这是我能想到的最好的标题，但是如果不看代码，我的问题有点难以解释。我添加了一些评论来解释发生了什么。这是我的代码:
private Movie APICallTwo(final String ID)
{
Movie found;//want to return this
OkHttpClient client2 = new OkHttpClient();
String urlforID = "https://movie-database-imdb-alternative.p.rapidapi.com/?i=" + ID + "&r=json";
final Request request = new Request.Builder()
.url(urlforID)
.get()
.addHeader("x-rapidapi-host", "movie-database-imdb-alternative.p.rapidapi.com")
.addHeader("x-rapidapi-key", "")//deleted the key for obvious reasons
.build();
client2.newCall(request).enqueue(new Callback()
{
@Override
public void onFailure(Call call, IOException e)
{
}
@Override
public void onResponse(Call call, Response response) throws IOException
{
if(response.isSuccessful())
{
String title = "";
String year = "";
String rated = "";
String released = "";
String runtime = "";
String genre = "";
String director = "";
String writer = "";
String actors = "";
String plot = "";
String imdbID = "";
String production = "";
String myResponse = response.body().string();
try
{
myObj = new JSONObject(myResponse);
}
catch (JSONException e)
{
e.printStackTrace();
}
try
{
title = myObj.getString("Title");
year = myObj.getString("Year");
rated = myObj.getString("Rated");
released = myObj.getString("Released");
runtime = myObj.getString("Runtime");
genre = myObj.getString("Genre");
director = myObj.getString("Director");
writer = myObj.getString("Writer");
actors = myObj.getString("Actors");
plot = myObj.getString("Plot");
imdbID = myObj.getString("imdbID");
try//not all movies have production companies listen on IMDb
{
production = myObj.getString("Production");
}
catch (JSONException e)
{
production = "N/A";
}
}
catch (JSONException e)
{
e.printStackTrace();
}
test = new Movie(title, year, rated, released, runtime, genre, director, writer, actors, plot, imdbID, production);//test is declared as "Movie test" up in the main
// Log.d("in ApiCall2", "The Details for: " + ID + " are: \n" + test.printMovie());//this check works, it returns what is in test properly
}
}
});
found = test;
Log.d("in ApiCall2", "The Details for: " + ID + " are: \n" + found.printMovie());//this check doesn't work, says that test is null
detailLatch.countDown();
return found;
}
我怎样才能得到它，以便保留抓取的信息并将其传递回调用它的地方？我知道，当在循环或另一个函数中声明某些内容时，当它离开该函数时，该函数会被遗忘，但这在这里没有意义。我没有其他想法如何解决这个问题。
最佳答案
如您所知，创建网络调用是异步的。因此，无论 View 是什么，从异步函数返回一个值到您的 Activity/fragment ，您需要创建一个适当的回调，一旦获取响应就会触发该回调。
第 1 步:创建界面
interface NetworkCallback{
void onNetworkResponse(Movie movie); //There could be any type of object you want to return.
}
第 2 步:将 NetworkCallback 引用传递给您异步函数
private void fetchMovie(final String ID,NetworkCallback networkCallback) {
Movie found;//want to return this
OkHttpClient client2=new OkHttpClient();
String urlforID="https://movie-database-imdb-alternative.p.rapidapi.com/?i="+ID+"&r=json";
final Request request=new Request.Builder()
.url(urlforID)
.get()
.addHeader("x-rapidapi-host","movie-database-imdb-alternative.p.rapidapi.com")
.addHeader("x-rapidapi-key","")//deleted the key for obvious reasons
.build();
client2.newCall(request).enqueue(new Callback()
{
@Override
public void onFailure(Call call,IOException e)
{
}
@Override
public void onResponse(Call call,Response response)throws IOException
{
if(response.isSuccessful())
{
String title="";
String year="";
String rated="";
String released="";
String runtime="";
String genre="";
String director="";
String writer="";
String actors="";
String plot="";
String imdbID="";
String production="";
String myResponse=response.body().string();
try
{
myObj=new JSONObject(myResponse);
}
catch(JSONException e)
{
e.printStackTrace();
}
try
{
title=myObj.getString("Title");
year=myObj.getString("Year");
rated=myObj.getString("Rated");
released=myObj.getString("Released");
runtime=myObj.getString("Runtime");
genre=myObj.getString("Genre");
director=myObj.getString("Director");
writer=myObj.getString("Writer");
actors=myObj.getString("Actors");
plot=myObj.getString("Plot");
imdbID=myObj.getString("imdbID");
try//not all movies have production companies listen on IMDb
{
production=myObj.getString("Production");
}
catch(JSONException e)
{
production="N/A";
}
}
catch(JSONException e)
{
e.printStackTrace();
}
Movie movie =new Movie(title,year,rated,released,runtime,genre,director,writer,actors,plot,imdbID,production);//test is declared as "Movie test" up in the main
networkCallback.onNetworkResponse(movie);
// Log.d("in ApiCall2", "The Details for: " + ID + " are: \n" + test.printMovie());//this check works, it returns what is in test properly
}
}
});
//Log.d("in ApiCall2","The Details for: "+ID+" are: \n"+found.printMovie());//this check doesn't work, says that test is null
// detailLatch.countDown();
}
第 3 步:从 Activity/fragment 调用 fetchMovie。并更新 onNetworkResponse 内的 UI
private void fetchMovie(){
APICallTwo("YOUR_ID", new NetworkCallback() {
@Override
public void onNetworkResponse(Movie movie) {
//NOW YOU RECEIVED YOUR FAVORITE MOVIE INFO
//UPDATE YOUR UI HERE
}
});
}
关于java - 如何保留我设置的值以便可以返回它？安卓，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/60837946/