在对象数组中填充缺失属性的最佳方法是什么,例如:
[
{
name: 'Tom',
number: '01234 567 890',
website: 'http://www.tom.com'
},
{
name: 'Richard',
number '07777 666 555'
},
{
name: 'Harry',
website: 'http://www.harry.com'
}
]
我需要使用空值添加缺少的属性,这样当我传递此数组以在HTML表格或CSV文件等内容中呈现时,所有内容都会正确排列。我想要两次传递数组,一次获取所有可能的属性,第二次将具有空值的缺少属性添加到它不存在的每个对象。有更好的方法吗?
编辑:在我获得数据之前,我不知道密钥是什么,它来自API并且并不总是明确请求密钥。谢谢大家,似乎两通方法确实是最好的方法。在我开始使用提供的示例开始编写之后,我意识到属性的顺序没有得到维护。这就是我如何填写缺少的道具,并保持正确的顺序。欢迎任何有关潜在改进的建议。
var fillMissingProps = function(arr) {
// build a list of keys in the correct order
var keys = [];
arr.forEach(function(obj) {
var lastIndex = -1;
Object.keys(obj).forEach(function(key, i) {
if (keys.includes(key)) {
// record the position of the existing key
lastIndex = keys.lastIndexOf(key);
if (lastIndex < i) {
// this key is in the wrong position so move it
keys.splice(i, 0, keys.splice(lastIndex, 1)[0]);
lastIndex = i;
}
} else {
// add the new key in the correct position
// after the previous existing key
lastIndex++;
keys.splice(lastIndex, 0, key);
}
});
});
// build a template object with all props set to null
// and in the correct position
var defaults = {};
keys.forEach(function(key) {
defaults[key] = null;
});
// and update the array by overwriting each element with a
// new object that's built from the template and the original object
arr.forEach(function(obj, i, arr) {
arr[i] = Object.assign({}, defaults, obj);
});
return arr;
};
/** TEST **/
var currentArray = [
{
website: 'http://www.unknown.com'
},
{
name: 'Tom',
number: '01234 567 890',
website: 'http://www.tom.com'
},
{
title: 'Mr',
name: 'Richard',
gender: 'Male',
number: '04321 666 555'
},
{
id: '003ABCDEFGHIJKL',
name: 'Harry',
website: 'http://www.harry.com',
mobile: '07890 123 456',
city: 'Brentwood',
county: 'Essex'
}
];
var newArray = fillMissingProps(currentArray);
for (var i = 0; i < newArray.length; i++) {
for (var prop in newArray[i]) {
console.log(prop + ": " + newArray[i][prop]);
}
console.log('---------');
}
&#13;
答案 0 :(得分:4)
鉴于您并不知道应该存在哪些密钥,您别无选择,只能对数组进行两次迭代:
// build a map of unique keys (with null values)
var keys = {}
array.forEach(el => Object.keys(el).forEach(k => keys[k] = null));
// and update the array by overwriting each element with a
// new object that's built from the null map and the original object
array.forEach((el, ix, a) => a[ix] = Object.assign({}, keys, el));
答案 1 :(得分:0)
const arr = [
{
name: 'Tom',
number: '01234 567 890',
website: 'http://www.tom.com',
},
{
name: 'Richard',
number: '07777 666 555',
},
{
name: 'Harry',
website: 'http://www.harry.com',
},
];
const newArr = arr.map(x => (
arr.map(x => Object.keys(x))
.reduce((a, b) =>
(b.forEach(z => a.includes(z) || a.push(z)), a)
)
.forEach(
y => (x[y] = x.hasOwnProperty(y) ? x[y] : null)
), x)
);
console.log(newArr);
答案 2 :(得分:0)
您可以使用for..of
循环获取所有密钥并设置所有密钥,.map()
以迭代所有Object.keys()
,重新定义原始数组
var arr = [{
name: 'Harry',
website: 'http://www.harry.com'
},{
name: 'Tom',
number: '01234 567 890',
website: 'http://www.tom.com'
}, {
name: 'Richard',
number: '07777 666 555'
}];
for (var obj of arr) {
for (var key of Object.keys(obj)) {
arr = arr.map(o => (o[key] = o[key] || null, o))
}
};
console.log(arr);
&#13;
答案 3 :(得分:0)
这是一个更有趣的答案,它有点有趣,但它会在新属性出现时动态构建您的对象:
var currentArray = [
{
name: 'Tom',
number: '01234 567 890',
website: 'http://www.tom.com'
},
{
name: 'Richard',
number: '07777 666 555'
},
{
name: 'Harry',
website: 'http://www.harry.com'
}
]
var newArray = []
function NewObject() {
}
for(var i = 0; i < currentArray.length; i++){
var nObj = new NewObject();
for(var prop in currentArray[i]){
if(!NewObject.hasOwnProperty(prop))
NewObject.prototype[prop] = null;
nObj[prop]=currentArray[i][prop];
}
newArray.push(nObj);
}
for(var i = 0; i < newArray.length; i++){
for(var prop in newArray[i]){
console.log(prop+ ": "+newArray[i][prop]);
}
console.log('---------');
}
&#13;
它会根据您提供的对象构建新对象,并在对象尚未存在的情况下为对象添加新属性。
这个想法更多的是出于好奇心,所以任何评论都会很有趣:)
答案 4 :(得分:-1)
这样的事情可行:
for (var i = 0; i < arrayLength; i++) {
yourArray[i].name = yourArray[i].name || null;
yourArray[i].number = yourArray[i].number || null;
yourArray[i].website= yourArray[i].website|| null;
}