javascript - 检测和修复 JavaScript 中的循环引用

Question

鉴于我在一个大型 JavaScript 对象中有一个循环引用

我尝试JSON.stringify(problematicObject)

浏览器抛出

“TypeError：将循环结构转换为 JSON”

（这是预期的）

那我想找这个循环引用的原因，最好用Chrome开发者工具？这可能吗？如何在大对象中查找和修复循环引用？

Answer 1

从http://blog.vjeux.com/2011/javascript/cyclic-object-detection.html中提取。添加了一行来检测循环在哪里。将其粘贴到 Chrome 开发工具中：

function isCyclic (obj) {
  var seenObjects = [];

  function detect (obj) {
    if (obj && typeof obj === 'object') {
      if (seenObjects.indexOf(obj) !== -1) {
        return true;
      }
      seenObjects.push(obj);
      for (var key in obj) {
        if (obj.hasOwnProperty(key) && detect(obj[key])) {
          console.log(obj, 'cycle at ' + key);
          return true;
        }
      }
    }
    return false;
  }

  return detect(obj);
}

这是测试：

> a = {}
> b = {}
> a.b = b; b.a = a;
> isCyclic(a)
  Object {a: Object}
   "cycle at a"
  Object {b: Object}
   "cycle at b"
  true

Answer 2

当我发现这个问题时，@tmack 的答案绝对是我想要的！

不幸的是，它会返回许多误报——如果在 JSON 中复制了一个对象，它会返回 true，这与循环性不同。圆形意味着一个对象是它自己的孩子，例如

obj.key1.key2.[...].keyX === obj

我修改了原始答案，这对我有用：

function isCyclic(obj) {
  var keys = [];
  var stack = [];
  var stackSet = new Set();
  var detected = false;

  function detect(obj, key) {
    if (obj && typeof obj != 'object') { return; }

    if (stackSet.has(obj)) { // it's cyclic! Print the object and its locations.
      var oldindex = stack.indexOf(obj);
      var l1 = keys.join('.') + '.' + key;
      var l2 = keys.slice(0, oldindex + 1).join('.');
      console.log('CIRCULAR: ' + l1 + ' = ' + l2 + ' = ' + obj);
      console.log(obj);
      detected = true;
      return;
    }

    keys.push(key);
    stack.push(obj);
    stackSet.add(obj);
    for (var k in obj) { //dive on the object's children
      if (Object.prototype.hasOwnProperty.call(obj, k)) { detect(obj[k], k); }
    }

    keys.pop();
    stack.pop();
    stackSet.delete(obj);
    return;
  }

  detect(obj, 'obj');
  return detected;
}

以下是一些非常简单的测试：

var root = {}
var leaf = {'isleaf':true};
var cycle2 = {l:leaf};
var cycle1 = {c2: cycle2, l:leaf};
cycle2.c1 = cycle1
root.leaf = leaf

isCyclic(cycle1); // returns true, logs "CIRCULAR: obj.c2.c1 = obj"
isCyclic(cycle2); // returns true, logs "CIRCULAR: obj.c1.c2 = obj"
isCyclic(leaf); // returns false
isCyclic(root); // returns false

Answer 3

JSON.stringify()以下是 MDN在循环对象上使用时检测和修复循环引用的方法： https ://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Errors/Cyclic_object_value ：

在如下的圆形结构中

var circularReference = {otherData: 123};
circularReference.myself = circularReference;

JSON.stringify()将失败：

JSON.stringify(circularReference);
// TypeError: cyclic object value

要序列化循环引用，您可以使用支持它们的库（例如cycle.js）或自己实现解决方案，这需要通过可序列化的值查找和替换（或删除）循环引用。

下面的代码片段说明了如何使用以下的 replacer 参数查找和过滤（从而导致数据丢失）循环引用：JSON.stringify()

const getCircularReplacer = () => {
      const seen = new WeakSet();
      return (key, value) => {
        if (typeof value === "object" && value !== null) {
          if (seen.has(value)) {
            return;
          }
          seen.add(value);
        }
        return value;
      };
    };

JSON.stringify(circularReference, getCircularReplacer());
// {"otherData":123}

Answer 4

您也可以使用JSON.stringifytry /catch

function hasCircularDependency(obj)
{
    try
    {
        JSON.stringify(obj);
    }
    catch(e)
    {
        return e.includes("Converting circular structure to JSON"); 
    }
    return false;
}

演示

function hasCircularDependency(obj) {
  try {
    JSON.stringify(obj);
  } catch (e) {
    return String(e).includes("Converting circular structure to JSON");
  }
  return false;
}

var a = {b:{c:{d:""}}};
console.log(hasCircularDependency(a));
a.b.c.d = a;
console.log(hasCircularDependency(a));

Answer 5

循环参考检测器

这是我的CircularReferenceDetector类，它输出循环引用值实际所在的所有属性堆栈信息，并显示罪魁祸首引用的位置。

这对于巨大的结构特别有用，在这些结构中，哪个值是伤害的来源不是很明显。

它输出字符串化的循环引用值，但所有对自身的引用都替换为“[Circular object --- fix me]”。

用法：
CircularReferenceDetector.detectCircularReferences(value);

注意： 如果您不想使用任何日志记录或没有可用的记录器，请删除 Logger.* 语句。

技术说明：
递归函数遍历对象的所有属性并测试 JSON.stringify 是否成功。如果它不成功（循环引用），那么它通过用一些常量字符串替换 value 本身来测试它是否成功。这意味着如果它成功使用这个替换器，这个值就是循环引用的值。如果不是，它将递归地遍历该对象的所有属性。

同时它还跟踪属性堆栈，为您提供罪魁祸首值所在的信息。

打字稿

import {Logger} from "../Logger";

export class CircularReferenceDetector {

    static detectCircularReferences(toBeStringifiedValue: any, serializationKeyStack: string[] = []) {
        Object.keys(toBeStringifiedValue).forEach(key => {
            var value = toBeStringifiedValue[key];

            var serializationKeyStackWithNewKey = serializationKeyStack.slice();
            serializationKeyStackWithNewKey.push(key);
            try {
                JSON.stringify(value);
                Logger.debug(`path "${Util.joinStrings(serializationKeyStack)}" is ok`);
            } catch (error) {
                Logger.debug(`path "${Util.joinStrings(serializationKeyStack)}" JSON.stringify results in error: ${error}`);

                var isCircularValue:boolean;
                var circularExcludingStringifyResult:string = "";
                try {
                    circularExcludingStringifyResult = JSON.stringify(value, CircularReferenceDetector.replaceRootStringifyReplacer(value), 2);
                    isCircularValue = true;
                } catch (error) {
                    Logger.debug(`path "${Util.joinStrings(serializationKeyStack)}" is not the circular source`);
                    CircularReferenceDetector.detectCircularReferences(value, serializationKeyStackWithNewKey);
                    isCircularValue = false;
                }
                if (isCircularValue) {
                    throw new Error(`Circular reference detected:\nCircularly referenced value is value under path "${Util.joinStrings(serializationKeyStackWithNewKey)}" of the given root object\n`+
                        `Calling stringify on this value but replacing itself with [Circular object --- fix me] ( <-- search for this string) results in:\n${circularExcludingStringifyResult}\n`);
                }
            }
        });
    }

    private static replaceRootStringifyReplacer(toBeStringifiedValue: any): any {
        var serializedObjectCounter = 0;

        return function (key: any, value: any) {
            if (serializedObjectCounter !== 0 && typeof(toBeStringifiedValue) === 'object' && toBeStringifiedValue === value) {
                Logger.error(`object serialization with key ${key} has circular reference to being stringified object`);
                return '[Circular object --- fix me]';
            }

            serializedObjectCounter++;

            return value;
        }
    }
}

export class Util {

    static joinStrings(arr: string[], separator: string = ":") {
        if (arr.length === 0) return "";
        return arr.reduce((v1, v2) => `${v1}${separator}${v2}`);
    }

}

从 TypeScript 编译的 JavaScript

"use strict";
const Logger_1 = require("../Logger");
class CircularReferenceDetector {
    static detectCircularReferences(toBeStringifiedValue, serializationKeyStack = []) {
        Object.keys(toBeStringifiedValue).forEach(key => {
            var value = toBeStringifiedValue[key];
            var serializationKeyStackWithNewKey = serializationKeyStack.slice();
            serializationKeyStackWithNewKey.push(key);
            try {
                JSON.stringify(value);
                Logger_1.Logger.debug(`path "${Util.joinStrings(serializationKeyStack)}" is ok`);
            }
            catch (error) {
                Logger_1.Logger.debug(`path "${Util.joinStrings(serializationKeyStack)}" JSON.stringify results in error: ${error}`);
                var isCircularValue;
                var circularExcludingStringifyResult = "";
                try {
                    circularExcludingStringifyResult = JSON.stringify(value, CircularReferenceDetector.replaceRootStringifyReplacer(value), 2);
                    isCircularValue = true;
                }
                catch (error) {
                    Logger_1.Logger.debug(`path "${Util.joinStrings(serializationKeyStack)}" is not the circular source`);
                    CircularReferenceDetector.detectCircularReferences(value, serializationKeyStackWithNewKey);
                    isCircularValue = false;
                }
                if (isCircularValue) {
                    throw new Error(`Circular reference detected:\nCircularly referenced value is value under path "${Util.joinStrings(serializationKeyStackWithNewKey)}" of the given root object\n` +
                        `Calling stringify on this value but replacing itself with [Circular object --- fix me] ( <-- search for this string) results in:\n${circularExcludingStringifyResult}\n`);
                }
            }
        });
    }
    static replaceRootStringifyReplacer(toBeStringifiedValue) {
        var serializedObjectCounter = 0;
        return function (key, value) {
            if (serializedObjectCounter !== 0 && typeof (toBeStringifiedValue) === 'object' && toBeStringifiedValue === value) {
                Logger_1.Logger.error(`object serialization with key ${key} has circular reference to being stringified object`);
                return '[Circular object --- fix me]';
            }
            serializedObjectCounter++;
            return value;
        };
    }
}
exports.CircularReferenceDetector = CircularReferenceDetector;
class Util {
    static joinStrings(arr, separator = ":") {
        if (arr.length === 0)
            return "";
        return arr.reduce((v1, v2) => `${v1}${separator}${v2}`);
    }
}
exports.Util = Util;

Answer 6

这是@Trey Mack和@Freddie Nfbnm在typeof obj != 'object'条件下的答案的修复。相反，它应该测试该obj值是否不是对象的实例，以便它在检查熟悉对象的值时也可以工作（例如，函数和符号（符号不是对象的实例，但仍然可以解决，顺便说一句。））。

我将其发布为答案，因为我还不能在此 StackExchange 帐户中发表评论。

PS.：随时要求我删除此答案。

function isCyclic(obj) {
  var keys = [];
  var stack = [];
  var stackSet = new Set();
  var detected = false;

  function detect(obj, key) {
    if (!(obj instanceof Object)) { return; } // Now works with other
                                              // kinds of object.

    if (stackSet.has(obj)) { // it's cyclic! Print the object and its locations.
      var oldindex = stack.indexOf(obj);
      var l1 = keys.join('.') + '.' + key;
      var l2 = keys.slice(0, oldindex + 1).join('.');
      console.log('CIRCULAR: ' + l1 + ' = ' + l2 + ' = ' + obj);
      console.log(obj);
      detected = true;
      return;
    }

    keys.push(key);
    stack.push(obj);
    stackSet.add(obj);
    for (var k in obj) { //dive on the object's children
      if (obj.hasOwnProperty(k)) { detect(obj[k], k); }
    }

    keys.pop();
    stack.pop();
    stackSet.delete(obj);
    return;
  }

  detect(obj, 'obj');
  return detected;
}

Answer 7

这是由@Aaron V和@user4976005的答案混合而成的Node ES6 版本，它解决了调用hasOwnProperty 的问题：

const isCyclic = (obj => {
  const keys = []
  const stack = []
  const stackSet = new Set()
  let detected = false

  const detect = ((object, key) => {
    if (!(object instanceof Object))
      return

    if (stackSet.has(object)) { // it's cyclic! Print the object and its locations.
      const oldindex = stack.indexOf(object)
      const l1 = `${keys.join('.')}.${key}`
      const l2 = keys.slice(0, oldindex + 1).join('.')
      console.log(`CIRCULAR: ${l1} = ${l2} = ${object}`)
      console.log(object)
      detected = true
      return
    }

    keys.push(key)
    stack.push(object)
    stackSet.add(object)
    Object.keys(object).forEach(k => { // dive on the object's children
      if (k && Object.prototype.hasOwnProperty.call(object, k))
        detect(object[k], k)
    })

    keys.pop()
    stack.pop()
    stackSet.delete(object)
  })

  detect(obj, 'obj')
  return detected
})

Answer 8

这里有很多答案，但我想我会将我的解决方案添加到组合中。它类似于@Trey Mack的答案，但该解决方案需要 O(n^2)。此版本使用WeakMap而不是数组，将时间缩短到 O(n)。

function isCyclic(object) {
   const seenObjects = new WeakMap(); // use to keep track of which objects have been seen.

   function detectCycle(obj) {
      // If 'obj' is an actual object (i.e., has the form of '{}'), check
      // if it's been seen already.
      if (Object.prototype.toString.call(obj) == '[object Object]') {

         if (seenObjects.has(obj)) {
            return true;
         }

         // If 'obj' hasn't been seen, add it to 'seenObjects'.
         // Since 'obj' is used as a key, the value of 'seenObjects[obj]'
         // is irrelevent and can be set as literally anything you want. I 
         // just went with 'undefined'.
         seenObjects.set(obj, undefined);

         // Recurse through the object, looking for more circular references.
         for (var key in obj) {
            if (detectCycle(obj[key])) {
               return true;
            }
         }

      // If 'obj' is an array, check if any of it's elements are
      // an object that has been seen already.
      } else if (Array.isArray(obj)) {
         for (var i in obj) {
            if (detectCycle(obj[i])) {
               return true;
            }
         }
      }

      return false;
   }

   return detectCycle(object);
}

这就是它在行动中的样子。

> var foo = {grault: {}};
> detectCycle(foo);
false
> foo.grault = foo;
> detectCycle(foo);
true
> var bar = {};
> detectCycle(bar);
false
> bar.plugh = [];
> bar.plugh.push(bar);
> detectCycle(bar);
true

Answer 9

您还可以使用符号 - 由于这种方法，您不必改变原始对象的属性，除了添加符号来标记访问节点。

它更干净，应该比收集节点属性并与对象进行比较更快。如果您不想序列化大的嵌套值，它还具有可选的深度限制：

// Symbol used to mark already visited nodes - helps with circular dependencies
const visitedMark = Symbol('VISITED_MARK');

const MAX_CLEANUP_DEPTH = 10;

function removeCirculars(obj, depth = 0) {
  if (!obj) {
    return obj;
  }

  // Skip condition - either object is falsy, was visited or we go too deep
  const shouldSkip = !obj || obj[visitedMark] || depth > MAX_CLEANUP_DEPTH;

  // Copy object (we copy properties from it and mark visited nodes)
  const originalObj = obj;
  let result = {};

  Object.keys(originalObj).forEach((entry) => {
    const val = originalObj[entry];

    if (!shouldSkip) {
      if (typeof val === 'object') { // Value is an object - run object sanitizer
        originalObj[visitedMark] = true; // Mark current node as "seen" - will stop from going deeper into circulars
        const nextDepth = depth + 1;
        result[entry] = removeCirculars(val, nextDepth);
      } else {
        result[entry] = val;
      }
    } else {
      result = 'CIRCULAR';
    }
  });

  return result;
}

这将导致对象的所有循环依赖项都被剥离，并且不会比给定的更深MAX_CLEANUP_DEPTH。

只要您不对对象执行任何元编程操作，使用符号是安全的 - 它们是透明的并且不可枚举，因此 - 它们不会显示在对象的任何标准操作中。

此外，如果您需要对其执行任何其他操作，则返回一个新的、已清理的对象具有不会改变原始对象的优点。

如果您不想CIRCULAR标记，您可以稍微修改代码，因此在实际执行操作之前跳过对象（在循环内）：

 originalObj[visitedMark] = true; // Mark current node as "seen" - will stop from going deeper into circulars
 const val = originalObj[entry];

 // Skip condition - either object is falsy, was visited or we go too deep
 const shouldSkip = val[visitedMark] || depth > MAX_SANITIZATION_DEPTH;

 if (!shouldSkip) {
   if (typeof val === 'object') { // Value is an object - run object sanitizer
    const nextDepth = depth + 1;
    result[entry] = removeCirculars(val, nextDepth);
  } else {
    result[entry] = val;
  }
 }

Answer 10

我刚做了这个。它可能很脏，但无论如何都可以使用...：P

function dump(orig){
  var inspectedObjects = [];
  console.log('== DUMP ==');
  (function _dump(o,t){
    console.log(t+' Type '+(typeof o));
    for(var i in o){
      if(o[i] === orig){
        console.log(t+' '+i+': [recursive]'); 
        continue;
      }
      var ind = 1+inspectedObjects.indexOf(o[i]);
      if(ind>0) console.log(t+' '+i+':  [already inspected ('+ind+')]');
      else{
        console.log(t+' '+i+': ('+inspectedObjects.push(o[i])+')');
        _dump(o[i],t+'>>');
      }
    }
  }(orig,'>'));
}

然后

var a = [1,2,3], b = [a,4,5,6], c = {'x':a,'y':b};
a.push(c); dump(c);

说

== DUMP ==
> Type object
> x: (1)
>>> Type object
>>> 0: (2)
>>>>> Type number
>>> 1: (3)
>>>>> Type number
>>> 2: (4)
>>>>> Type number
>>> 3: [recursive]
> y: (5)
>>> Type object
>>> 0:  [already inspected (1)]
>>> 1: (6)
>>>>> Type number
>>> 2: (7)
>>>>> Type number
>>> 3: (8)
>>>>> Type number

这表明 cx[3] 等于 c，并且 cx = cy[0]。

或者，对这个函数进行一点编辑可以告诉你你需要什么......

function findRecursive(orig){
  var inspectedObjects = [];
  (function _find(o,s){
    for(var i in o){
      if(o[i] === orig){
        console.log('Found: obj.'+s.join('.')+'.'+i); 
        return;
      }
      if(inspectedObjects.indexOf(o[i])>=0) continue;
      else{
        inspectedObjects.push(o[i]);
        s.push(i); _find(o[i],s); s.pop(i);
      }
    }
  }(orig,[]));
}

Answer 11

我将 Freddie Nfbnm 的答案转换为 TypeScript：

export class JsonUtil {

    static isCyclic(json) {
        const keys = [];
        const stack = [];
        const stackSet = new Set();
        let detected = false;

        function detect(obj, key) {
            if (typeof obj !== 'object') {
                return;
            }

            if (stackSet.has(obj)) { // it's cyclic! Print the object and its locations.
                const oldIndex = stack.indexOf(obj);
                const l1 = keys.join('.') + '.' + key;
                const l2 = keys.slice(0, oldIndex + 1).join('.');
                console.log('CIRCULAR: ' + l1 + ' = ' + l2 + ' = ' + obj);
                console.log(obj);
                detected = true;
                return;
            }

            keys.push(key);
            stack.push(obj);
            stackSet.add(obj);
            for (const k in obj) { // dive on the object's children
                if (obj.hasOwnProperty(k)) {
                    detect(obj[k], k);
                }
            }

            keys.pop();
            stack.pop();
            stackSet.delete(obj);
            return;
        }

        detect(json, 'obj');
        return detected;
    }

}

Answer 12

只是为了把我的版本混在一起......下面是@dkurzaj 代码的混音（它本身是@Aaron V、@user4976005、@Trey Mack 和最后@Freddie Nfbnm 的混音） s [删除？] 代码）加上@darksinge 的WeakMap想法。所以......我猜这个线程的Megamix :)

在我的版本中，可以选择将报告（而不是console.log'ed 条目）作为对象数组返回。如果不需要报告，则在第一次看到循环引用（a'la @darksinge 的代码）时停止测试。

此外，hasOwnProperty已被删除为Object.keys仅返回hasOwnProperty属性（请参阅：https ://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Object/keys ）。

function isCyclic(x, bReturnReport) {
    var a_sKeys = [],
        a_oStack = [],
        wm_oSeenObjects = new WeakMap(), //# see: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/WeakMap
        oReturnVal = {
            found: false,
            report: []
        }
    ;

    //# Setup the recursive logic to locate any circular references while kicking off the initial call
    (function doIsCyclic(oTarget, sKey) {
        var a_sTargetKeys, sCurrentKey, i;

        //# If we've seen this oTarget before, flip our .found to true
        if (wm_oSeenObjects.has(oTarget)) {
            oReturnVal.found = true;

            //# If we are to bReturnReport, add the entries into our .report
            if (bReturnReport) {
                oReturnVal.report.push({
                    instance: oTarget,
                    source: a_sKeys.slice(0, a_oStack.indexOf(oTarget) + 1).join('.'),
                    duplicate: a_sKeys.join('.') + "." + sKey
                });
            }
        }
        //# Else if oTarget is an instanceof Object, determine the a_sTargetKeys and .set our oTarget into the wm_oSeenObjects
        else if (oTarget instanceof Object) {
            a_sTargetKeys = Object.keys(oTarget);
            wm_oSeenObjects.set(oTarget /*, undefined*/);

            //# If we are to bReturnReport, .push the  current level's/call's items onto our stacks
            if (bReturnReport) {
                if (sKey) { a_sKeys.push(sKey) };
                a_oStack.push(oTarget);
            }

            //# Traverse the a_sTargetKeys, pulling each into sCurrentKey as we go
            //#     NOTE: If you want all properties, even non-enumerables, see Object.getOwnPropertyNames() so there is no need to call .hasOwnProperty (per: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Object/keys)
            for (i = 0; i < a_sTargetKeys.length; i++) {
                sCurrentKey = a_sTargetKeys[i];

                //# If we've already .found a circular reference and we're not bReturnReport, fall from the loop
                if (oReturnVal.found && !bReturnReport) {
                    break;
                }
                //# Else if the sCurrentKey is an instanceof Object, recurse to test
                else if (oTarget[sCurrentKey] instanceof Object) {
                    doIsCyclic(oTarget[sCurrentKey], sCurrentKey);
                }
            }

            //# .delete our oTarget into the wm_oSeenObjects
            wm_oSeenObjects.delete(oTarget);

            //# If we are to bReturnReport, .pop the current level's/call's items off our stacks
            if (bReturnReport) {
                if (sKey) { a_sKeys.pop() };
                a_oStack.pop();
            }
        }
    }(x, '')); //# doIsCyclic

    return (bReturnReport ? oReturnVal.report : oReturnVal.found);
}

Answer 13

这是适用于节点的@Thomas 答案：

const {logger} = require("../logger")
// Or: const logger = {debug: (...args) => console.log.call(console.log, args) }

const joinStrings = (arr, separator) => {
  if (arr.length === 0) return "";
  return arr.reduce((v1, v2) => `${v1}${separator}${v2}`);
}

exports.CircularReferenceDetector = class CircularReferenceDetector {

  detectCircularReferences(toBeStringifiedValue, serializationKeyStack = []) {
    Object.keys(toBeStringifiedValue).forEach(key => {
      let value = toBeStringifiedValue[key];

      let serializationKeyStackWithNewKey = serializationKeyStack.slice();
      serializationKeyStackWithNewKey.push(key);
      try {
        JSON.stringify(value);
        logger.debug(`path "${joinStrings(serializationKeyStack)}" is ok`);
      } catch (error) {
        logger.debug(`path "${joinStrings(serializationKeyStack)}" JSON.stringify results in error: ${error}`);

        let isCircularValue;
        let circularExcludingStringifyResult = "";
        try {
          circularExcludingStringifyResult = JSON.stringify(value, this.replaceRootStringifyReplacer(value), 2);
          isCircularValue = true;
        } catch (error) {
          logger.debug(`path "${joinStrings(serializationKeyStack)}" is not the circular source`);
          this.detectCircularReferences(value, serializationKeyStackWithNewKey);
          isCircularValue = false;
        }
        if (isCircularValue) {
          throw new Error(`Circular reference detected:\nCircularly referenced value is value under path "${joinStrings(serializationKeyStackWithNewKey)}" of the given root object\n`+
              `Calling stringify on this value but replacing itself with [Circular object --- fix me] ( <-- search for this string) results in:\n${circularExcludingStringifyResult}\n`);
        }
      }
    });
  }

  replaceRootStringifyReplacer(toBeStringifiedValue) {
    let serializedObjectCounter = 0;

    return function (key, value) {
      if (serializedObjectCounter !== 0 && typeof(toBeStringifiedValue) === 'object' && toBeStringifiedValue === value) {
        logger.error(`object serialization with key ${key} has circular reference to being stringified object`);
        return '[Circular object --- fix me]';
      }

      serializedObjectCounter++;

      return value;
    }
  }
}

Answer 14

大多数其他答案仅显示如何检测对象树具有循环引用 - 他们没有告诉您如何修复这些循环引用（即用例如替换循环引用值undefined）。

以下是我用来替换所有循环引用的函数undefined：

export const specialTypeHandlers_default = [
    // Set and Map are included by default, since JSON.stringify tries (and fails) to serialize them by default
    {type: Set, keys: a=>a.keys(), get: (a, key)=>key, delete: (a, key)=>a.delete(key)},
    {type: Map, keys: a=>a.keys(), get: (a, key)=>a.get(key), delete: (a, key)=>a.set(key, undefined)},
];
export function RemoveCircularLinks(node, specialTypeHandlers = specialTypeHandlers_default, nodeStack_set = new Set()) {
    nodeStack_set.add(node);

    const specialHandler = specialTypeHandlers.find(a=>node instanceof a.type);
    for (const key of specialHandler ? specialHandler.keys(node) : Object.keys(node)) {
        const value = specialHandler ? specialHandler.get(node, key) : node[key];
        // if the value is already part of visited-stack, delete the value (and don't tunnel into it)
        if (nodeStack_set.has(value)) {
            if (specialHandler) specialHandler.delete(node, key);
            else node[key] = undefined;
        }
        // else, tunnel into it, looking for circular-links at deeper levels
        else if (typeof value == "object" && value != null) {
            RemoveCircularLinks(value, specialTypeHandlers, nodeStack_set);
        }
    }

    nodeStack_set.delete(node);
}

具体而言JSON.stringify，只需在字符串化之前调用上面的函数（注意它确实会改变传入的对象）：

const objTree = {normalProp: true};
objTree.selfReference = objTree;
RemoveCircularLinks(objTree); // without this line, the JSON.stringify call errors
console.log(JSON.stringify(objTree));

Answer 15

如果您只需要查看该圆形对象的内容，只需使用 console.table(circularObj)

Answer 16

尝试console.log()在 chrome/firefox 浏览器上使用以确定问题所在。

在使用 Firebug 插件的 Firefox 上，您可以逐行调试您的 javascript。

更新：

请参阅下面的循环引用问题示例并已处理：-

// JSON.stringify, avoid TypeError: Converting circular structure to JSON
// Demo: Circular reference
var o = {};
o.o = o;

var cache = [];
JSON.stringify(o, function(key, value) {
    if (typeof value === 'object' && value !== null) {
        if (cache.indexOf(value) !== -1) {
            // Circular reference found, discard key
            alert("Circular reference found, discard key");
            return;
        }
        alert("value = '" + value + "'");
        // Store value in our collection
        cache.push(value);
    }
    return value;
});
cache = null; // Enable garbage collection

var a = {b:1};
var o = {};
o.one = a;
o.two = a;
// one and two point to the same object, but two is discarded:
JSON.stringify(o);

var obj = {
  a: "foo",
  b: obj
};

var replacement = {"b":undefined};

alert("Result : " + JSON.stringify(obj,replacement));

参考示例LIVE DEMO