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

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
 }

其他回答

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

注意:与Android捆绑的Apache HTTP客户端现在已弃用,转而支持HttpURLConnection。请参阅Android开发者博客了解更多细节。

添加<uses-permission android:name="android.permission. "INTERNET" />到您的舱单。

然后你会像这样检索一个网页:

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,请参阅文档。

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

看看这个很棒的新库,它可以通过gradle获得:)

构建。Gradle:编译'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

我做了这个webservice请求URL,使用一个Gson库:

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;
}