在后台线程中执行OkHttp网络操作

时间:2015-12-08 02:54:44

标签: java android okhttp

我正在使用OKHttp对服务器执行Post请求,如下所示:

public class NetworkManager {
    public static final MediaType JSON = MediaType.parse("application/json; charset=utf-8");
    OkHttpClient client = new OkHttpClient();

    String post(String url, JSONObject json) throws IOException {
        try {
            JSONArray array = json.getJSONArray("d");
            RequestBody body = new FormEncodingBuilder()
                    .add("m", json.getString("m"))
                    .add("d", array.toString())
                    .build();
            Request request = new Request.Builder()
                    .url(url)
                    .post(body)
                    .build();
            Response response = client.newCall(request).execute();
            return response.body().string();
        } catch (JSONException jsone) {
            return "ERROR: " + jsone.getMessage();
        }
    }
}

并将其命名为:

NetworkManager manager = new NetworkManager();
String response = manager.post("http://www.example.com/api/", jsonObject);

当我尝试运行App时,它会在logcat中提示错误:

  

android.os.NetworkOnMainThreadException
  在   android.os.StrictMode $ AndroidBlockGuardPolicy.onNetwork(StrictMode.java:1273)

参考SO中的其他问题,我添加了这个以覆盖政策:

if (android.os.Build.VERSION.SDK_INT > 9)
{
        StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
        StrictMode.setThreadPolicy(policy);
}

但我认为这是不健康的,我想将NetworkManager行动置于背景之下。我怎么能这样做?

1 个答案:

答案 0 :(得分:7)

由于OkHttp也支持异步方式,因此IMO可以参考以下GET请求示例,然后申请POST请求:

        OkHttpClient client = new OkHttpClient();
        // GET request
        Request request = new Request.Builder()
                .url("http://google.com")
                .build();
        client.newCall(request).enqueue(new Callback() {
            @Override
            public void onFailure(Request request, IOException e) {
                Log.e(LOG_TAG, e.toString());
            }
            @Override
            public void onResponse(Response response) throws IOException {
                Log.w(LOG_TAG, response.body().string());
                Log.i(LOG_TAG, response.toString());
            }
        });

希望它有所帮助!