我到处都找过了,但我找不到我的答案,有没有一种方法可以做一个简单的HTTP请求?我想在我的一个网站上请求一个PHP页面/脚本,但我不想显示网页。

如果可能的话,我甚至想在后台(在一个BroadcastReceiver)做它


当前回答

有一根线:

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);
    }
}

其他回答

更新

这是一个非常古老的答案。我绝对不会再推荐Apache的客户端了。你可以用任意一种:

改造 OkHttp 截击 HttpUrlConnection

原来的答案

首先,申请访问网络的权限,在您的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Apache http客户端与Android捆绑:

    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,你可以通过以下方式提出请求:

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

除非你有明确的理由选择Apache HttpClient,否则你应该选择java.net.URLConnection。你可以在网上找到很多如何使用它的例子。

我们也改进了Android文档,因为你原来的帖子:http://developer.android.com/reference/java/net/HttpURLConnection.html

我们已经在官方博客http://android-developers.blogspot.com/2011/09/androids-http-clients.html上讨论了这些权衡

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接口,其中包含我们的请求方法,参数,我们还可以为每个请求定制头部:

    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方法轻松地进行异步操作

这是android中HTTP Get/POST请求的新代码。HTTPClient已被废弃,可能无法使用,因为在我的情况下。

首先在build.gradle中添加两个依赖:

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

然后在ASyncTask in doBackground方法中编写此代码。

 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
 }

有一根线:

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);
    }
}