Question

可能是一个愚蠢的问题，假设我有一个像下面这样的对象，

{"Country":"country","Continent":"continent","Province":"","District":"","State":"state","City":""}"

在不使用循环的情况下，检查哪些属性保存上述对象中的值的最快方法是什么？

在for循环中执行此操作时，

if(typeof someUndefVar == whatever) -- works

预期输出

国家，大陆和州

Answer 1

var a = {"Country":"country","Continent":"continent","Province":"","District":"","State":"state","City":""};

Object.keys(a).filter( prop => a[prop] );

这还取决于您希望如何处理0，null，undefined值。

Answer 2

您可以执行以下操作;

var obj  = {"Country":"country","Continent":"continent","Province":"","District":"","State":"state","City":""};
for (var prop in obj) !!obj[prop] && console.log(prop);

Answer 3

您需要使用object.keys()功能

像这样，

var jsonData= {"Country":"country","Continent":"continent","Province":"","District":"","State":"state","City":""};
console.log(Object.keys(jsonData));

希望这有帮助！

Answer 4

对不起，你说没有循环，我很兴奋，我很无聊，我有功能。

首先，我缩短了一些名字，我猜他们有点自我解释：

head：数组的第一个元素
tail：所有元素，但第一个
isEmpty：检查数组是否长度为0

transform使用回调执行实际工作，如果keys数组为空则如果不递归则停止。

var data = {"Country":"country","Continent":"continent","Province":"","District":"","State":"state","City":""}

// plumbing
var head = (a) => a[0]
var tail = (a) => a.slice(1)
var isEmpty = (a) => a.length === 0

// actual stuff happens here
var transform = (obj, callback) => {

    var withval = (keys) => {

      if(isEmpty(keys)) return

      var p = head(keys)

      if(!!obj[p]) callback(p)

      withval(tail(keys))
    }

    withval(Object.keys(obj))
}

// Logs
var log = (p) => console.log(p)
transform(data, log)

// Array making
var arr = [];
var toArr = (p) => arr.push(p)
transform(data, toArr)
console.log(arr)

// Object making
var obj = {}
var toObj = (p) => obj[p] = data[p]
transform(data, toObj)
console.log(obj)

结果：

Country
Continent
State
[ 'Country', 'Continent', 'State' ]
{ Country: 'country', Continent: 'continent', State: 'state' }

检查对象中哪个属性保存值的最快方法？

4 个答案: