如何使用observable的输出来过滤另一个

Question

所以我有两个observable,一个返回当前类别,其他产品.我希望根据类别过滤产品.

这是在Angular 2中,所以我真的希望我的ng2-view成为订阅者(通过异步管道).

像这个简单的例子:

let category$ = Observable.of({id: 1});
let products$ = Observable.from([{name: 'will be included', cat_ids: [1, 5]}, {name: 'nope', cat_ids: [2, 3]}, {name: 'also yep', cat_ids: [1, 7]}]);

return products$
  .toArray()
  .filter(prod => {
    return prod.cat_id.some(id => id === <how do I get the value of the category observable here?>)
  });

也许答案很简单,但它让我望而却步.

Answer 1

你需要加入这两个流,例如使用combineLatest:

let category$ = Observable.of({id: 1});
let products$ = Observable.from([{name: 'will be included', cat_ids: [1, 5]}, {name: 'nope', cat_ids: [2, 3]}, {name: 'also yep', cat_ids: [1, 7]}]);

return Observable.combineLatest(products$, category$)
  .map(([products, category]) => {
    return products.filter(prod => prod.cat_id.some(id => id === category.id);
  });

更新

正如@olsn指出的那样Observable.from,事件流不是事件数组的流.因此解决方案应该是:

let category$ = Observable.of({id: 1});
let products$ = Observable.from([{name: 'will be included', cat_ids: [1, 5]}, {name: 'nope', cat_ids: [2, 3]}, {name: 'also yep', cat_ids: [1, 7]}]);

return Observable.combineLatest(products$, category$)
  .filter(([product, category]) => {
    return product.cat_id.some(id => id === category.id);
  })
  .map(([product, category]) => product);