Question

我有几个可观察者。我需要知道哪一个触发了订阅。

Observable.combineLatest(
      this.tournamentsService.getUpcoming(),
      this.favoriteService.getFavoriteTournaments(),
      this.teamsService.getTeamRanking(),
(tournament, favorite, team) => {
//what triggered combinelatest to run?
}).subscribe()

Answer 1

简短的回答是：你不知道。你可以实现一些解决方法，但这真的很难看，我建议你重新考虑用法为什么需要这个，也许你可以改变架构。还要记住，在所有三个observable都发出至少1个值之后，你的函数的第一次执行将会发生。

无论如何 - 可能的解决方法可能是：

let trigger = "";
Observable.combineLatest(
      this.tournamentsService.getUpcoming().do(() => trigger = "tournament"),
      this.favoriteService.getFavoriteTournaments().do(() => trigger = "favTournament"),
      this.teamsService.getTeamRanking().do(() => trigger = "teamRanking"),
(tournament, favorite, team) => {
   console.log(`triggered by ${trigger}`);
}).subscribe();

如果你想根据触发的observable执行一个特定的操作，你应该使用每个observable并将它们作为单独的触发器使用，切换到组合的触发器，它可能稍微多一些代码，但它更清晰，你不会在一个丑陋的if / else，switch / case-mess中遇到一些hacky变通方法 - plus 你甚至会有机会使用async - 管道而不是手动订阅所有内容并更新局部变量（无论如何这都是一种不好的做法）：

以下是一些示例代码：

let upcoming$ = this.tournamentsService.getUpcoming();
let favorite$ = this.favoriteService.getFavoriteTournaments();
let rankings$ = this.teamsService.getTeamRanking();

let allData$ = Observable.combineLatest(
    upcoming$, favorite$, rankings$,
    (tournament, favorite, team) => {
        return {tournament, favorite, team};
    }
);

// initial call -> this SHOULD be redundant,
// but since I don't know your code in detail
// i've put this in - if you can remove it or not
// depends on the order your data coming in
allData$
    .take(1)
    .do(({tournament, favorite, team}) => {
        this.displayMatches(...);
        this.sortByFavorites(...);
        this.fillWithRanking(...);
    })
    .subscribe();

// individual update triggers
upcoming$
    .switchMapTo(allData$.take(1))
    .do(({tournament, favorite, team}) => this.displayMatches(...))
    .subscribe();

favorite$
    .switchMapTo(allData$.take(1))
    .do(({tournament, favorite, team}) => this.sortByFavorites(...))
    .subscribe();

rankings$
    .switchMapTo(allData$.take(1))
    .do(({tournament, favorite, team}) => this.fillWithRanking(...))
    .subscribe();

Answer 2

您可以使用scan运算符将发出的值与任何先前发出的值进行比较，并且可以包含指示组合的observable的组件是否实际已更改的其他数据。例如：

let combined = Observable
  .combineLatest(
    this.tournamentsService.getUpcoming(),
    this.favoriteService.getFavoriteTournaments(),
    this.teamsService.getTeamRanking()
  )
  .scan((acc, values) => [
    ...values,
    acc[0] !== values[0],
    acc[1] !== values[1],
    acc[2] !== values[2]
  ], []);

combined.subscribe(
  ([tournament, favorite, team, tournamentChanged, favoriteChanged, teamChanged]) => {
    console.log(`tournament = ${tournament}; changed = ${tournamentChanged}`);
    console.log(`favorite = ${favorite}; changed = ${favoriteChanged}`);
    console.log(`team = ${team}; changed = ${teamChanged}`);
  }
);

Answer 3

通过使用时间戳运算符http://reactivex.io/documentation/operators/timestamp.html

，实现这一点非常简洁和“rx”

示例代码

sourceObservable
  .pipe(
    timestamp(),  // wraps the source items in object with timestamp of emit
    combineLatest( otherObservable.pipe( timestamp() ), function( source, other ) {

      if( source.timestamp > other.timestamp ) {

        // source emitted and triggered combineLatest
        return source.value;
      }
      else {

        // other emitted and triggered combineLatest
        return other.value;
      }

    } ),
  )

如果combineLatest()中有两个以上的可观察对象按时间戳排序，则可以检测哪一个触发了combineLatest()。

Answer 4

我正在Flutter中使用RxJava，我想使用操作符CombineLatest组合12种不同的可观察对象。

我看到了一个函数原型，该原型带有一个可观察对象列表和一个实现。但是我不确定该怎么做，我在实现call方法时遇到了麻烦。请检查我的代码并做必要的事情。

Stream>获取城市=> _citiesController.stream;

Stream get city => _cityController.stream;

Stream get agentcity => _agentcityController.stream;

Stream get userpackages => _packagesController.stream;

流获取电子邮件=> _emailController.stream.transform（validateEmail）;

Stream get firstName => _firstNameController.stream.transform（validateFirstName）;

Stream get lastName => _lastNameController.stream.transform（validateLastName）;

Stream get mobileNumber => _mobileNumberController.stream.transform（validateMobile）;

Stream get dob => _dobController.stream.transform（validatedob）;

流获取约会日期=> _appointmentdateController.stream.transform（validateappointmentDate）;

Stream get pincode => _pincodeController.stream.transform（validatePincode）;

流获取性别=> _genderController.stream;

流获取地址=> _addressController.stream.transform（validateAddress）;

Stream get agentname => _agentnameController.stream.transform（validateAgentName）;

流获取有效的提交=> Observable.combineLatest9（

email,
firstName,
mobileNumber,
pincode,
dob,
address,
agentname,
_genderController.stream,
_cityController.stream,
_agentcityController.stream,
_packagesController.stream,
_appointmentdateController.stream,
(e, f, m, p, d, a, an, g, c, ac, pc, ad) => true,

）;

请让我知道如何在Flutter代码中使用CombineLatest。

Answer 5

解决此问题的更好方法是使用可区分的联合类型。如果您的语言没有内置可区分联合，您可以通过将构造函数设为私有并公开 n 可为空的公共静态属性来创建一个联合，每个可观察对象一个。用 C# 编写这个，因为我更喜欢这种语言，但它应该很容易翻译。注意可为空的字符串。如果您的语言不支持可空值，请使用其他一些机制来指示是否设置了值。

private class DiscriminatedUnion
{
    private DiscriminatedUnion(string? property1, string? property2)
    {
        Property1 = property1;
        Property2 = property2;
    }

    public string? Property1 { get; }
    public string? Property2 { get; }

    public static DiscrimintatedUnion FromObservable1(string property1)
    {
        return new DiscriminatedUnion(property1, null);
    }

    public static DiscrimintatedUnion FromObservable2(string property2)
    {
        return new DiscriminatedUnion(null, property2);
    }

}

private IObservable<DiscriminatedUnion> CreateCombination()
{
    var firstObservable = tournamentsService
        .getUpcoming()
        .Select(x => DiscriminatedUnion.FromObservable1(x));

    var secondObservable = favoriteService
        .getFavoriteTournaments()
        .Select(x => DiscriminatedUnion.FromObservable2(x));

    return Observable
        CombineLatest(firstObservable, secondObservable);
}

所以现在你可以质疑从 CreateCombination() 返回的可区分联合，哪个 observable 发出了一个值。

什么引发了联合最新？

5 个答案: