Question

我试图从我的Android应用程序向Yelp发送API请求。我从这里获得了用Java发送请求的示例代码https://github.com/Yelp/yelp-api/tree/master/v2/java

当我运行存储库中提供的脚本时，它工作正常，但是，当我将代码添加到我的Android应用程序时，我收到了错误。

这是我的Android应用代码：

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    YelpAPI yelp = new YelpAPI(
            getString(R.string.yelp_consumer_key),
            getString(R.string.yelp_consumer_secret),
            getString(R.string.yelp_token),
            getString(R.string.yelp_token_secret));

    String result = yelp.searchForBusinessesByLocation("bar", "San Jose, CA");
    System.out.println(result);
}

和堆栈跟踪：

--------- beginning of crash
07-23 12:50:23.848    E/AndroidRuntime﹕ FATAL EXCEPTION: main
    Process: PID: 2328
    java.lang.RuntimeException: Unable to start activity ComponentInfo{.MainActivity}: org.scribe.exceptions.OAuthConnectionException: There was a problem while creating a connection to the remote service.
            at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2298)
            at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2360)
            at android.app.ActivityThread.access$800(ActivityThread.java:144)                           ...

我在互联网上发现了类似的问题，我认为这是因为我无法使用主线程发送请求。如何使用单独的线程发送请求？

Answer 1

如果您只想创建一个新线程，可以这样做：

 new Thread(new Runnable() {
    public void run() {
        YelpAPI yelp = new YelpAPI(
        getString(R.string.yelp_consumer_key),
        getString(R.string.yelp_consumer_secret),
        getString(R.string.yelp_token),
        getString(R.string.yelp_token_secret));

        String result = yelp.searchForBusinessesByLocation("bar", "San Jose, CA");
        System.out.println(result);
    }
 }).start();

有关Android线程的信息，请访问： http://developer.android.com/guide/components/processes-and-threads.html

使用单独的线程发送请求

1 个答案: