Ale*_*der 4 java reactive-programming rx-java
我有两个可观察量:
Observable<String[]>
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和
Observable<KalturaVideo>
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我想检查每个KalturaVideo.getId(),它是否包含在String []中,它返回第一个observable,如果它包含,则不要在sourceList中发出它.
所以,第一个可观察到的emmits String []只有一次,但是第二个可观察的emmits项目是17次.如何恰当地组合这两个可观测量?这是我的解决方案:
private Observable<KalturaVideo> getSourceListObservalbe(String kalturaPlaylistId){
Observable<String[]> targetIdList = Observable.just(targetList)
.map((List<KalturaVideo> kalturaVideos) -> {
if(kalturaVideos == null || kalturaVideos.isEmpty()){
return new String[]{""};
}
String[] result = new String[kalturaVideos.size()];
int index = 0;
for (KalturaVideo item : kalturaVideos) {
result[index] = item.getId();
}
return result;
});
return Observable.zip(
targetIdList,
KalturaVideoRetriver.getVideoList(BuildPlaylistStep2Activity.this, kalturaPlaylistId),
(String[] idListOfTarget, KalturaVideo kalturaVideo) -> {
for (String item :idListOfTarget){
if(item.equals(kalturaVideo.getId())){
return null;
}
}
return kalturaVideo;
})
.filter(kalturaVideo -> {
return kalturaVideo != null;
});
}
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targetList - 我想从sourceList中排除的KalturaVideo列表,KalturaVideoRetriver.getVideoList - 返回sourceList
这个解决方案只给我一个KalturaVideo,但我需要17个
你可以使用combineLatest而不是zip:
return Observable.combineLatest(
targetIdList,
KalturaVideoRetriver.getVideoList(BuildPlaylistStep2Activity.this, kalturaPlaylistId),
(String[] idListOfTarget, KalturaVideo kalturaVideo) -> {
...
return kalturaVideo;
})
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combineLatest有点像zip,但它结合了每个最近的排放.
所以在你的情况下,你的targetIdListobservable发出一个单一的String[].combineLatest将用你的其他可观察物的每一次发射拉上那个.