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我在for循環中訂閱了一個for循環,它將JSON數據從外部源獲取爲一系列「類別數據」,然後根據用戶篩選該數據當前位置。我需要的是等待所有訂閱完成,然後才能繼續執行我的應用。現在,它不會等待所有訂閱完成,它只完成一些訂閱,然後繼續,而其他訂閱在後臺繼續。如何在繼續執行之前等待for循環中的訂閱
我試過下面的「蠻力」方法,我知道一定數量的類別將被添加到過濾的數組中,並且它可以工作,但我不知道如何使它適用於任何情況。
這裏是我的代碼:
getMultipleCategoryData(categoryIds: string[]) {
for (let i = 0; i < this.waypointIds.length; i++) {
//after user selects their categories, its added to waypointNames, and occurences() gets the # of occurences of each waypoint name
let occurences = this.occurences(this.waypointNames[i], this.waypointNames);
this.categoryApi.getCategoryData(this.waypointIds[i]).toPromise().then(data => {
let filteredLocs = data.locations.filter(loc => this.distanceTo(loc, this.hbLoc) < MAX_RADIUS);
let foundLocs = [];
//fill filteredLocs with first n, n = occurences, entries of data
for (let n = 0; n < occurences; n++) {
if (filteredLocs[n] != undefined) {
foundLocs[n] = filteredLocs[n];
}
}
//find locations closest to hbLoc (users current location), and add them to waypoints array
for (let j = 0; j < foundLocs.length; j++) {
for (let k = 0; k < filteredLocs.length; k++) {
if (this.distanceTo(this.hbLoc, filteredLocs[k]) < this.distanceTo(this.hbLoc, foundLocs[j]) && foundLocs.indexOf(filteredLocs[k]) < 0) {
foundLocs[j] = filteredLocs[k];
}
}
}
if (foundLocs.length > 0 && foundLocs.indexOf(undefined) < 0) {
for (let m = 0; m < foundLocs.length; m++) {
this.waypointLocs.push(foundLocs[m]);
}
}
}).then(() => {
//this hardcoded, brute force method works, but i'd need it to be more elegant and dynamic
if (this.waypointLocs.length >= 5) {
let params = { waypointLocs: this.waypointLocs, hbLoc: this.hbLoc };
this.navCtrl.push(MapPage, params);
}
});
}
}
而且categoryApi.getCategoryData方法:
getCategoryData(categoryId): Observable<any> {
// don't have data yet
return this.http.get(`${this.baseUrl}/category-data/${categoryId}.json`)
.map(response => {
this.categoryData[categoryId] = response.json();
this.currentCategory = this.categoryData[categoryId];
return this.currentCategory;
});
}
一切工作正常,除了等待訂閱完成,我真的很喜歡的方式來確定當所有訂閱都完成時。任何幫助表示讚賞!
大和簡單的解決方案!像魅力一樣工作。謝謝! – AlexT