QueryFirebase的唯一键的组合和过滤器可观察量

时间:2017-03-09 21:52:23

标签: firebase rxjs observer-pattern geofire

我正在开发一个应用程序,承包商可以说它们在特定日期“可用”,每个承包商都有一个“位置”。雇主可以根据位置和可用性搜索可用性。

该位置基于GeoFire。这将返回可用承包商的$ key。

看起来像这样:

geoQueryContractor(radius, lat, lng) {

    const subject = new Subject();

    this.fbGeoRef = firebase.database().ref('geofire')

    this.geoFire = new GeoFire(this.fbGeoRef);

    this.geoFire.ref();

    this.geoQuery = this.geoFire.query({
        center: [lat, lng],
        radius: radius
    });

    this.geoQuery.on("key_entered", function(key, location, distance) {
        subject.next(key);
    });

    return subject.asObservable();

}

接下来,我可以通过搜索看起来像这样的“/ availForContractor / $ {timestamp} / $ uid:true”的firebase节点来获取可用性

这是如何工作的,它会返回他们的个人资料:

getAvailablitybyContractor(timestamp) {

    const availContractorKeys$ = this.db.list(`/AvailForContractor/${timestamp}`);

    const AvailContractors$ = availContractorKeys$

        //maping each key
        .map(keysPerContractor => keysPerContractor

        //once we have each key, we can map it and create an fb object observable
        .map(keyPerContractor => this.db.object(`/users/${keyPerContractor.$key}`)))

        //now we got back an array of firebase object observables (fbojs) and we need to combine them in to one observable
        .mergeMap(fbojs => Observable.combineLatest(fbojs))
        .do(console.log)

    AvailContractors$.subscribe();
}

我让这两个人彼此独立工作。我真的需要知道第二个函数中返回的所有$ key,哪些是第一个函数中的位置。我只需要返回符合这两个条件的配置文件。

我一直在搞乱CombineLatest,mergeMap,withLatestFrom和Filter,但我无法弄清楚如何以正确的方式做到这一点。

我的想法是,一旦我从第二个功能获得键,将它与GeoFire observable结合并过滤唯一键,然后执行此部分:

    //once we have each key, we can map it and create an fb object observable
    .map(keyPerContractor => this.db.object(`/users/${keyPerContractor.$key}`)))

    //now we got back an array of firebase object observables (fbojs) and we need to combine them in to one observable
    .mergeMap(fbojs => Observable.combineLatest(fbojs))
    .do(console.log)

这不完整,但尝试很差......

getAvailablitybyContractor(timestamp, radius, lat, lng) {

    const availContractorKeys$ = this.db.list(`/AvailForContractor/${timestamp}`);

    //when we get back the keys, we are going to switch to another obeservables
    const AvailContractors$ = availContractorKeys$

        //maping each key
        .map(keysPerContractor => keysPerContractor
            .map(keyPerContractor => keyPerContractor.$key))
            .combineLatest(this.geoQueryContractor(radius, lat, lng))

            //  .withLatestFrom(this.geoQueryContractor(radius, lat, lng), (keysPerContractor, geo) => ( [keysPerContractor, geo] ))
            //once we have each key, we can map it and create an fb object observable

        //     .map(keyPerContractor => this.db.object(`/users/${keyPerContractor.$key}`)))
        // //now we got back an array of firebase object observables (fbojs) and we need to combine them in to one observable
        // .mergeMap(fbojs => Observable.combineLatest(fbojs))

        .do(console.log)

    AvailContractors$.subscribe();
}

GeoFire顺便推出了这样的各个键:

3vAWWHaxHRZ94tc8yY08CH3QNQy3H74INXgYWIMrUcAtZloFGkwJ6Qd2

Firebase将推出一系列密钥:

[3vAWWHaxHRZ94tc8yY08CH3QNQy3, H74INXgYWIMrUcAtZloFGkwJ6Qd2, J9DHhg5VQrMpNyAN8ElCWyMWh8i2, fdZYKqqiL0bSVF66zGjBhQVu9Hf1  ]

最终结果将是我用于获取配置文件的RX方式的唯一组合。

有人可以帮忙吗?谢谢!

1 个答案:

答案 0 :(得分:0)

这是我的解决方案。我确信有更好的方法可以做到这一点。更新:下面更好的解决方案。

static geoArray: Array<string> = [];

constructor(private af: AngularFire, private db: AngularFireDatabase) {

}

getAvailablitybyContractor(timestamp, radius, lat, lng) {

    const availContractorKeys$ = this.db.list(`/AvailForContractor/${timestamp}`);

    const AvailContractors$ = availContractorKeys$

        //maping each key
        .map(keysPerContractor => keysPerContractor.map(keyPerContractor => keyPerContractor.$key)
        .filter(key => ContractorService.geoArray.indexOf(key) > -1)

         //once we have each key, we can map it and create an fb object observable
        .map(keyPerContractor => this.db.object(`/users/${keyPerContractor}`)))

        //now we got back an array of firebase object observables (fbojs) and we need to combine them in to one observable
        .mergeMap(fbojs => Observable.combineLatest(fbojs))
        .do(console.log)

    AvailContractors$.subscribe();
}

geoQueryContractor(radius, lat, lng) {


    this.fbGeoRef = firebase.database().ref('geofire')

    this.geoFire = new GeoFire(this.fbGeoRef);

    this.geoFire.ref();

    this.geoQuery = this.geoFire.query({
        center: [lat, lng],
        radius: radius
    });

    this.geoQuery.on("key_entered", function(key, location, distance) {

        ContractorService.geoArray.push(key);

    });

}


}

这个解决方案要好得多。上面的那个真的很麻烦。它与清除阵列有关。在一天结束时,我将不得不执行搜索2x以获得正确的结果。不能接受。

这是一种更好的方法,它更具反应性,并且没有上面的错误。我确信有更好的方法来折射或改进它。

getAvailablitybyContractor(timestamp) {

    let availContractorKeys$ = this.db.list(`/AvailForContractor/${timestamp}`);

    this.AvailContractors$ = availContractorKeys$

        //maping each key
        .map(keysPerContractor => keysPerContractor.map(keyPerContractor => keyPerContractor.$key))
        //Combine observable from GeoQuery
        .combineLatest(this.keys$, (fb, geo) => ([fb, geo]))
        // fb, geo are accessible individually
        // .filter method creates a new array with all elements that pass the test implemented by the provided function
        // key is now iteriable through geo.indexOf
        .map(([fb, geo]) => {
            return fb.filter(key => geo.indexOf(key) > -1)
        })
        .map(filteredKeys => filteredKeys.map(keyPerContractor => this.db.object(`/users/${keyPerContractor}`)))
        //now we got back an array of firebase object observables (fbojs) and we need to combine them in to one observable
        .mergeMap(fbojs => {
            return Observable.combineLatest(fbojs)
        })
        .do(console.log)

}

getGeoQuery(radius, lat, lng) {

    this.geoQuery = this.geoFire.query({
        center: [lat, lng],
        radius: radius
    });

}

geoQueryContractor() {

    return this.keys$ = Observable.create(observer => {

        var keys = new Array();

        this.geoQuery.on("key_entered", (key, location, distance) => {
            keys.push(key);
            observer.next(keys);
        });

    });

}