Question

在以下代码中，return不返回等待的值。我怎样才能使诺言在返回之前得到解决。因此，我希望result成为SUCCESS而不是承诺。

const foo = async()=>{
    try{
        let a = await new Promise((resolve)=>{
            resolve('SUCCESS')
        })
        console.log("this is inside the try block");
        return a
    }catch{
        console.log('error')
    }
}
let result = foo();
console.log(result);

Answer 1

public class TaskFragment extends Fragment { private static SectionsPagerAdapter sectionsPagerAdapter; private ArrayList<Task> todo; private ArrayList<Task> inp; private ArrayList<Task> done; @Nullable @Override public View onCreateView(@NonNull LayoutInflater inflater, @Nullable ViewGroup container, @Nullable Bundle savedInstanceState) { View view = inflater.inflate(R.layout.fragment_task, container, false); sectionsPagerAdapter = new SectionsPagerAdapter(getChildFragmentManager()); sectionsPagerAdapter.addFragments(new FragmentToDo(), "To Do"); sectionsPagerAdapter.addFragments(new FragmentInProgress(), "In Progress"); sectionsPagerAdapter.addFragments(new FragmentDone(), "Done"); ViewPager viewPager = view.findViewById(R.id.view_pager); viewPager.setAdapter(sectionsPagerAdapter); TabLayout tabs = view.findViewById(R.id.tabs); tabs.setupWithViewPager(viewPager); return view; }是一个异步函数，它将返回一个Promise。要获得承诺的结果，您将需要链接一个foo方法：

then

更新：

要使用返回的值，可以在const foo = async()=>{ try{ let a = await new Promise((resolve)=>{ resolve('SUCCESS') }) console.log("this is inside the try block"); return a }catch{ console.log('error') } } foo().then(result => console.log(result));方法内使用它，或使用结果调用另一个函数。

then

OR：

foo().then(result => {
  console.log(result);
  //do what you want with the result here
});

返回不等待异步功能

1 个答案: