根据@Stephane LaFlèche的回答，我提出了另一个剧本版本。

JSFiddle演示

var obj = {"a":{"b":{"c":"Hello World"}},"resTest":"potato","success":"This path exists"};
checkForPathInObject = function(object,path,value) {
        var pathParts   = path.split("."),
            result      = false;
        // Check if required parameters are set; if not, return false
        if(!object || typeof object == 'undefined' || !path || typeof path != 'string')
            return false;
        /* Loop through object keys to find a way to the path or check for value
         * If the property does not exist, set result to false
         * If the property is an object, update @object
         * Otherwise, update result */
        for(var i=0;i<pathParts.length;i++){
            var currentPathPart = pathParts[i];
            if(!object.hasOwnProperty( currentPathPart )) {
                result = false;
            } else if (object[ currentPathPart ] && path == pathParts[i]) {
                result = pathParts[i];
                break;
            } else if(typeof object[ currentPathPart ] == 'object') {
                object = object[ currentPathPart ];
            } else {
                result = object[ currentPathPart ];
            }
        }
        /* */
        if(typeof value != 'undefined' && value == result)
            return true;
        return result;
};
// Uncomment the lines below to test the script
// alert( checkForPathInObject(obj,'a.b.c') ); // Results "Hello World"
// alert( checkForPathInObject(obj,'a.success') ); // Returns false
// alert( checkForPathInObject(obj,'resTest', 'potato') ); // Returns true

您可以使用递归函数来实现这一点。即使您不知道所有嵌套的对象键名称，也可以使用此方法。

function FetchKeys(obj) {
    let objKeys = [];
    let keyValues = Object.entries(obj);
    for (let i in keyValues) {
        objKeys.push(keyValues[i][0]);
        if (typeof keyValues[i][1] == "object") {
            var keys = FetchKeys(keyValues[i][1])
            objKeys = objKeys.concat(keys);
        }
    }
    return objKeys;
}

let test = { level1: { level2: { level3: "level3" } } };
let keyToCheck = "level2";
let keys = FetchKeys(test); //Will return an array of Keys

if (keys.indexOf(keyToCheck) != -1) {
    //Key Exists logic;
}
else {
    //Key Not Found logic;
}

我也遇到了同样的问题，我想看看是否能找到自己的解决方案。这接受要检查的路径作为字符串。

function checkPathForTruthy(obj, path) {
  if (/\[[a-zA-Z_]/.test(path)) {
    console.log("Cannot resolve variables in property accessors");
    return false;
  }

  path = path.replace(/\[/g, ".");
  path = path.replace(/]|'|"/g, "");
  path = path.split(".");

  var steps = 0;
  var lastRef = obj;
  var exists = path.every(key => {
    var currentItem = lastRef[path[steps]];
    if (currentItem) {
      lastRef = currentItem;
      steps++;
      return true;
    } else {
      return false;
    }
  });

  return exists;
}

下面是一些日志记录和测试用例的片段：

console.clear（）；var测试案例=[[“data.Messages[0].Code”，true]，[“data.Messages[1].Code”，true]，[“data.Messages[0]['Code']”，true]，['data.Messages[0][“Code”]'，true]，[“data[Messages][0]['Code']”，错误]，[“data['Messages'][0]['Code']”，真]];var path=“data.Messages[0].Code”；变量obj={数据：{消息：[{代码：“0”}, {代码：“1”}]}}函数checkPathForTruthy（obj，路径）{if（/\[[a-zA-Z_]/.test（路径））{console.log（“无法解析属性访问器中的变量”）；return false；}path=路径替换（/\[/g，“.”）；path=路径替换（/]|'|“/g，”“）；path=路径拆分（“.”）；var步数=0；var lastRef=obj；var logOutput=[]；var exists=path.every（key=>{var currentItem=lastRef[path[steps]]；if（currentItem）{logOutput.push（currentItem）；lastRef=当前项；步骤++；返回true；}其他{return false；}});console.log（存在，logOutput）；返回存在；}testCase.forEach（testCase=>{如果（checkPathForTruthy（obj，testCase[0]）==testCase[1]）{console.log（“通过：”+testCase[0]）；}其他{console.log（“失败：”+testCase[0]+“预期”+testCase[1]）；}});

您可以尝试可选链接（但要注意浏览器兼容性）。

let test = {level1: {level2: {level3: 'level3'}}};

let level3 = test?.level1?.level2?.level3;
console.log(level3); // level3

level3 = test?.level0?.level1?.level2?.level3;
console.log(level3); // undefined

有一个babel插件（@babel/plugin建议可选链接）用于光学通道。所以，如果需要，请升级您的babel。

另一个选项（接近这个答案）：

function resolve(root, path){
    try {
        return (new Function(
            'root', 'return root.' + path + ';'
        ))(root);
    } catch (e) {}
}

var tree = { level1: [{ key: 'value' }] };
resolve(tree, 'level1[0].key'); // "value"
resolve(tree, 'level1[1].key'); // undefined

更多信息：https://stackoverflow.com/a/18381564/1636522

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 );
}