RXJS过滤另一个可观察的

时间:2017-03-10 19:45:32

标签: javascript typescript rxjs observable

在我的代码中,我从MS Graph获取了一个用户列表,但这列出了所有用户。我想只发出那些属于指定组成员的用户。所以我设置了一个过滤器:

private loadUsers() {
    console.log(`Loading users...`);
    this.graph.get$(`users`)
        .map(resp => resp.json().value)
        .flatMap(user => user)
        .do(user => {
            console.log(`Got user: ${user['displayName']}`);
        })
        .filter(user => {
            // we only want users who are members of the TrackerContact group
            const userId: string = user['id'];
            console.log(`Checking user ${userId}`);
            return this.userIsTrackerContact(user); // <<= PROBLEM HERE!
        })
        .subscribe(_ => {
            console.log(`Completed user fetch!`);
        });
}

问题是为了找出用户所属的组,我必须将另一个REST回调给MS Graph

private userIsTrackerContact(user): Observable<boolean> {
    // get the groups this user is a member of
    return this.graph.get$(`users/${user['id']}/memberOf`)
        .map(resp => resp.json().value)
        .do(groups => {
            console.log(`${user['displayName']} is a member of ${groups.length} groups:`);
            groups.forEach((group, idx) => {
                console.log(`   [${idx+1}]: ${group['displayName']}`);
            })
        })
        .map(groups => {
            const found = groups.find((group) => group['id'] === this.authConfig.azureGroups.trackerContact)
            return !isUndefined(found);
        })
}

在我看来,filter基于另一个可观测量的Observable是可能的。谓词测试必须是一个简单的真/假测试。

如何根据另一个Observable的结果过滤Observable?

1 个答案:

答案 0 :(得分:1)

.concatAll()可能会更好:

this.graph.get$(`users`)
    .map(resp => resp.json().value)
    .flatMap(user => user)
    .do(user => {
        console.log(`Got user: ${user['displayName']}`);
    })
    .filter(user => {
        // we only want users who are members of the TrackerContact group
        const userId: string = user['id'];
        console.log(`Checking user ${userId}`);
        return this.userIsTrackerContact(user); // <<= PROBLEM HERE!
    })
    .concatAll()
    .subscribe(...)

我相信你应该能够merge async observables然后过滤。

Rx.Observable.merge(this.graph.get$(`users`)
    .map(resp => resp.json().value)
    .flatMap(user => user)
    .do(user => {
        console.log(`Got user: ${user['displayName']}`);
    })
    .map(user => {
        // we only want users who are members of the TrackerContact group
        const userId: string = user['id'];
        console.log(`Checking user ${userId}`);
        return this.userIsTrackerContact(user); // <<= PROBLEM HERE!
    }))
    .filter(x => x)
    .subscribe(_ => console.info('success'));

您可以在用户isGroupMember上更改谓词集属性。然后是filter(x => x.isGroupMember)。然后,您在订阅方法中拥有用户对象。