I want to convert them to ArrayList and I will store them. After that, I have to convert them into old values. Do you have any suggestions? Thanks in advance.
public List<List<String>> phones = new ArrayList<>();
public List<List<Restaurant.Menu>> menus = new ArrayList<>();
public ArrayList<String> phones = ?
public ArrayList<String> menus = ?
For the first scenario, you can flatten the phones nested list into a single list and then collect into an ArrayList.
ArrayList<String> result =
phones.stream()
.flatMap(Collection::stream)
.collect(Collectors.toCollection(ArrayList::new));
For the second scenario, you will need to extract the string representation of the Menu objects given you've overridden toString, otherwise you'll need to extract some type of property from the Menu objects in order to project from Menu to String.
Given you've overridden toString, then do it this way:
ArrayList<String> menuResult =
menus.stream()
.flatMap(Collection::stream)
.map(Menu::toString)
.collect(Collectors.toCollection(ArrayList::new));
Given you need to extract some property from menus, then do it this way:
ArrayList<String> menuResult =
menus.stream()
.flatMap(Collection::stream)
.map(Menu::getName)
.collect(Collectors.toCollection(ArrayList::new));
If your API level doesn't support these features then you can use:
// flatten List<List<String>> to ArrayList<String>
ArrayList<String> phonesAccumulator = new ArrayList<>();
for (List<String> temp : phones) {
phonesAccumulator.addAll(temp);
}
// flatten List<List<Restaurant.Menu>> to ArrayList<String>
ArrayList<String> menusAccumulator = new ArrayList<>();
for (List<Restaurant.Menu> temp : menus) {
for(Restaurant.Menu m : temp){
menusAccumulator.add(m.toString());
// or m.getName();
}
}
List<List<String>> menusAccumulator = new ArrayList<>(); for (List<Restaurant.Menu> temp : menus) { List<String> tempAccumulator = new ArrayList<>(); for(Restaurant.Menu m : temp){ tempAccumulator.add(m.toString()); // or m.getName(); } menusAccumulator.add(tempAccumulator); } ?If you're using Java 8, flatMap can be useful here:
ArrayList<String> phoneList = phones.stream()
.flatMap(List::stream)
.collect(Collectors.toCollection(ArrayList::new));
I am proving my resolution as below said, so you can change as per your own -
List<String> nonvegList = new ArrayList<String>();
nonvegList.add("Mutton Keema");
nonvegList.add("Chicken Keema");
nonvegList.add("Korma Veg Keema");
nonvegList.add("Pulaav Biryaani");
nonvegList.add("Mutton Biryaani");
nonvegList.add("Chicken Biryaani");
List<List<String>> menuList = new ArrayList<List<String>>();
menuList.add(nonvegList);
ArrayList<String> resultList = new ArrayList<String>(menuList.get(0));
System.out.println(resultList);
sweet, simple and in understadable format, compatible after Java 6+,
hope this will help you, thanks.
I used gson to solve it. I share a sample.
Gson gson = new Gson();
ArrayList<String> gsonString = new ArrayList<>();
for(int i=0; i<restaurants.size(); i++)
gsonString.add(gson.toJson(restaurants.get(i)));
// Store it
tinydb.putListString("tinyRestaurant",gsonString);
And convert again
Gson gson = new Gson();
for(int i=0; i<tinydb.getListString("tinyRestaurant").size(); i++)
restaurants.add(gson.fromJson(tinydb.getListString("tinyRestaurant").get(i), Restaurant.class));