Question

我有一个大约30个对象的json数组。这是数组中的一个示例对象：

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

我正在基于用户输入在页面上构建另一个对象。用户将在单选按钮之间进行选择，在他们做出选择之后，我将有一个对象，例如：

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

我正在使用jQuery的$.each()方法遍历每个对象，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;
    }
});

这将返回一个包含所有四个匹配项的对象，但是如何将具有最多匹配项的json对象返回给用户构建的对象？如果两个匹配，我想查看“priority”属性并返回优先级最低的属性。

Answer 1

您可以使用Array#map并使用匹配属性的总和构建一个新数组。

稍后您可以sort with map并使用结果进行排序并获取第一个元素。

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;
});

单循环解决方案

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

只需跟踪匹配的数量，并根据是否有更多匹配项更新您选择的匹配项。

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简化初始计数：

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;
  }
});

基于匹配数据返回Array中的Object

2 个答案: