Question

我想使用ljoop库来发出http请求。但是，当我执行get方法或post时，它只是跳过它，不会进入回调。

ip地址成功地在浏览器中提供了json结果。所以它可以使用10.0.2.2我知道这一点，但它也适用于192 ....在浏览器中。我已经添加了互联网权限，但它似乎似乎没有发送获取请求。带有express的nodejs服务器只有在模拟器上使用chrome时才会收到任何内容。

import android.app.Activity;

import com.loopj.android.http.AsyncHttpClient;
import com.loopj.android.http.AsyncHttpResponseHandler;
import com.loopj.android.http.RequestHandle;

import java.io.IOException;
import java.io.UnsupportedEncodingException;
import java.net.SocketTimeoutException;
import java.util.List;

import cz.msebera.android.httpclient.Header;
import cz.msebera.android.httpclient.conn.ConnectTimeoutException;
import cz.msebera.android.httpclient.entity.StringEntity;
import cz.msebera.android.httpclient.message.BasicHeader;
import cz.msebera.android.httpclient.protocol.HTTP;
import schaschinger.eggermode.Restservice.Models.Item;
import schaschinger.eggermode.Restservice.Models.Stammkunde;


public class HttpRequestTask {

    public static final String URL_STAMMKUNDEN = "http://192.168.1.190:9999/stammkunden/stammkunden";
    public static final String URL_ITEMS = "http://192.168.1.190:9999/item/items";

    Activity activity;
    final AsyncHttpClient asyncHttpClient;
    HttpResponseHandler httpResponseHandler;


    public HttpRequestTask(Activity activity) {
        this.activity = activity;
        this.asyncHttpClient = new AsyncHttpClient();
        httpResponseHandler = new HttpResponseHandler();

        this.asyncHttpClient.allowRetryExceptionClass(IOException.class);
        this.asyncHttpClient.allowRetryExceptionClass(SocketTimeoutException.class);
        this.asyncHttpClient.allowRetryExceptionClass(ConnectTimeoutException.class);
    }

    public boolean insertStammkunde(Stammkunde stammkunde) {

        String stammkundenJson = stammkunde.getJson();

        StringEntity stringEntity = null;
        try {
            stringEntity = new StringEntity(stammkundenJson);
        } catch (UnsupportedEncodingException e) {
            return false;
        }
        stringEntity.setContentType(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));

        RequestHandle requestHandle = asyncHttpClient.post(activity.getApplicationContext()
                , HttpRequestTask.URL_STAMMKUNDEN, stringEntity, "application/json", httpResponseHandler);

        return requestHandle.isFinished();
    }

    public List<Item> getItems(){

        asyncHttpClient.setConnectTimeout(1000);

        asyncHttpClient.get("https://www.google.com", new AsyncHttpResponseHandler() {

            @Override
            public void onStart() {
                // called before request is started
            }

            @Override
            public void onSuccess(int statusCode, Header[] headers, byte[] response) {
                // called when response HTTP status is "200 OK"
            }

            @Override
            public void onFailure(int statusCode, Header[] headers, byte[] errorResponse, Throwable e) {
                // called when response HTTP status is "4XX" (eg. 401, 403, 404)
            }

            @Override
            public void onRetry(int retryNo) {
                // called when request is retried
            }
        });

        return null;
    }


}

Answer 1

当控制到达具有异步调用的行时，您的代码不应该调用回调。相反，当服务器向您的请求发送响应时，稍后会调用回调。这就是“异步”的含义。

您可以返回Future中包含的项目，例如，将getItems的返回类型更改为CompletableFuture<List<Item>>并返回CompletableFuture。在complete回调中调用未来onSuccess，在completeExceptionally回调中调用onFailure。

然后，您可以选择调用将来的get来同步获取结果（这将阻止直到请求完成或超时），或者使用handle（或{{分配回调）来自CompletionStage的{}}和thenAccept} exceptionally实现。

Ljoop http get - 回调未执行

1 个答案: