Question

我正在努力从加入中返回一个值。我认为问题在于放置＆＃34; return＆＃34;一句话，但我找不到任何文件。

我在代码中添加了一些注释来更好地解释我的问题。

this.databaseService.getAlbumName(0).then(result =>{
    return result;
}).then(snap => {
    //snap return what I want so albumName.
    //Problem start when getPhotosFromAlbum(snap) kicks is.
    return this.databaseService.getPhotosFromAlbum(snap).then(photoResult => {
        //Am I returning value(array) from getPhotosFromAlbum(snap) in correct place?
        return photoResult;
    }).then(snapp => {
        //this part if just to display photoResult which is currently equal
        //to [Object object]
        console.log('array of photos: ' + snapp);
    })

和

    public getPhotosFromAlbum(albumName: string): any{
        //getting access to all Photos in album
        let photosInAlbum = firebase.database().ref('PhotosInAlbum' + '/' + albumName)
        return photosInAlbum.once('value', snap => {
            let photoRef = this.databaseRef.child('Photo');
            photoRef.once('value', photoSnap => {
                let photos = photoSnap.val();
                //creating an accessible array
                let keys = Object.keys(photos);
                var k = [];
                for(var i = 0; i < keys.length; i++){
                    k.push(keys[i]);
                }
                console.log('From get Function: ' + k);
                return k;
            })    
        })
    }

Answer 1

就我所见，你的getPhotosFromAlbum并没有回复承诺。

public getPhotosFromAlbum(albumName: string): any{
    let photosInAlbum = firebase.database().ref('PhotosInAlbum' + '/' + albumName)
    return photosInAlbum.once('value').then(snap => {
        let photoRef = this.databaseRef.child('Photo');
        return photoRef.once('value').then(photoSnap => {
            let photos = photoSnap.val();
            //creating an accessible array
            let keys = Object.keys(photos);
            var k = [];
            for(var i = 0; i < keys.length; i++){
                k.push(keys[i]);
            }
            return k;
        })    
    })
}

最内心的倾听者似乎是一个非常复杂的方式：

        return photoRef.once('value').then(photoSnap => {
            return Object.keys(photoSnap.val());
        })

从join中返回结果

1 个答案: