从JSON属性中选择随机值 [英] Pick random value from JSON property

问题描述

我有一个标题中描述的任务.我一定不要在我的代码中的任何地方使用数组，循环和Object.keys.我只能使用Math.random.

I have got a task that I described in title. I MUST NOT to use arrays, loops and Object.keys somewhere in my code. I can only use Math.random.

JSON文件:

{
  "key": "item 1, item 2, item 3",
  "key2": "abc",
  "key3": "123"
}

如果没有我描述的技术，我什至无法想象如何做到这一点.如果该任务支持Object.keys，它将非常简单(但我无法将其写为答案).

I can't even imagine how to do this without technologies that i described. If the task supported Object.keys it will be very simple (but i can't write this as answer).

我可以从JSON获取的对象:

let obj = {
           key: "item 1, item 2, item 3",
           key2: "abc",
           key3: "123"
       };

我使用Object.keys(我无法使用)选择随机值的函数:

 let randomProperty = function (obj) { 
 let keys = Object.keys(obj);    
 return obj[keys[ keys.length * Math.random() << 0]]; 
};

可以使用JSON.stringify和parse，因为它必须是JSON文件而不是我编写的对象.

JSON.stringify and parse can be used, because it must be JSON file instead object that i wrote.

请帮助！

推荐答案

使用更少的代码来改变Ivan的答案. Stringify替换器将第一个参数作为参数时间，然后第二次选择一个.

Variation on Ivan's answer with much less code. Stringify replacer counts arguments the first time, then picks one the second time.

let obj = {
  key: "item 1, item 2, item 3",
  key2: "abc",
  key3: "123"
};

let k, n = 0;
JSON.stringify(obj, (key, value) => (key && ++n || value));
n *= Math.random();
JSON.stringify(obj, (key, value) => (key && --n | 0 || (k = key) || value));
console.log(obj[k]);

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