假设您有以下结构:
shopping-carts (collection)
- shopping-cart 1(doc)
-- dateCreated (field)
-- items (collection)
- shopping-cart 2(doc
-- dateCreated
-- items
.
.
.那么,我们如何获取整个购物车(文档)作为我们定义为的ShoppingCart对象呢
export interface ShoppingCart {
items: ShoppingCartItem[]
dateCreated: string
}afs.doc('shopping-cart/id').valueChanges()仅返回dateCreated
afs.doc('shopping-cart/id').collection('items').valueChanges()返回这些项。
一种简单的方法,把所有的东西都放在一起,并将其创建为一个对象?
发布于 2017-11-17 06:09:33
可能不是最好的解决方案,但还找不到更好的解决方案。
this.shoppingCartsCollection = afs.collection<Servicio>('shopping-carts',
ref => ref.orderBy('dateCreated', 'asc'));
this.shoppingCarts = this.shoppingCartsCollection
.snapshotChanges()
.map(values => {
return values.map(value => {
const shopping-cart = value.payload.doc.data();
const itemsArray = shopping-cart.items;
shopping-cart.observableItems = [];
for (var i = 0, len = itemsArray.length; i < len; i++) {
itemDocRef = itemsArray[i];
value.observableItems.push(afs.doc('items/'+ itemDocRef.id).valueChanges());
}
return shopping-cart;
});
});像访问任何其他可观察对象一样,访问您的项目数据
<div *ngFor="let shoppingCart of shoppingCarts | async">
<p *ngFor="let observableItem of shoppingCart.observableItems">
{{(observableItem | async)?.displayName}}
</p>
</div>https://stackoverflow.com/questions/46865124
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