【发布时间】:2020-07-02 16:13:28
【问题描述】:
我尝试调用 IMDb API 2 次,第一次调用并获取该电影/节目的 ID,第二次使用该 ID 获取有关该电影/节目的所有信息,我还需要该 ID 用于应用程序的另一部分,这就是我这样做的原因。问题是第二个调用没有等待第一个调用完成。我认为这就是为什么当我尝试使用它们时变量没有更新的原因。这是我的 onCreate 方法,这一切都发生了,出于显而易见的原因,我取出了一些 API 密钥:
@Override
protected void onCreate(Bundle savedInstanceState)
{
super.onCreate(savedInstanceState);
setContentView(R.layout.imdb_activity);
mTextViewResult = findViewById(R.id.text_view_result);
OkHttpClient client1 = new OkHttpClient();
//change the url to be generic and usable for user input
String urlforID = "https://movie-database-imdb-alternative.p.rapidapi.com/?page=1&r=json&s=Avengers";
final Request request = new Request.Builder()
.url(urlforID)
.get()
.addHeader("x-rapidapi-host", "movie-database-imdb-alternative.p.rapidapi.com")
.addHeader("x-rapidapi-key", "KEYGOESHERE")
.build();
client1.newCall(request).enqueue(new Callback()
{
@Override
public void onFailure(Call call, IOException e)
{
e.printStackTrace();
}
@Override
public void onResponse(Call call, Response response) throws IOException
{
if(response.isSuccessful())
{
String myResponse = response.body().string();
try
{
myObj = new JSONObject(myResponse);
}
catch (JSONException e)
{
e.printStackTrace();
}
try
{
myArray = myObj.getJSONArray("Search");
// responseID = new String[myArray.length()];//might have to subtract 1
for(int i = 0; i < myArray.length(); i++)
{
JSONObject obj1 = myArray.getJSONObject(i);
responseID[i] = obj1.getString("imdbID");
// Log.d("id","the id that was just put in was: " + responseID[i]);
}
}
catch(JSONException e)
{
e.printStackTrace();
}
imdb_activity.this.runOnUiThread(new Runnable()
{
@Override
public void run()
{
//new ReadJsonForID().execute();
Log.d("id", "The id is: " + responseID[0]);
mTextViewResult.setText(responseID[0]);
}
});
}
}
});
//this is where call 2 would happen but it is not saving the variable how it should
Log.d("id", "The id is after finish: " + mTextViewResult.getText());
【问题讨论】:
-
如何执行第二次调用?代码-sn-p 只显示一个调用。
-
@J.Gerbershagen 我实际上做了同样的事情,比如进行调用,然后在第一个调用的主体之外执行它,这样它就在 onCreate() 内部。我应该从第一个内部拨打第二个电话吗?
标签: java api android-studio okhttp