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。


当前回答

一个简单的方法是:

try {
    alert(test.level1.level2.level3);
} catch(e) {
    alert("undefined");    // this is optional to put any output here
}

try/catch捕捉未定义任何更高级别对象(如test、test.level1、test.level1.level2)时的情况。

其他回答

我以以下方式使用函数。

var a = {};
a.b = {};
a.b.c = {};
a.b.c.d = "abcdabcd";

function isDefined(objectChainString) {
    try {
        var properties = objectChainString.split('.');
        var currentLevel = properties[0];
        if (currentLevel in window) {
            var consolidatedLevel = window[currentLevel];
            for (var i in properties) {
                if (i == 0) {
                    continue;
                } else {
                    consolidatedLevel = consolidatedLevel[properties[i]];
                }
            }
            if (typeof consolidatedLevel != 'undefined') {
                return true;
            } else {
                return false;
            }
        } else {
            return false;
        }
    } catch (e) {
        return false;
    }
}

// defined
console.log(checkUndefined("a.b.x.d"));
//undefined
console.log(checkUndefined("a.b.c.x"));
console.log(checkUndefined("a.b.x.d"));
console.log(checkUndefined("x.b.c.d"));

今天刚刚编写了这个函数,它对嵌套对象中的属性进行了深入搜索,如果找到了,则返回该属性的值。

/**
 * Performs a deep search looking for the existence of a property in a 
 * nested object. Supports namespaced search: Passing a string with
 * a parent sub-object where the property key may exist speeds up
 * search, for instance: Say you have a nested object and you know for 
 * certain the property/literal you're looking for is within a certain
 * sub-object, you can speed the search up by passing "level2Obj.targetProp"
 * @param {object} obj Object to search
 * @param {object} key Key to search for
 * @return {*} Returns the value (if any) located at the key
 */
var getPropByKey = function( obj, key ) {
    var ret = false, ns = key.split("."),
        args = arguments,
        alen = args.length;

    // Search starting with provided namespace
    if ( ns.length > 1 ) {
        obj = (libName).getPropByKey( obj, ns[0] );
        key = ns[1];
    }

    // Look for a property in the object
    if ( key in obj ) {
        return obj[key];
    } else {
        for ( var o in obj ) {
            if ( (libName).isPlainObject( obj[o] ) ) {
                ret = (libName).getPropByKey( obj[o], key );
                if ( ret === 0 || ret === undefined || ret ) {
                    return ret;
                }
            }
        }
    }

    return false;
}
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

您可以使用“.”分隔对象和路径

函数checkPathExist(obj,路径){var pathArray=路径.split(“.”)for(pathArray的var i){if(反射get(obj,i)){obj=obj[i];}其他{return false;}}返回true;}var测试={level1:{level2:{level3:‘level3‘}}}};console.log('level.level.level3=>',checkPathExist(测试,'level.level 2.level3'));//真的console.log('level.level.foo=>',checkPathExist(测试,'level.level 2.foo'));//假的

我编写了一个名为l33teral的库来帮助测试嵌套财产。您可以这样使用:

var myObj = {/*...*/};
var hasNestedProperties = leet(myObj).probe('prop1.prop2.prop3');

我也很喜欢这里的ES5/6解决方案。