[英]More efficient way to search an array of javascript objects?
這是直截了當的做法。 如果您需要多次快速訪問,則應創建一個由您正在搜索的屬性名稱鍵入的地圖。
這是一個采用數組和構建鍵控映射的函數。 這不是萬能的,但你應該能夠修改它以供自己使用。
/**
* 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;
};
@jondavidjohn - 你可以使用這個javascript lib,DefiantJS( http://defiantjs.com ),你可以使用它在JSON結構上使用XPath過濾匹配。 把它放在JS代碼中:
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
這是一個工作小提琴;
http://jsfiddle.net/hbi99/weKVL/
DefiantJS使用方法“search”擴展全局對象,並返回一個匹配的數組(如果沒有找到匹配,則返回空數組)。 您可以在此處使用XPath Evaluator嘗試lib和XPath查詢:
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