用android发出HTTP请求

Question

我到处搜索但我找不到答案,有没有办法发出简单的HTTP请求？我想在我的某个网站上请求PHP页面/脚本,但我不想显示该网页.

如果可能的话我甚至想在后台(在BroadcastReceiver中)这样做

Answer 1

UPDATE

这是一个非常古老的答案.我绝对不会再推荐Apache的客户了.而是使用:

• OkHttp
• HttpURLConnection的

原始答案

首先,请求访问网络的权限,在清单中添加以下内容:

<uses-permission android:name="android.permission.INTERNET" />

那么最简单的方法是使用与Android捆绑的Apache http客户端:

    HttpClient httpclient = new DefaultHttpClient();
    HttpResponse response = httpclient.execute(new HttpGet(URL));
    StatusLine statusLine = response.getStatusLine();
    if(statusLine.getStatusCode() == HttpStatus.SC_OK){
        ByteArrayOutputStream out = new ByteArrayOutputStream();
        response.getEntity().writeTo(out);
        String responseString = out.toString();
        out.close();
        //..more logic
    } else{
        //Closes the connection.
        response.getEntity().getContent().close();
        throw new IOException(statusLine.getReasonPhrase());
    }

如果你想让它在单独的线程上运行,我建议扩展AsyncTask:

class RequestTask extends AsyncTask<String, String, String>{

    @Override
    protected String doInBackground(String... uri) {
        HttpClient httpclient = new DefaultHttpClient();
        HttpResponse response;
        String responseString = null;
        try {
            response = httpclient.execute(new HttpGet(uri[0]));
            StatusLine statusLine = response.getStatusLine();
            if(statusLine.getStatusCode() == HttpStatus.SC_OK){
                ByteArrayOutputStream out = new ByteArrayOutputStream();
                response.getEntity().writeTo(out);
                responseString = out.toString();
                out.close();
            } else{
                //Closes the connection.
                response.getEntity().getContent().close();
                throw new IOException(statusLine.getReasonPhrase());
            }
        } catch (ClientProtocolException e) {
            //TODO Handle problems..
        } catch (IOException e) {
            //TODO Handle problems..
        }
        return responseString;
    }

    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);
        //Do anything with response..
    }
}

然后您可以通过以下方式提出请求:

   new RequestTask().execute("http://stackoverflow.com");

Answer 2

除非您有明确的理由选择Apache HttpClient,否则您应该更喜欢java.net.URLConnection.你可以找到很多关于如何在网上使用它的例子.

自您的原始帖子以来,我们还改进了Android文档:http://developer.android.com/reference/java/net/HttpURLConnection.html

我们在官方博客上谈到了权衡取舍:http://android-developers.blogspot.com/2011/09/androids-http-clients.html

Answer 3

注意:现在不推荐使用与Android捆绑在一起的Apache HTTP Client,而使用HttpURLConnection.有关详细信息,请参阅Android开发人员博客.

添加<uses-permission android:name="android.permission.INTERNET" />到您的清单.

然后,您将检索如下所示的网页:

URL url = new URL("http://www.android.com/");
HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
try {
     InputStream in = new BufferedInputStream(urlConnection.getInputStream());
     readStream(in);
}
finally {
     urlConnection.disconnect();
}

我还建议在一个单独的线程上运行它:

class RequestTask extends AsyncTask<String, String, String>{

@Override
protected String doInBackground(String... uri) {
    String responseString = null;
    try {
        URL url = new URL(myurl);
        HttpURLConnection conn = (HttpURLConnection) url.openConnection();
        if(conn.getResponseCode() == HttpsURLConnection.HTTP_OK){
            // Do normal input or output stream reading
        }
        else {
            response = "FAILED"; // See documentation for more info on response handling
        }
    } catch (ClientProtocolException e) {
        //TODO Handle problems..
    } catch (IOException e) {
        //TODO Handle problems..
    }
    return responseString;
}

@Override
protected void onPostExecute(String result) {
    super.onPostExecute(result);
    //Do anything with response..
}
}

有关响应处理和POST请求的详细信息,请参阅文档.

Answer 4

最简单的方法是使用名为Volley的Android库

Volley提供以下好处:

自动调度网络请求.多个并发网络连接.具有标准HTTP缓存一致性的透明磁盘和内存响应缓存.支持请求优先级.取消请求API.您可以取消单个请求,也可以设置要取消的请求块或范围.易于定制,例如,重试和退避.强大的排序功能,可以使用从网络异步获取的数据轻松正确填充UI.调试和跟踪工具.

您可以像这样简单地发送http/https请求:

        // Instantiate the RequestQueue.
        RequestQueue queue = Volley.newRequestQueue(this);
        String url ="http://www.yourapi.com";
        JsonObjectRequest request = new JsonObjectRequest(url, null,
            new Response.Listener<JSONObject>() {
                @Override
                public void onResponse(JSONObject response) {
                    if (null != response) {
                         try {
                             //handle your response
                         } catch (JSONException e) {
                             e.printStackTrace();
                         }
                    }
                }
            }, new Response.ErrorListener() {

            @Override
            public void onErrorResponse(VolleyError error) {

            }
        });
        queue.add(request);

在这种情况下,您不必考虑自己"在后台运行"或"使用缓存",因为所有这些都已由Volley完成.

Answer 5

private String getToServer(String service) throws IOException {
    HttpGet httpget = new HttpGet(service);
    ResponseHandler<String> responseHandler = new BasicResponseHandler();
    return new DefaultHttpClient().execute(httpget, responseHandler);

}

问候

Answer 6

按照上面的建议使用 Volley。将以下内容添加到 build.gradle（模块：app）

implementation 'com.android.volley:volley:1.1.1'

将以下内容添加到 AndroidManifest.xml 中：

<uses-permission android:name="android.permission.INTERNET" />

并将以下内容添加到您的活动代码中：

public void httpCall(String url) {

    RequestQueue queue = Volley.newRequestQueue(this);

    StringRequest stringRequest = new StringRequest(Request.Method.GET, url,
            new Response.Listener<String>() {
                @Override
                public void onResponse(String response) {
                    // enjoy your response
                }
            }, new Response.ErrorListener() {
                @Override
                public void onErrorResponse(VolleyError error) {
                    // enjoy your error status
                }
    });

    queue.add(stringRequest);
}

它取代了http客户端，非常简单。

Answer 7

用线程：

private class LoadingThread extends Thread {
    Handler handler;

    LoadingThread(Handler h) {
        handler = h;
    }
    @Override
    public void run() {
        Message m = handler.obtainMessage();
        try {
            BufferedReader in = 
                new BufferedReader(new InputStreamReader(url.openStream()));
            String page = "";
            String inLine;

            while ((inLine = in.readLine()) != null) {
                page += inLine;
            }

            in.close();
            Bundle b = new Bundle();
            b.putString("result", page);
            m.setData(b);
        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

        handler.sendMessage(m);
    }
}

Answer 8

看看这个很棒的新库，可以通过gradle获取它：)

build.gradle： compile 'com.apptakk.http_request:http-request:0.1.2'

用法：

new HttpRequestTask(
    new HttpRequest("http://httpbin.org/post", HttpRequest.POST, "{ \"some\": \"data\" }"),
    new HttpRequest.Handler() {
      @Override
      public void response(HttpResponse response) {
        if (response.code == 200) {
          Log.d(this.getClass().toString(), "Request successful!");
        } else {
          Log.e(this.getClass().toString(), "Request unsuccessful: " + response);
        }
      }
    }).execute();

https://github.com/erf/http-request

Answer 9

由于没有一个答案描述了使用OkHttp执行请求的方法，这是目前非常流行的 Android 和 Java 的 http 客户端，我将提供一个简单的例子：

//get an instance of the client
OkHttpClient client = new OkHttpClient();

//add parameters
HttpUrl.Builder urlBuilder = HttpUrl.parse("https://www.example.com").newBuilder();
urlBuilder.addQueryParameter("query", "stack-overflow");


String url = urlBuilder.build().toString();

//build the request
Request request = new Request.Builder().url(url).build();

//execute
Response response = client.newCall(request).execute();

这个库的明显优势在于，它将我们从一些低级细节中抽象出来，提供更友好和安全的方式与它们交互。语法也得到了简化，可以编写漂亮的代码。