        return null;
    }

    @Override
    protected void onPostExecute(String s) {
        super.onPostExecute(s);

        Log.e("Response", "" + server_response);


    }
}

// Converting InputStream to String

private String readStream(InputStream in) {
        BufferedReader reader = null;
        StringBuffer response = new StringBuffer();
        try {
            reader = new BufferedReader(new InputStreamReader(in));
            String line = "";
            while ((line = reader.readLine()) != null) {
                response.append(line);
            }
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            if (reader != null) {
                try {
                    reader.close();
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        }
        return response.toString();
    }

调用这个AsyncTask类

new GetMethodDemo().execute("your web-service url");
于 2016-07-11T18:03:59.723 回答
19

我已经创建了对 Activity 类的 callBack(delegate) 响应。

public class WebService extends AsyncTask<String, Void, String> {

    private Context mContext;
    private OnTaskDoneListener onTaskDoneListener;
    private String urlStr = "";

    public WebService(Context context, String url, OnTaskDoneListener onTaskDoneListener) {
        this.mContext = context;
        this.urlStr = url;
        this.onTaskDoneListener = onTaskDoneListener;
    }

    @Override
    protected String doInBackground(String... params) {
        try {

            URL mUrl = new URL(urlStr);
            HttpURLConnection httpConnection = (HttpURLConnection) mUrl.openConnection();
            httpConnection.setRequestMethod("GET");
            httpConnection.setRequestProperty("Content-length", "0");
            httpConnection.setUseCaches(false);
            httpConnection.setAllowUserInteraction(false);
            httpConnection.setConnectTimeout(100000);
            httpConnection.setReadTimeout(100000);

            httpConnection.connect();

            int responseCode = httpConnection.getResponseCode();

            if (responseCode == HttpURLConnection.HTTP_OK) {
                BufferedReader br = new BufferedReader(new InputStreamReader(httpConnection.getInputStream()));
                StringBuilder sb = new StringBuilder();
                String line;
                while ((line = br.readLine()) != null) {
                    sb.append(line + "\n");
                }
                br.close();
                return sb.toString();
            }
        } catch (IOException e) {
            e.printStackTrace();
        } catch (Exception ex) {
            ex.printStackTrace();
        }
        return null;
    }

    @Override
    protected void onPostExecute(String s) {
        super.onPostExecute(s);

        if (onTaskDoneListener != null && s != null) {
            onTaskDoneListener.onTaskDone(s);
        } else
            onTaskDoneListener.onError();
    }
}

在哪里

public interface OnTaskDoneListener {
    void onTaskDone(String responseData);

    void onError();
}

您可以根据自己的需要进行修改。是为了得到

于 2016-08-31T06:33:33.580 回答
4

如果你只需要一个非常简单的调用，你可以直接使用 URL：

import java.net.URL;

    new URL("http://wheredatapp.com").openStream();
于 2015-09-19T16:27:36.877 回答
1

简单有效的解决方案：使用Volley

 StringRequest stringRequest = new StringRequest(Request.Method.GET, finalUrl ,
           new Response.Listener<String>() {
                    @Override
                    public void onResponse(String){
                        try {
                            JSONObject jsonObject = new JSONObject(response);
                            HashMap<String, Object> responseHashMap = new HashMap<>(Utility.toMap(jsonObject)) ;
                        } catch (JSONException e) {
                            e.printStackTrace();
                        }
                    }
                }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {
                Log.d("api", error.getMessage().toString());
            }
        });

        RequestQueue queue = Volley.newRequestQueue(context) ;
        queue.add(stringRequest) ;
于 2018-07-01T07:32:37.537 回答
0

使用 Tasks 和 Kotlin 在单独的线程上执行此操作的更现代的方式

private val mExecutor: Executor = Executors.newSingleThreadExecutor()

private fun createHttpTask(u:String): Task<String> {
    return Tasks.call(mExecutor, Callable<String>{
        val url = URL(u)
        val conn: HttpURLConnection = url.openConnection() as HttpURLConnection
        conn.requestMethod = "GET"
        conn.connectTimeout = 3000
        conn.readTimeout = 3000
        val rc = conn.responseCode
        if ( rc != HttpURLConnection.HTTP_OK) {
            throw java.lang.Exception("Error: ${rc}")
        }
        val inp: InputStream = BufferedInputStream(conn.inputStream)
        val resp: String = inp.bufferedReader(UTF_8).use{ it.readText() }
        return@Callable resp
    })
}

现在你可以在很多地方像下面这样使用它：

            createHttpTask("https://google.com")
                    .addOnSuccessListener {
                        Log.d("HTTP", "Response: ${it}") // 'it' is a response string here
                    }
                    .addOnFailureListener {
                        Log.d("HTTP", "Error: ${it.message}") // 'it' is an Exception object here
                    }
于 2020-11-21T05:36:58.303 回答
-5

URL url = 新 URL(" https://www.google.com ");

//如果你正在使用

URLConnection conn = url.openConnection();

//把它改成

HttpURLConnection conn =(HttpURLConnection )url.openConnection();

于 2016-11-02T08:06:22.847 回答