我需要使用流过滤列表。我已经有一个流可以拉动所有客户端,现在我想将它们过滤掉。
我有这段代码可以正常工作,它拉动客户端并在列表视图中显示我。我只想过滤从主流获得的数据。有知识的人可以帮助我吗?
class ClientesControles extends BlocBase {
final ClienteService clienteService;
ClientesControles(this.clienteService);
BuildContext _context;
init(BuildContext context) {
_context = context;
}
Observable<List<ClienteModel>> get clientesStream => clienteService.clientes;
final _stringFiltroController = BehaviorSubject<String>();
Observable<String> get stringFiltroFluxo => _stringFiltroController.stream;
Sink<String> get stringFiltroEvent => _stringFiltroController.sink;
@mustCallSuper
void dispose() {
_cadnomecliente.close();
_cadnomefcliente.close();
_cadtelcliente.close();
_cademailcliente.close();
_cadidcliente.close();
_stringFiltroController.close();
}
}
Class ClienteService{
Observable<List<ClienteModel>> get clientes =>
Observable(collection.snapshots().map((item) => item.documents
.map<ClienteModel>((item) => ClienteModel.fromJson(item.data))
.toList()));
}
答案 0 :(得分:0)
要过滤流,请添加where子句:
Observable<List<ClienteModel>> get clientes =>
Observable(collection.snapshots().map((item) => item.documents
.where((item) => hasWhatIWant(item))
.map<ClienteModel>((item) => ClienteModel.fromJson(item.data))
.toList()));
bool hasWhatIWant(item){
//some check
}
答案 1 :(得分:0)
我遇到了类似的问题,并花了3天时间才找到这篇文章...谢谢...我的有点不同
Observable<List<Contact>> get contacts => _contactsListFetcher.stream
.map((s) => s.where((c) => hasFilteredContent(c)).map((i) => i).toList());
bool hasFilteredContent(Contact contact) {
return contact.firstName.toLowerCase().contains(currentFilter.toLowerCase()) || contact.lastName.toLowerCase().contains(currentFilter.toLowerCase());
}