Flutter: filter list as per some condition

Question

I'm having a list of movies. That contains all animated and non-animated movies. To identify whether it's Animated or not there is one flag called isAnimated.

I want to show only Animated movies. How can I do that?

Mild

Answer 1

toList() is missing to materializer the result

_AnimatedMovies = AllMovies.where((i) => i.isAnimated).toList();

Eppes

No work here :( . It says type 'Where Iterable<>' is not a subtype of type 'List – Against

If you need a List add .toList() – Kalagher

It worked like charm actually I was creating an app in which I have some static data (list of a model) and when I am navigating to a single view page I am sending the id of the model and I am finding the model by id and showing it's data in the single view page. I am doing like this CompanyModel company = companies.where((c) => c.id == companyId).toList().first; – Hunsinger

What would be the null safe version of this? – Royer

@JoelGMathew I think AllMovies.whereNotNull().where((i) => i.isAnimated).toList(); (not tested) – Kalagher

Answer 2

The Solution is here

Just try with this Function getCategoryList(),
Here the condition will be catogory_id == '1' from the list

List<dynamic> getCategoryList(List<dynamic> inputlist) {
    List outputList = inputlist.where((o) => o['category_id'] == '1').toList();
    return outputList;
  }

Spread

Answer 3

You can use this for specific condition

List<String> strings = ['one', 'two', 'three', 'four', 'five'];
List<String> filteredStrings  = strings.where((item) {
   return item.length == 3;
});

Ravelment

Answer 4

where function on a List returns Iterable, you have to convert it to List using the function List.from(Iterable).

So in the above scenario, you should use the following code snippet.

Iterable _AnimatedMoviesIterable = AllMovies.where((i) => i.isAnimated);

_AnimatedMovies = List.from(_AnimatedMoviesIterable);

Edited:

As per Günter Zöchbauer solution, we can use a single line instead of multiple. So the code is

_AnimatedMovies = AllMovies.where((i) => i.isAnimated).toList();

Lynnette

Iterables have a toList() method which is much easier to read IMHO. – Kalagher

Agreed @GünterZöchbauer. We can use AllMovies.where((i) => i.isAnimated).toList(); as you mentioned – Lynnette