Question

我有一个Aurelia网络应用程序，可以按名称搜索艺术家，然后返回艺术家的所有艺术作品。艺术家搜索返回的json看起来类似于：

{
"source": "Museum",
"language": "EN",
"resultsCount": 1,
"objects": [
  {
  "objectNumber": "125.1988",
  "objectID": 981,
  "title": "Governor's Palace (Raj Bhaven) project, Chandigarh, India",
  "displayName": "Le Corbusier (Charles-Édouard Jeanneret)",
  "alphaSort": "Le Corbusier (Charles-Édouard Jeanneret)",
  "artistID": 3426,
  "displayDate": "French, born Switzerland. 1887–1965",
  "dated": "1951–1976",
  "dateBegin": 1951,
  "dateEnd": 1976,
  "medium": "Wood, cardboard, and plexiglass",
  "dimensions": "33 x 71 1/4 x 65\" (83.8 x 181 x 165.1 cm)",
  "department": "Architecture & Design",
  "classification": "A&D Architectural Model",
  "onView": 0,
  "provenance": "",

  etc....

完整的json在这里：

https://github.com/smoore4moma/tms-api/blob/master/object.json

我遇到的问题是，艺术家搜索可以返回多个具有相似名称的艺术家（例如“Serra”返回Richard Serra和Daniel Serra-Badué）。因此，我可以循环搜索并调用API两次以获得艺术作品，但我还没有找到一种方法来组合两个不同的API调用，而不是Aurelia。这是我想要做的代码片段。我已经尝试过jQuery，.extend，.concat，.push，创建数组，字典....我只是卡住了。

    getByConstituentId(id) {
    return this.http.fetch(baseUrl + "/artists/" + id + token)
    .then(response => response.json())
    .then(response => {
        return response.objects;
    });
    }

// This is not the actual looping code.  But the idea is to combine these two function calls into one result and return that.

     getBySearchTerms(searchCreator) {

     let obj0 = this.getByConstituentId(5350); 
     let obj1 = this.getByConstituentId(5349); 

     return ???  // This is where I am looking for help.
                 // just for ex. "return this.getByConstituentId(5349);" 
                 // works just fine

     }

关于如何做到这一点的任何建议？否则，我将构建一个采用艺术家ID数组的API方法，但我宁愿使用简单的API方法并在需要时组合json。感谢。

后续

按照@Tomalak的建议，这最终会起作用：

   return Promise.all([
   this.getByConstituentId(5349),
   this.getByConstituentId(5350)])
   .then(response => {

     return response[0].concat(response[1]);

    });

Answer 1

请参阅Promise.all() function。

[...]返回一个promise，它在iterable参数中的所有promise都已解析时解析，或者拒绝第一个传递的拒绝承诺的原因。

(defn only-those-actors? [db movie actors]
  (->> (datoms db :eavt movie :film/cast) seq
    (every? (fn [[_ _ actor]]
                (contains? actors actor)))
    ))

(defn find-movies-with-exact-cast [db actors-names]
  (let [actors (set (d/q '[:find [?actor ...] :in $ [?name ...] ?only-those-actors :where
                           [?actor :actor/name ?name]]
                      db actors-names))
        query {:find '[[?movie ...]]
               :in '[$ ?actors ?db]
               :where
               (concat
                 (for [actor actors]
                   ['?movie :film/cast actor])
                 [['(only-those-actors? ?db ?movie ?actors)]])}]
    (d/q query db actors db only-those-actors?)))

如果jQuery实际上执行了Ajax调用（即如果function getByConstituentId(id) { return this.http.fetch(baseUrl + "/artists/" + id + token) .then(response => response.json().objects); }; function getBySearchTerms(searchCreator) { return Promise.all([ this.getByConstituentId(5350), this.getByConstituentId(5349) ]); }是this.http.fetch的包装器），则可以使用jQuery＆＃39; s jquery.when()，它可以实现相同的目的，但会让你独立于the browser's native support for the Promise API。

$.ajax

请注意不同的呼叫约定。 function getBySearchTerms(searchCreator) { return $.when( this.getByConstituentId(5350), this.getByConstituentId(5349) ); }期望参数列表，而$.when期望迭代（如数组）。

您可以通过Promise.all()使用$.when数组参数。

.apply()

Aurelia - 将两个API调用中的json合并为一个

1 个答案: