我是android编程语言的新手。我指的是link,但是无法理解这段代码中的MyAPI是什么。我在stack overflow上发现了类似的问题。但是,我什至无法解析该MyAPI库,他最初可以找到它,但后来它停止工作。在哪里可以找到它?
import android.content.Context;
import android.os.AsyncTask;
import android.util.Pair;
import android.widget.Toast;
import com.google.api.client.extensions.android.http.AndroidHttp;
import com.google.api.client.extensions.android.json.AndroidJsonFactory;
import com.google.api.client.googleapis.services.AbstractGoogleClientRequest;
import com.google.api.client.googleapis.services.GoogleClientRequestInitializer;
import java.io.IOException;
class SendRegisterInfo extends AsyncTask<Pair<Context, String>, Void, String> {
private static MyApi myApiService = null;
private Context context;
@Override
protected String doInBackground(Pair<Context, String>... params) {
if(myApiService == null) { // Only do this once
MyApi.Builder builder = new MyApi.Builder(AndroidHttp.newCompatibleTransport(),
new AndroidJsonFactory(), null)
// options for running against local devappserver
// - 10.0.2.2 is localhost's IP address in Android emulator
// - turn off compression when running against local devappserver
.setRootUrl("http://10.0.2.2:8080/_ah/api/")
.setGoogleClientRequestInitializer(new GoogleClientRequestInitializer() {
@Override
public void initialize(AbstractGoogleClientRequest<?>
abstractGoogleClientRequest)
throws IOException {
abstractGoogleClientRequest.setDisableGZipContent(true);
}
});
// end options for devappserver
myApiService = builder.build();
}
context = params[0].first;
String name = params[0].second;
try {
return myApiService.sayHi(name).execute().getData();
} catch (IOException e) {
return e.getMessage();
}
}
@Override
protected void onPostExecute(String result) {
Toast.makeText(context, result, Toast.LENGTH_LONG).show();
}
}
最佳答案