Question

以下是正在使用的JSON：

origJSON

{
  "2014": {
    "grade": "A-",
    "className": "Geography 101"
  },
  "_id": "564d098e2e59e72412e4d795",
  "purl": "3",
  "firstName": "Jon",
  "finalGrade": "B",
  "template": "User.html",
  "sensitive": "{"ssn":\"123-00-6789\",\"acc\":\"987654300\",\"password\":\"zxcvbn!\"}"
}

newJSON

{
  "purl": "3",
  "firstName": "Jon",
  "sensitive": {
    "password": "qazwx",
    "phone": "1234567890"
  }
}

合并origJSON

{
  "2014": {
    "grade": "A-",
    "className": "Geography 101"
  },
  "_id": "564d098e2e59e72412e4d795",
  "purl": "3",
  "firstName": "Jon",
  "finalGrade": "B",
  "template": "User.html",
  "sensitive": {
    "password": "qazwx",
    "phone": "1234567890"
  }
}

所需的origJSON

{
  "2014": {
    "grade": "A-",
    "className": "Geography 101"
  },
  "_id": "564d098e2e59e72412e4d795",
  "purl": "3",
  "firstName": "Jon",
  "finalGrade": "B",
  "template": "User.html",
  "sensitive": {
    "ssn": "123-00-6789",
    "acc": "987654300",
    "password": "qazwx",
    "phone": "1234567890"
  }
}

以下是我的代码：

for(var key in newJSON) origJSON[key]=newJSON[key];

有关更新嵌套对象而不是替换嵌套对象的任何帮助吗？

Answer 1

使用lodash库的 _.merge() 方法来获得所需的结果。以下是可用方法的各种示例，但 _.merge() 是您要使用的方法：

   var origJSON = {
        "2014": {
            "grade": "A-",
            "className": "Geography 101"
        },
        "_id": "564d098e2e59e72412e4d795",
        "purl": "3",
        "firstName": "Jon",
        "finalGrade": "B",
        "template": "User.html",
        "sensitive": {"ssn":"123-00-6789","acc":"987654300","password":"zxcvbn!"}
    };

    var newJSON = {
        "purl": "3",
        "firstName": "Jon",
        "sensitive": {
            "password": "qazwx",
            "phone": "1234567890"
        }
    };

使用 _.assign() ：

    var assigned = _.clone(newJSON);
    _.assign(assigned, origJSON);
    console.log("assign:", JSON.stringify(assigned, null, 4));

<强>输出

assign: {
    "2014": {
        "grade": "A-",
        "className": "Geography 101"
    },
    "purl": "3",
    "firstName": "Jon",
    "sensitive": {
        "ssn": "123-00-6789",
        "acc": "987654300",
        "password": "zxcvbn!"
    },
    "_id": "564d098e2e59e72412e4d795",
    "finalGrade": "B",
    "template": "User.html"
}

使用 _.merge()

var merged = _.clone(newJSON);
_.merge(merged, origJSON);
console.log("merge:", JSON.stringify(merged, null, 4));

<强>输出

merge: {
        "2014": {
            "grade": "A-",
            "className": "Geography 101"
        },
        "purl": "3",
        "firstName": "Jon",
        "sensitive": {
            "password": "zxcvbn!",
            "phone": "1234567890",
            "ssn": "123-00-6789",
            "acc": "987654300"
        },
        "_id": "564d098e2e59e72412e4d795",
        "finalGrade": "B",
        "template": "User.html"
    }

使用 _.defaults()

var defaulted = _.clone(newJSON);
_.defaults(defaulted, origJSON);
console.log("defaults:", JSON.stringify(defaulted, null, 4));

输出

defaults: {
    "2014": {
        "grade": "A-",
        "className": "Geography 101"
    },
    "purl": "3",
    "firstName": "Jon",
    "sensitive": {
        "password": "zxcvbn!",
        "phone": "1234567890",
        "ssn": "123-00-6789",
        "acc": "987654300"
    },
    "_id": "564d098e2e59e72412e4d795",
    "finalGrade": "B",
    "template": "User.html"
}

查看下面的演示。

var origJSON = {
	"2014": {
		"grade": "A-",
		"className": "Geography 101"
	},
	"_id": "564d098e2e59e72412e4d795",
	"purl": "3",
	"firstName": "Jon",
	"finalGrade": "B",
	"template": "User.html",
	"sensitive": {"ssn":"123-00-6789","acc":"987654300","password":"zxcvbn!"}
};

var newJSON = {
	"purl": "3",
	"firstName": "Jon",
	"sensitive": {
		"password": "qazwx",
		"phone": "1234567890"
	}
};


var assigned = _.clone(newJSON);
_.assign(assigned, origJSON);
console.log("assign:", assigned);

var merged = _.clone(newJSON);
_.merge(merged, origJSON);
console.log("merge:", merged);

var defaulted = _.clone(newJSON);
_.defaults(defaulted, origJSON);
console.log("defaults:", defaulted);

pre.innerHTML = "assign: " + JSON.stringify(assigned, null, 4) + "</br>merge: " + JSON.stringify(merged, null, 4) + "</br>defaults: "+ JSON.stringify(defaulted, null, 4);

<script src="//cdnjs.cloudflare.com/ajax/libs/lodash.js/3.10.1/lodash.min.js"></script>
<pre id="pre"></pre>

Answer 2

使用lodash等实用程序库以及ES6解构（可能），有很多选项可以做到这一点。

但是，imo，最易维护的方式，在使用其中之一时，将使用对象工厂。假设对象结构将是全部相同的...比你不了解的库或复杂的克隆/合并/重复数据删除算法更容易包围你。我敢打赌，JS引擎可以更好地优化工厂。

function outputFactory(input_1, input_2){
    // some conditions and/ or data handling here
    var desiredProperty = (input_1.someProperty >= input_2.someProperty) ? input_1.someProperty : input_2.someProperty;
    return {
        desiredProperty: desiredProperty
    }
}

var merged = outputFactory({someProperty:1}, {someProperty:2})
console.log(merged)

希望这能为您提供一个选择。

将2个嵌套的JSON对象与nodejs中的add或update字段合并

2 个答案: