如果我有对象的引用:
var test = {};
可能(但不是立即)具有嵌套对象,例如:
{level1: {level2: {level3: "level3"}}};
检查深度嵌套对象中是否存在属性的最佳方法是什么?
警报(测试级别1);生成未定义,但警告(test.level1.level2.level3);失败。
我目前正在做这样的事情:
if(test.level1 && test.level1.level2 && test.level1.level2.level3) {
alert(test.level1.level2.level3);
}
但我想知道是否有更好的方法。
还有一个非常紧凑的:
function ifSet(object, path) {
return path.split('.').reduce((obj, part) => obj && obj[part], object)
}
打电话:
let a = {b:{c:{d:{e:'found!'}}}}
ifSet(a, 'b.c.d.e') == 'found!'
ifSet(a, 'a.a.a.a.a.a') == undefined
它的性能不会很好,因为它拆分了一个字符串(但增加了调用的可读性),并遍历所有内容,即使已经很明显找不到任何内容(但提高了函数本身的可读性。
至少比get快http://jsben.ch/aAtmc
function getValue(base, strValue) {
if(base == null) return;
let currentKey = base;
const keys = strValue.split(".");
let parts;
for(let i=1; i < keys.length; i++) {
parts = keys[i].split("[");
if(parts == null || parts[0] == null) return;
let idx;
if(parts.length > 1) { // if array
idx = parseInt(parts[1].split("]")[0]);
currentKey = currentKey[parts[0]][idx];
} else {
currentKey = currentKey[parts[0]];
}
if(currentKey == null) return;
}
return currentKey;
}
如果结果在嵌套或值本身的任何地方失败,则调用函数返回undefined
const a = {
b: {
c: [
{
d: 25
}
]
}
}
console.log(getValue(a, 'a.b.c[1].d'))
// output
25
这是我使用的一个小助手函数,对我来说,它非常简单明了。希望这对一些人有帮助:)。
static issetFromIndices(param, indices, throwException = false) {
var temp = param;
try {
if (!param) {
throw "Parameter is null.";
}
if(!Array.isArray(indices)) {
throw "Indices parameter must be an array.";
}
for (var i = 0; i < indices.length; i++) {
var index = indices[i];
if (typeof temp[index] === "undefined") {
throw "'" + index + "' index is undefined.";
}
temp = temp[index];
}
} catch (e) {
if (throwException) {
throw new Error(e);
} else {
return false;
}
}
return temp;
}
var person = {
hobbies: {
guitar: {
type: "electric"
}
}
};
var indices = ["hobbies", "guitar", "type"];
var throwException = true;
try {
var hobbyGuitarType = issetFromIndices(person, indices, throwException);
console.log("Yay, found index: " + hobbyGuitarType);
} catch(e) {
console.log(e);
}
基于这个答案,我使用ES2015提出了一个通用函数,可以解决这个问题
function validChain( object, ...keys ) {
return keys.reduce( ( a, b ) => ( a || { } )[ b ], object ) !== undefined;
}
var test = {
first: {
second: {
third: "This is not the key your are looking for"
}
}
}
if ( validChain( test, "first", "second", "third" ) ) {
console.log( test.first.second.third );
}
这个功能怎么样?它不需要单独列出每个嵌套属性,而是保持“dot”语法(尽管是字符串),使其更具可读性。如果未找到属性,则返回undefined或指定的默认值,如果找到,则返回属性的值。
val(obj, element, default_value)
// Recursively checks whether a property of an object exists. Supports multiple-level nested properties separated with '.' characters.
// obj = the object to test
// element = (string or array) the name of the element to test for. To test for a multi-level nested property, separate properties with '.' characters or pass as array)
// default_value = optional default value to return if the item is not found. Returns undefined if no default_value is specified.
// Returns the element if it exists, or undefined or optional default_value if not found.
// Examples: val(obj1, 'prop1.subprop1.subsubprop2');
// val(obj2, 'p.r.o.p', 'default_value');
{
// If no element is being requested, return obj. (ends recursion - exists)
if (!element || element.length == 0) { return obj; }
// if the element isn't an object, then it can't have properties. (ends recursion - does not exist)
if (typeof obj != 'object') { return default_value; }
// Convert element to array.
if (typeof element == 'string') { element = element.split('.') }; // Split on dot (.)
// Recurse into the list of nested properties:
let first = element.shift();
return val(obj[first], element, default_value);
}