搜索JavaScript对象数组的更高效方法？

Question

不确定发布规则，但我会告诉你，这是重复的问题，但我想问这是否是最佳实践的方法？

- jondavidjohn

Answer 1

这很容易实现。如果您需要多次快速访问，请创建一个由属性名称为键的映射。

下面是一个函数，它接受数组并构建键控映射。它不是通用的，但你应该能够修改它以符合你自己的使用。

/**
 * Given an array and a property name to key by, returns a map that is keyed by each array element's chosen property
 * This method supports nested lists
 * Sample input: list = [{a: 1, b:2}, {a:5, b:7}, [{a:8, b:6}, {a:7, b:7}]]; prop = 'a'
 * Sample output: {'1': {a: 1, b:2}, '5': {a:5, b:7}, '8': {a:8, b:6}, '7':{a:7, b:7}}
 * @param {object[]} list of objects to be transformed into a keyed object
 * @param {string} keyByProp The name of the property to key by
 * @return {object} Map keyed by the given property's values
 */
function mapFromArray (list , keyByProp) {
  var map = {};
  for (var i=0, item; item = list[i]; i++) {
    if (item instanceof Array) {
      // Ext.apply just copies all properties from one object to another,
      // you'll have to use something else. this is only required to support nested arrays.
      Ext.apply(map, mapFromArray(item, keyByProp));
    } else {
      map[item[keyByProp]] = item;
    }
  }
  return map;
};

- Ruan Mendes

Answer 2

@jondavidjohn - 您可以使用这个JavaScript库，DefiantJS（http://defiantjs.com），通过XPath在JSON结构中进行过滤匹配。将其放入JS代码中：

var data = [
   {
      "restaurant": {
         "name": "McDonald's",
         "food": "burger"
      }
   },
   {
      "restaurant": {
         "name": "KFC",
         "food": "chicken"
      }
   },
   {
      "restaurant": {
         "name": "Pizza Hut",
         "food": "pizza"
      }
   }
].
res = JSON.search( data, '//*[food="pizza"]' );

console.log( res[0].name );
// Pizza Hut

这里有一个可用的 fiddle；
http://jsfiddle.net/hbi99/weKVL/

DefiantJS 通过方法 "search" 扩展全局对象，并返回匹配项数组（如果没有找到匹配项，则为空数组）。您可以使用 XPath 评估器在此处尝试该库和 XPath 查询：

http://www.defiantjs.com/#xpath_evaluator

- Hakan Bilgin