根據匹配的數據返回數組中的對象

Question

我有大約30個對象的json數組。 這是數組中的一個示例對象：

{
    "id": 0,
    "name": "Valle",
    "activities": "night-life",
    "food": "fancy-food",
    "sport" : "baseball",
    "geoProfile": "artsy",
    "priority": 2
}

我正在根據用戶輸入在頁面上構建另一個對象。 用戶將在單選按鈕之間進行選擇，並且在做出選擇之后，我將擁有一個對象，例如：

{geoProfile: "artsy", activities: "nature", food: "fancy-food", sport: "cricket"}

我正在使用jQuery的$.each()方法來遍歷每個對象，如下所示：

$.each(self.data(), function (i, s) {
    if (s.geoProfile == self.prefLocation() &&
        s.activities == self.prefActivity() &&
        s.food == self.prefFood() &&
        s.sport == self.prefSport()) {
        optionMatched = s;
        return false;
    }
});

這將返回具有所有四個匹配項的對象，但是如何返回與用戶構建的對象匹配度最高的json對象？ 如果兩個匹配，那么我想查看“優先級”屬性並返回優先級最低的屬性。

Answer 1

您可以使用Array#map並使用匹配屬性的總和構建一個新數組。

稍后，您可以使用map進行排序，並將結果用於排序並獲取第一個元素。

var data = [/* your data here */],
    search = { geoProfile: "artsy", activities: "nature", food: "fancy-food", sport: "cricket" },
    result = data.map(function (a, i) {
        return {
            count: Object.keys(search).reduce(function (r, k) { return r + +(a[k] === search[k]); }, 0),
            priority: a.priority,
            index: i
        }
    });

result.sort(function (a, b) {
    return b.count - a.count || a.priority - b.priority;
});

單回路解決方案

var data = [/* your data here */],
    search = { geoProfile: "artsy", activities: "nature", food: "fancy-food", sport: "cricket" },
    result = data.reduce(function (r, a, i) {
        document.write('<pre>' + JSON.stringify(r, 0, 4) + '</pre><hr>');
        var o = {
            count: Object.keys(search).reduce(function (q, k) { return q + +(a[k] === search[k]); }, 0),
            priority: a.priority,
            index: i
        };
        if (!i || o.count > r[0].count || o.count === r[0].count && o.priority < r[0].priority) {
            return [o];
        }
        o.count === r[0].count && o.priority === r[0].priority && r.push(o);
        return r;
    }, []);

Answer 2

只需跟蹤匹配數並根據是否有更多匹配項來更新您選擇的匹配項即可。

var numOfMatches = 0;
$.each(self.data(), function(i, s) {
  var matchingProperties = 0;
  if (s.geoProfile == self.prefLocation()) {
    matchingProperties++;
  }
  if (s.activities == self.prefActivity()) {
    matchingProperties++;
  }
  if (s.food == self.prefFood()) {
    matchingProperties++;
  }
  if (s.sport == self.prefSport()) {
    matchingProperties++;
  }

  if (matchingProperties === 0 || matchingProperties < numOfMatches) {
    return;
  }

  if (!optionMatched // Grab the first match
      || matchingProperties > numOfMatches // or if more properties match
      || s.priority < optionMatched.priority) { // or the one with a lower priority
    optionMatched = s;
    numOfMatches = matchingProperties;
  }
});

或者您可以使用filter簡化初始計數：

var numOfMatches = 0;
$.each(self.data(), function(i, s) {
  var matchingProperties = [
    s.geoProfile == self.prefLocation(),
    s.activities == self.prefActivity(),
    s.food == self.prefFood(),
    s.sport == self.prefSport()
  ].filter(function(val) { return val; }).length;

  if (matchingProperties === 0 || matchingProperties < numOfMatches) {
    return;
  }

  if (!optionMatched // Grab the first match
      || matchingProperties > numOfMatches // or if more properties match
      || s.priority < optionMatched.priority) { // or the one with a lower priority
    optionMatched = s;
    numOfMatches = matchingProperties;
  }
});