big*_*ato 1 rxjs redux-observable
我有一个Observable,我想继续执行直到:
1)uploadActions.MARK_UPLOAD_AS_COMPLETE用一定的有效载荷调用动作
要么
2)uploadActions.UPLOAD_FAILURE使用任何有效负载调用该动作
这是我所能得到的(并且不起作用):
return Observable.interval(5000)
.takeUntil(
action$
.ofType(
uploadActions.UPLOAD_FAILURE,
uploadActions.MARK_UPLOAD_AS_COMPLETE
)
.filter(a => { // <---- this filter only applies to uploadActions.MARK_UPLOAD_AS_COMPLETE
const completedFileHandle = a.payload;
return handle === completedFileHandle;
})
)
.mergeMap(action =>
...
);
Run Code Online (Sandbox Code Playgroud)
有没有一种干净的方法可以实现这一目标?
我将两个条件拆分为单独的流,然后像这样合并它们:
const action$ = new Rx.Subject();
const uploadActions = {
UPLOAD_FAILURE: "UPLOAD_FAILURE",
MARK_UPLOAD_AS_COMPLETE: "MARK_UPLOAD_AS_COMPLETE"
};
const handle = 42;
window.setTimeout(() => action$.next({
type: uploadActions.MARK_UPLOAD_AS_COMPLETE,
payload: handle
}), 1200);
Rx.Observable.interval(500)
.takeUntil(
Rx.Observable.merge(
action$.filter(x => x.type === uploadActions.UPLOAD_FAILURE),
action$.filter(x => x.type === uploadActions.MARK_UPLOAD_AS_COMPLETE)
.filter(x => x.payload === handle)
)
).subscribe(
x => { console.log('Next: ', x); },
e => { console.log('Error: ', e); },
() => { console.log('Completed'); }
);Run Code Online (Sandbox Code Playgroud)
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对于该示例,我不得不使用filter运算符,而不是ofType因为这ofType是一个redux问题。
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