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

11 回答

更新

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：

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");

回复于 2024-05-09T02:49:48+08:00

0

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：http://android-developers.blogspot.com/2011/09/androids-http-clients.html

回复于 2024-05-09T02:49:48+08:00

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Blog .

将 <uses-permission android:name="android.permission.INTERNET" /> 添加到清单中 .

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请求的更多信息，请参阅documentation .

回复于 2024-05-09T02:49:48+08:00

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 .

回复于 2024-05-09T02:49:48+08:00

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

回复于 2024-05-09T02:49:48+08:00

有一个线程：

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);
    }
}

回复于 2024-05-09T02:49:48+08:00

看看这个可以通过gradle获得的真棒新库:)

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

回复于 2024-05-09T02:49:48+08:00

我使用Gson lib为web服务创建了这个URL：

客户：

public EstabelecimentoList getListaEstabelecimentoPorPromocao(){

        EstabelecimentoList estabelecimentoList  = new EstabelecimentoList();
        try{
            URL url = new URL("http://" +  Conexao.getSERVIDOR()+ "/cardapio.online/rest/recursos/busca_estabelecimento_promocao_android");
            HttpURLConnection con = (HttpURLConnection) url.openConnection();

            if (con.getResponseCode() != 200) {
                    throw new RuntimeException("HTTP error code : "+ con.getResponseCode());
            }

            BufferedReader br = new BufferedReader(new InputStreamReader((con.getInputStream())));
            estabelecimentoList = new Gson().fromJson(br, EstabelecimentoList.class);
            con.disconnect();

        } catch (IOException e) {
            e.printStackTrace();
        }
        return estabelecimentoList;
}

回复于 2024-05-09T02:49:48+08:00

这是android中HTTP Get / POST请求的新代码 . HTTPClient 已被删除，可能无法使用，就像我的情况一样 .

首先在build.gradle中添加两个依赖项：

compile 'org.apache.httpcomponents:httpcore:4.4.1'
compile 'org.apache.httpcomponents:httpclient:4.5'

然后在 ASyncTask 中以 doBackground 方法编写此代码 .

URL url = new URL("http://localhost:8080/web/get?key=value");
 HttpURLConnection urlConnection = (HttpURLConnection)url.openConnection();
 urlConnection.setRequestMethod("GET");
 int statusCode = urlConnection.getResponseCode();
 if (statusCode ==  200) {
      InputStream it = new BufferedInputStream(urlConnection.getInputStream());
      InputStreamReader read = new InputStreamReader(it);
      BufferedReader buff = new BufferedReader(read);
      StringBuilder dta = new StringBuilder();
      String chunks ;
      while((chunks = buff.readLine()) != null)
      {
         dta.append(chunks);
      }
 }
 else
 {
     //Handle else
 }

回复于 2024-05-09T02:49:48+08:00

对我来说，最简单的方法是使用名为Retrofit2的库

我们只需要创建一个包含我们的请求方法，参数的接口，并且我们也可以为每个请求创建自定义标头：

public interface MyService {

      @GET("users/{user}/repos")
      Call<List<Repo>> listRepos(@Path("user") String user);

      @GET("user")
      Call<UserDetails> getUserDetails(@Header("Authorization") String   credentials);

      @POST("users/new")
      Call<User> createUser(@Body User user);

      @FormUrlEncoded
      @POST("user/edit")
      Call<User> updateUser(@Field("first_name") String first, 
                            @Field("last_name") String last);

      @Multipart
      @PUT("user/photo")
      Call<User> updateUser(@Part("photo") RequestBody photo, 
                            @Part("description") RequestBody description);

      @Headers({
        "Accept: application/vnd.github.v3.full+json",
        "User-Agent: Retrofit-Sample-App"
      })
      @GET("users/{username}")
      Call<User> getUser(@Path("username") String username);    

    }

最好的是，我们可以使用enqueue方法轻松地异步进行

回复于 2024-05-09T02:49:48+08:00

459

由于没有一个答案描述了使用OkHttp执行请求的方法，这对于Android和Java来说是非常流行的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从一些低级细节中抽象出我们，提供了更友好和安全的方式来与它们进行交互 . 语法也简化了，允许编写漂亮的代码 .

回复于 2024-05-09T02:49:48+08:00

用android发出HTTP请求

11 回答

更新

原始答案

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