我有两个对象:oldObj和newObj。

oldObj中的数据用于填充表单,而newObj是用户更改该表单中的数据并提交的结果。

这两个物体都很深。它们具有对象或对象数组等属性-它们可以有n层深,因此diff算法需要递归。

现在我不仅需要弄清楚从oldObj到newObj更改了什么(如添加/更新/删除),而且还需要知道如何最好地表示它。

到目前为止,我的想法只是构建一个通用的deepdiffbetweenobjects方法,该方法将返回窗体上的对象{add:{…},upd:{…},del:{…但我转念一想:以前一定有人需要它。

所以…有没有人知道一个库或一段代码可以做到这一点,并且可能有更好的方式来表示差异(以一种仍然是JSON可序列化的方式)?

更新:

我想到了一个更好的方法来表示更新的数据,通过使用与newObj相同的对象结构,但将所有属性值转换为窗体上的对象:

{type: '<update|create|delete>', data: <propertyValue>}

如果newObj。prop1 = 'new value'和oldObj。prop1 = 'old value'它将设置returnbj。Prop1 = {type: 'update', data: 'new value'}

更新2:

当我们处理数组的属性时,它会变得非常麻烦,因为数组[1,2,3]应该被计算为等于[2,3,1],这对于基于值的类型的数组(如string, int和bool)来说足够简单,但是当涉及到引用类型的数组(如对象和数组)时就很难处理了。

数组的示例应该是相等的:

[1,[{c: 1},2,3],{a:'hey'}] and [{a:'hey'},1,[3,{c: 1},2]]

不仅要检查这种类型的深度值相等相当复杂,而且要找出一种表示可能发生的变化的好方法。


当前回答

这将把[1,2,3]和[3,2,1]视为相等(deep object) 因为我需要把它们之间的区别形象化:

[
  {
    "a":1,
    "b":1
  },
  {
    "a":1,
    "b":1
  }
]

and

[
  {
    "a":1,
    "b":1
  },
  {
    "a":"OH NO",
    "b":"an insertion"
  },
  {
    "a":1,
    "b":1
  }
]

所以我想看看它们的碰撞,这是剩下的:

[]
and
[
  {
    "a":"OH NO",
    "b":"an insertion"
  }
]

在我看来,这是最好的表现方式。 {添加:{…},乌利希期刊指南:{…},德尔:{…}}很难读懂


我提供了2个函数:objectcollision (obj1,obj2)和arraycollision (arr1,arr2)

console.log(ArrayCollide([1,2,3],[3,2,1]))
// false
//everything collided -> false
console.log(ArrayCollide([1],[2,1]))
// [ [], [ 2 ] ]
//1 and 1 collided, even if they are on different indices

//array of objects
const arr1 = 
[
  {
    "a":1,
    "b":1
  },
  {
    "a":1,
    "b":1
  }
]
const arr2 = 
[
  {
    "a":1,
    "b":1
  },
  {
    "a":"OH NO",
    "b":"an insertion"
  },
  {
    "a":1,
    "b":1
  }
]
const newArrays = ArrayCollide(arr1, arr2)
console.log(newArrays[0])
console.log(newArrays[1])
console.log('\n')
// []
// [ { a: 'OH NO', b: 'an insertion' } ]
// everything collided until this is left

//ObjectCollide
const obj1 = { a: '111', c: { q: 'no', a: '333' } }
const obj2 = { a: '111', p: 'ok', c: { a: '333' } }
ObjectCollide(obj1, obj2) //in place
console.log(obj1)
console.log(obj2)
console.log('\n')
// { c: { q: 'no' } }
// { p: 'ok', c: {} }
// obj["a"] collided and obj["c"]["a"] collided

//testing empty array
const a1 = { a: [] }
const a2 = { a: [], b: '2' }
ObjectCollide(a1, a2) //in place
console.log(a1)
console.log(a2)
console.log('\n')
// {}
// { b: '2' }
// obj["a"] collided

//DIFFERENT TYPES
const b1 = {a:true}
const b2 = {a:[1,2]}
ObjectCollide(b1,b2) //in place
console.log(b1)
console.log(b2)
// { a: true }
// { a: [ 1, 2 ] }

function ObjectCollide(obj1, obj2) {
  //in place, returns true if same

  // delete same
  const keys = Object.keys(obj1)
  const len = keys.length
  let howManyDeleted = 0
  for (let i = 0; i < len; i++) {
    const key = keys[i]

    const type1 = Array.isArray(obj1[key]) === true ? 'array' : typeof obj1[key]
    const type2 = Array.isArray(obj2[key]) === true ? 'array' : typeof obj2[key]
    if (type1!==type2) {
      continue
    }
    switch (type1) {
      case 'object':
        if (ObjectCollide(obj1[key], obj2[key])) {
          delete obj1[key]
          delete obj2[key]
          howManyDeleted++
        }
        continue
      case 'array':
        const newArrays = ArrayCollide(obj1[key], obj2[key])
        if (newArrays) {
          obj1[key] = newArrays[0]
          obj2[key] = newArrays[1]
        } else {
          delete obj1[key]
          delete obj2[key]
          howManyDeleted++
        }
        continue
      default:
        //string, number, I hope it covers everything else
        if (obj1[key] === obj2[key]) {
          delete obj1[key]
          delete obj2[key]
          howManyDeleted++
        }
    }
  }


  if (howManyDeleted === len && Object.keys(obj2).length === 0) {
    // return 'delete the stuff'
    // same. we've deleted everything!
    return true
  }

}
function ArrayCollide(arr1, arr2) {
  // returns [newArr1, newArr2] or false if same arrays (ignore order)
  const stringifyObj = {}

  const newArr1 = []
  const newArr2 = []
  for (let i = 0, len = arr1.length; i < len; i++) {
    const value = arr1[i]
    const stringified = JSON.stringify(value)
    stringifyObj[stringified]
    // arr = [count, ...]
    const arr = stringifyObj[stringified] || (stringifyObj[stringified] = [0])
    arr[0]++
    arr.push(value)
  }
  //in 2 but not in 1
  for (let i = 0, len = arr2.length; i < len; i++) {
    const value = arr2[i]
    const stringified = JSON.stringify(value)
    const arr = stringifyObj[stringified]
    if (arr === undefined) {
      newArr2.push(value)
    } else {
      if (arr[0] === 0) {
        newArr2.push(value)
      } else {
        arr[0]--
      }
    }
  }
  //in 1 but not in 2
  stringifyKeys = Object.keys(stringifyObj)
  for (let i = 0, len = stringifyKeys.length; i < len; i++) {
    const arr = stringifyObj[stringifyKeys[i]]

    for (let i = 1, len = arr[0] + 1; i < len; i++) {
      newArr1.push(arr[i])
    }
  }
  if (newArr1.length || newArr2.length) {
    return [newArr1, newArr2]
  } else {
    return false
  }

}

我想解决的问题是:

JSON文件不断重新排序,我想恢复JSON,如果它是等效的:像{a:1,b:2}和{b:2,a:1} 但是因为我不相信我的代码(我犯了一个错误),我想看到diff并自己检查它,我可以Ctrl+F到原始文件使用这个diff。

其他回答

这将把[1,2,3]和[3,2,1]视为相等(deep object) 因为我需要把它们之间的区别形象化:

[
  {
    "a":1,
    "b":1
  },
  {
    "a":1,
    "b":1
  }
]

and

[
  {
    "a":1,
    "b":1
  },
  {
    "a":"OH NO",
    "b":"an insertion"
  },
  {
    "a":1,
    "b":1
  }
]

所以我想看看它们的碰撞,这是剩下的:

[]
and
[
  {
    "a":"OH NO",
    "b":"an insertion"
  }
]

在我看来,这是最好的表现方式。 {添加:{…},乌利希期刊指南:{…},德尔:{…}}很难读懂


我提供了2个函数:objectcollision (obj1,obj2)和arraycollision (arr1,arr2)

console.log(ArrayCollide([1,2,3],[3,2,1]))
// false
//everything collided -> false
console.log(ArrayCollide([1],[2,1]))
// [ [], [ 2 ] ]
//1 and 1 collided, even if they are on different indices

//array of objects
const arr1 = 
[
  {
    "a":1,
    "b":1
  },
  {
    "a":1,
    "b":1
  }
]
const arr2 = 
[
  {
    "a":1,
    "b":1
  },
  {
    "a":"OH NO",
    "b":"an insertion"
  },
  {
    "a":1,
    "b":1
  }
]
const newArrays = ArrayCollide(arr1, arr2)
console.log(newArrays[0])
console.log(newArrays[1])
console.log('\n')
// []
// [ { a: 'OH NO', b: 'an insertion' } ]
// everything collided until this is left

//ObjectCollide
const obj1 = { a: '111', c: { q: 'no', a: '333' } }
const obj2 = { a: '111', p: 'ok', c: { a: '333' } }
ObjectCollide(obj1, obj2) //in place
console.log(obj1)
console.log(obj2)
console.log('\n')
// { c: { q: 'no' } }
// { p: 'ok', c: {} }
// obj["a"] collided and obj["c"]["a"] collided

//testing empty array
const a1 = { a: [] }
const a2 = { a: [], b: '2' }
ObjectCollide(a1, a2) //in place
console.log(a1)
console.log(a2)
console.log('\n')
// {}
// { b: '2' }
// obj["a"] collided

//DIFFERENT TYPES
const b1 = {a:true}
const b2 = {a:[1,2]}
ObjectCollide(b1,b2) //in place
console.log(b1)
console.log(b2)
// { a: true }
// { a: [ 1, 2 ] }

function ObjectCollide(obj1, obj2) {
  //in place, returns true if same

  // delete same
  const keys = Object.keys(obj1)
  const len = keys.length
  let howManyDeleted = 0
  for (let i = 0; i < len; i++) {
    const key = keys[i]

    const type1 = Array.isArray(obj1[key]) === true ? 'array' : typeof obj1[key]
    const type2 = Array.isArray(obj2[key]) === true ? 'array' : typeof obj2[key]
    if (type1!==type2) {
      continue
    }
    switch (type1) {
      case 'object':
        if (ObjectCollide(obj1[key], obj2[key])) {
          delete obj1[key]
          delete obj2[key]
          howManyDeleted++
        }
        continue
      case 'array':
        const newArrays = ArrayCollide(obj1[key], obj2[key])
        if (newArrays) {
          obj1[key] = newArrays[0]
          obj2[key] = newArrays[1]
        } else {
          delete obj1[key]
          delete obj2[key]
          howManyDeleted++
        }
        continue
      default:
        //string, number, I hope it covers everything else
        if (obj1[key] === obj2[key]) {
          delete obj1[key]
          delete obj2[key]
          howManyDeleted++
        }
    }
  }


  if (howManyDeleted === len && Object.keys(obj2).length === 0) {
    // return 'delete the stuff'
    // same. we've deleted everything!
    return true
  }

}
function ArrayCollide(arr1, arr2) {
  // returns [newArr1, newArr2] or false if same arrays (ignore order)
  const stringifyObj = {}

  const newArr1 = []
  const newArr2 = []
  for (let i = 0, len = arr1.length; i < len; i++) {
    const value = arr1[i]
    const stringified = JSON.stringify(value)
    stringifyObj[stringified]
    // arr = [count, ...]
    const arr = stringifyObj[stringified] || (stringifyObj[stringified] = [0])
    arr[0]++
    arr.push(value)
  }
  //in 2 but not in 1
  for (let i = 0, len = arr2.length; i < len; i++) {
    const value = arr2[i]
    const stringified = JSON.stringify(value)
    const arr = stringifyObj[stringified]
    if (arr === undefined) {
      newArr2.push(value)
    } else {
      if (arr[0] === 0) {
        newArr2.push(value)
      } else {
        arr[0]--
      }
    }
  }
  //in 1 but not in 2
  stringifyKeys = Object.keys(stringifyObj)
  for (let i = 0, len = stringifyKeys.length; i < len; i++) {
    const arr = stringifyObj[stringifyKeys[i]]

    for (let i = 1, len = arr[0] + 1; i < len; i++) {
      newArr1.push(arr[i])
    }
  }
  if (newArr1.length || newArr2.length) {
    return [newArr1, newArr2]
  } else {
    return false
  }

}

我想解决的问题是:

JSON文件不断重新排序,我想恢复JSON,如果它是等效的:像{a:1,b:2}和{b:2,a:1} 但是因为我不相信我的代码(我犯了一个错误),我想看到diff并自己检查它,我可以Ctrl+F到原始文件使用这个diff。

我已经使用这段代码来完成你所描述的任务:

function mergeRecursive(obj1, obj2) {
    for (var p in obj2) {
        try {
            if(obj2[p].constructor == Object) {
                obj1[p] = mergeRecursive(obj1[p], obj2[p]);
            }
            // Property in destination object set; update its value.
            else if (Ext.isArray(obj2[p])) {
                // obj1[p] = [];
                if (obj2[p].length < 1) {
                    obj1[p] = obj2[p];
                }
                else {
                    obj1[p] = mergeRecursive(obj1[p], obj2[p]);
                }

            }else{
                obj1[p] = obj2[p];
            }
        } catch (e) {
            // Property in destination object not set; create it and set its value.
            obj1[p] = obj2[p];
        }
    }
    return obj1;
}

这将为您提供一个新对象,该对象将合并窗体中的旧对象和新对象之间的所有更改

我已经为我的一个项目写了一个函数,它将对象作为用户选项与其内部克隆进行比较。 它还可以验证,甚至替换默认值,如果用户输入坏类型的数据或删除,在纯javascript。

在IE8中100%有效。测试成功。

//  ObjectKey: ["DataType, DefaultValue"]
reference = { 
    a : ["string", 'Defaul value for "a"'],
    b : ["number", 300],
    c : ["boolean", true],
    d : {
        da : ["boolean", true],
        db : ["string", 'Defaul value for "db"'],
        dc : {
            dca : ["number", 200],
            dcb : ["string", 'Default value for "dcb"'],
            dcc : ["number", 500],
            dcd : ["boolean", true]
      },
      dce : ["string", 'Default value for "dce"'],
    },
    e : ["number", 200],
    f : ["boolean", 0],
    g : ["", 'This is an internal extra parameter']
};

userOptions = { 
    a : 999, //Only string allowed
  //b : ["number", 400], //User missed this parameter
    c: "Hi", //Only lower case or case insitive in quotes true/false allowed.
    d : {
        da : false,
        db : "HelloWorld",
        dc : {
            dca : 10,
            dcb : "My String", //Space is not allowed for ID attr
            dcc: "3thString", //Should not start with numbers
            dcd : false
      },
      dce: "ANOTHER STRING",
    },
    e: 40,
    f: true,
};


function compare(ref, obj) {

    var validation = {
        number: function (defaultValue, userValue) {
          if(/^[0-9]+$/.test(userValue))
            return userValue;
          else return defaultValue;
        },
        string: function (defaultValue, userValue) {
          if(/^[a-z][a-z0-9-_.:]{1,51}[^-_.:]$/i.test(userValue)) //This Regex is validating HTML tag "ID" attributes
            return userValue;
          else return defaultValue;
        },
        boolean: function (defaultValue, userValue) {
          if (typeof userValue === 'boolean')
            return userValue;
          else return defaultValue;
        }
    };

    for (var key in ref)
        if (obj[key] && obj[key].constructor && obj[key].constructor === Object)
          ref[key] = compare(ref[key], obj[key]);
        else if(obj.hasOwnProperty(key))
          ref[key] = validation[ref[key][0]](ref[key][1], obj[key]); //or without validation on user enties => ref[key] = obj[key]
        else ref[key] = ref[key][1];
    return ref;
}

//console.log(
    alert(JSON.stringify( compare(reference, userOptions),null,2 ))
//);

/ *结果

{
  "a": "Defaul value for \"a\"",
  "b": 300,
  "c": true,
  "d": {
    "da": false,
    "db": "Defaul value for \"db\"",
    "dc": {
      "dca": 10,
      "dcb": "Default value for \"dcb\"",
      "dcc": 500,
      "dcd": false
    },
    "dce": "Default value for \"dce\""
  },
  "e": 40,
  "f": true,
  "g": "This is an internal extra parameter"
}

*/

我只是使用ramda,为了解决同样的问题,我需要知道在新对象中有什么改变。这是我的设计。

const oldState = {id:'170',name:'Ivab',secondName:'Ivanov',weight:45};
const newState = {id:'170',name:'Ivanko',secondName:'Ivanov',age:29};

const keysObj1 = R.keys(newState)

const filterFunc = key => {
  const value = R.eqProps(key,oldState,newState)
  return {[key]:value}
}

const result = R.map(filterFunc, keysObj1)

结果是属性的名称和状态。

[{"id":true}, {"name":false}, {"secondName":true}, {"age":false}]

我为我自己的用例(es5环境)编写了这个,认为这可能对某些人有用,所以它是:

function deepCompare(obj1, obj2) {
    var diffObj = Array.isArray(obj2) ? [] : {}
    Object.getOwnPropertyNames(obj2).forEach(function(prop) {
        if (typeof obj2[prop] === 'object') {
            diffObj[prop] = deepCompare(obj1[prop], obj2[prop])
            // if it's an array with only length property => empty array => delete
            // or if it's an object with no own properties => delete
            if (Array.isArray(diffObj[prop]) && Object.getOwnPropertyNames(diffObj[prop]).length === 1 || Object.getOwnPropertyNames(diffObj[prop]).length === 0) {
                delete diffObj[prop]
            }
        } else if(obj1[prop] !== obj2[prop]) {
            diffObj[prop] = obj2[prop]
        }
    });
    return diffObj
}

这可能不是很有效,但是会根据第二个Obj输出一个只有不同道具的对象。