我试图通过使用forkjoin
来避免嵌套的observable。当前(嵌套)版本如下所示:
this.http.get('https://testdb1.firebaseio.com/.json').map(res => res.json()).subscribe(data_changes => {
this.http.get('https://testdb2.firebaseio.com/.json').map(res => res.json()).subscribe(data_all => {
/* Do this once resolved */
this.platform.ready().then(() => {
this.storage.set('data_changes', data_changes);
this.storage.set('data_all', data_all);
document.getElementById("chart").innerHTML = "";
this.createChart();
});
});
},
err => {
this.platform.ready().then(() => {
console.log("server error 2");
document.getElementById("chart").innerHTML = "";
this.createChart();
});
});
}
我可以将第一部分重写为:
Observable.forkJoin(
this.http.get('https://testdb1.firebaseio.com/.json').map((res: Response) => res.json()),
this.http.get('https://testdb2.firebaseio.com/.json').map((res: Response) => res.json())
)
但我不确定如何添加.subscribe
方法来同时访问data_changes
和data_all
。
再看另一个例子,我知道它应该看起来像.subscribe(res => this.combined = {friends:res[0].friends, customer:res[1]});
,但我不知道如何将它改编为我的例子。
答案 0 :(得分:36)
尝试使用combineLatest
代替forkJoin
:
使用combineLatest
:
const combined = Observable.combineLatest(
this.http.get('https://testdb1.firebaseio.com/.json').map((res: Response) => res.json()),
this.http.get('https://testdb2.firebaseio.com/.json').map((res: Response) => res.json())
)
combined.subscribe(latestValues => {
const [ data_changes , data_all ] = latestValues;
console.log( "data_changes" , data_changes);
console.log( "data_all" , data_all);
});
您也可以通过forkJoin处理它,但forkJoin将在所有调用完成后返回数据并返回结果,但在combineLatest中当任何observable发出值时,从每个值发出最新值。
使用forkJoin
:
const combined = Observable.forkJoin(
this.http.get('https://testdb1.firebaseio.com/.json').map((res: Response) => res.json()),
this.http.get('https://testdb2.firebaseio.com/.json').map((res: Response) => res.json())
)
combined.subscribe(latestValues => {
const [ data_changes , data_all ] = latestValues;
console.log( "data_changes" , data_changes);
console.log( "data_all" , data_all);
});
调用两个并检查控制台日志,你会明白的。