单例模式确保只创建一个类的实例。我如何在达特建立这个?
当前回答
你好,这样的东西怎么样?非常简单的实现,Injector本身是单例的,并且还添加了类。当然可以很容易地扩展。如果您正在寻找更复杂的东西,请检查这个包:https://pub.dartlang.org/packages/flutter_simple_dependency_injection
void main() {
Injector injector = Injector();
injector.add(() => Person('Filip'));
injector.add(() => City('New York'));
Person person = injector.get<Person>();
City city = injector.get<City>();
print(person.name);
print(city.name);
}
class Person {
String name;
Person(this.name);
}
class City {
String name;
City(this.name);
}
typedef T CreateInstanceFn<T>();
class Injector {
static final Injector _singleton = Injector._internal();
final _factories = Map<String, dynamic>();
factory Injector() {
return _singleton;
}
Injector._internal();
String _generateKey<T>(T type) {
return '${type.toString()}_instance';
}
void add<T>(CreateInstanceFn<T> createInstance) {
final typeKey = _generateKey(T);
_factories[typeKey] = createInstance();
}
T get<T>() {
final typeKey = _generateKey(T);
T instance = _factories[typeKey];
if (instance == null) {
print('Cannot find instance for type $typeKey');
}
return instance;
}
}
其他回答
由const构造函数和工厂组成的Dart单例
class Singleton {
factory Singleton() =>
Singleton._internal_();
Singleton._internal_();
}
void main() {
print(new Singleton() == new Singleton());
print(identical(new Singleton() , new Singleton()));
}
你可以只使用Constant构造函数。
class Singleton {
const Singleton(); //Constant constructor
void hello() { print('Hello world'); }
}
例子:
Singleton s = const Singleton();
s.hello(); //Hello world
根据文件:
恒定的构造函数 如果类生成永不更改的对象,则可以使这些对象成为编译时常量。为此,定义一个const构造函数,并确保所有实例变量都是final变量。
这是我做单例的方式,接受参数(你可以直接粘贴到https://dartpad.dev/):
void main() {
Logger x = Logger('asd');
Logger y = Logger('xyz');
x.display('Hello');
y.display('Hello There');
}
class Logger{
Logger._(this.message);
final String message;
static Logger _instance = Logger._('??!?*');
factory Logger(String message){
if(_instance.message=='??!?*'){
_instance = Logger._(message);
}
return _instance;
}
void display(String prefix){
print(prefix+' '+message);
}
}
输入:
Hello asd
Hello There asd
“? ? ! ?*'你看到的只是一个工作区,我做了初始化_instance变量暂时没有使它成为一个Logger?类型(空安全)。
这就是我如何在我的项目中实现单例
灵感来自flutter firebase => FirebaseFirestore.instance.collection('collectionName')
class FooAPI {
foo() {
// some async func to api
}
}
class SingletonService {
FooAPI _fooAPI;
static final SingletonService _instance = SingletonService._internal();
static SingletonService instance = SingletonService();
factory SingletonService() {
return _instance;
}
SingletonService._internal() {
// TODO: add init logic if needed
// FOR EXAMPLE API parameters
}
void foo() async {
await _fooAPI.foo();
}
}
void main(){
SingletonService.instance.foo();
}
来自我的项目的例子
class FirebaseLessonRepository implements LessonRepository {
FirebaseLessonRepository._internal();
static final _instance = FirebaseLessonRepository._internal();
static final instance = FirebaseLessonRepository();
factory FirebaseLessonRepository() => _instance;
var lessonsCollection = fb.firestore().collection('lessons');
// ... other code for crud etc ...
}
// then in my widgets
FirebaseLessonRepository.instance.someMethod(someParams);
使用空安全操作符和工厂构造函数可以更好地创建单例对象。
class Singleton {
static Singleton? _instance;
Singleton._internal();
factory Singleton() => _instance ??= Singleton._internal();
void someMethod() {
print("someMethod Called");
}
}
用法:
void main() {
Singleton object = Singleton();
object.someMethod(); /// Output: someMethod Called
}
注意:? ?是一个Null感知操作符,如果左边值为Null,它将返回右边的值,这意味着在我们的例子中_instance ??Singleton._internal();, Singleton._internal()将在第一次调用object时返回,rest _instance将返回。