Question

family = {
  'person1':[{"id":"1111","name":"adam", "sex":"male", "born":"USA"}],
  'person2':[{"id":"2222","name":"sarah", "sex":"female", "born":"Canada"}],
  'person3':[{"id":"3333","name":"adam", "sex":"male", "born":"USA"}]
};

鉴于上面的家庭对象，我如何提取具有特定id（或名称）值的一个人物对象的所有属性（id，name，sex，born）？例如id = 1111

理想情况下，我可以获得一个我可以操作的新对象personInQuestion，其中：

personInQuestion = {"id":"1111","name":"adam", "sex":"male", "born":"USA"}

Answer 1

遍历对象，并抓取匹配的元素。

var search = 1111;

var personInQuestion = {};
for(var x in family){
    var item = family[x][0];
    if(item.id == search){
        personInQuestion = item;
        break;
    }
}

Answer 2

我不认为jQuery是最好的工具，相反我建议你看看at the Where method that the Backbone library offers。不确定它是否有点矫枉过正。

用法是这样的：

var friends = new Backbone.Collection([
  {name: "Athos",      job: "Musketeer"},
  {name: "Porthos",    job: "Musketeer"},
  {name: "Aramis",     job: "Musketeer"},
  {name: "d'Artagnan", job: "Guard"},
]);

var musketeers = friends.where({job: "Musketeer"});

alert(musketeers.length);

Answer 3

jQuery不为JSON（JavaScript Object）数据提供选择器，因此您需要迭代该对象。例如：

result =  null;
$.each(family, function(i, v) {
  if (v.id === "1111" && v.name === "adam" ...) {
    result = v;
    return;
  }
});

Answer 4

以下是Rob Hruska的suggestion（在评论中）使用Underscore的示例。

family = {
  'person1':[{"id":"1111","name":"adam", "sex":"male", "born":"USA"}],
  'person2':[{"id":"2222","name":"sarah", "sex":"female", "born":"Canada"}],
  'person3':[{"id":"3333","name":"adam", "sex":"male", "born":"USA"}]
};

var searchId = 1111;

var person = _.find(family, function(item) { return item[0].id == searchId; })[0];

使用javascript / jquery在更大的对象中查找对象及其属性

4 个答案: