使用jQuery将对象转换为另一个对象

时间:2016-09-27 16:12:18

标签: javascript jquery

我在我的JavaScript文件中得到了一个结果,我希望将其转换为另一个对象。

我原来的结果

    [
  {
    "SName": "Set1",
    "Elements": [
      {
        "Id": "3",
        "Name": "Name1"
      },
      {
        "Id": "5",
        "Name": "Name2"
      }
    ]
  },
  {
    "SName": "Set2",
    "Elements": [
      {
        "Id": "7",
        "Name": "Name3"
      },
      {
        "Id": "8",
        "Name": "Name4"
      }
    ]
  }
]

使用jQuery或JavaScript将其转换为类似于对象的数组。我怎样才能做到这一点?

[
  {
    "SName": "Set1",
    "Id": 3,
    "Name": "Name1"
  },
  {
    "SName": "Set1",
    "Id": 5,
    "Name": "Name2"
  },
  {
    "SName": "Set2",
    "Id": 7,
    "Name": "Name3"
  },
  {
    "SName": "Set2",
    "Id": 8,
    "Name": "Name4"
  }
]

5 个答案:

答案 0 :(得分:1)

您可以迭代数组Elements和属性并构建一个新对象并将其推送到结果集。

var array = [{ "SName": "Set1", "Elements": [{ "Id": "3", "Name": "Name1" }, { "Id": "5", "Name": "Name2" }] }, { "SName": "Set2", "Elements": [{ "Id": "7", "Name": "Name3" }, { "Id": "8", "Name": "Name4" }] }],
    result = [];

array.forEach(function (a) {
    a.Elements.forEach(function (b) {
        var o = { SName: a.SName };
        Object.keys(b).forEach(function (k) {
            o[k] = b[k];
        });
        result.push(o);
    });
});

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

ES6

var array = [{ "SName": "Set1", "Elements": [{ "Id": "3", "Name": "Name1" }, { "Id": "5", "Name": "Name2" }] }, { "SName": "Set2", "Elements": [{ "Id": "7", "Name": "Name3" }, { "Id": "8", "Name": "Name4" }] }],
    result = [];

array.forEach(a => a.Elements.forEach(b => result.push(Object.assign({ SName: a.SName }, b))));

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

答案 1 :(得分:1)

您可以使用reduce()forEach()Object.assign()

执行此操作

var data = [{
  "SName": "Set1",
  "Elements": [{
    "Id": "3",
    "Name": "Name1"
  }, {
    "Id": "5",
    "Name": "Name2"
  }]
}, {
  "SName": "Set2",
  "Elements": [{
    "Id": "7",
    "Name": "Name3"
  }, {
    "Id": "8",
    "Name": "Name4"
  }]
}]

var result = data.reduce(function(r, e) {
  e.Elements.forEach(function(o) {
    r.push(Object.assign({SName: e.SName}, o))
  })
  return r;
}, [])

console.log(result)

答案 2 :(得分:1)

var data =     [
  {
    "SName": "Set1",
    "Elements": [
      {
        "Id": "3",
        "Name": "Name1"
      },
      {
        "Id": "5",
        "Name": "Name2"
      }
    ]
  },
  {
    "SName": "Set2",
    "Elements": [
      {
        "Id": "7",
        "Name": "Name3"
      },
      {
        "Id": "8",
        "Name": "Name4"
      }
    ]
  }
];
console.log(data);

var newData = data.reduce(function (newArray, currentSet) {
    return newArray.concat(currentSet.Elements.map(function (element) {
        return Object.assign( { SName: currentSet.SName }, element);
    }));
}, []);

console.log(newData);

这里的关键是reduce功能。我们正在做的是通过循环遍历外部数组的每个值来创建一个全新的数组。我们不断地concat使用我们从内部数组映射的值来使用我们的新数组。

答案 3 :(得分:0)

这是使用jQuery的解决方案,这里是jsfiddle:

https://jsfiddle.net/noitse/3uk9qjnf/

我希望你知道所有的关键名称,所以修复它不会有问题。

var json = [
  {
    "SName": "Set1",
    "Elements": [
      {
        "Id": "3",
        "Name": "Name1"
      },
      {
        "Id": "5",
        "Name": "Name2"
      }
    ]
  },
  {
    "SName": "Set2",
    "Elements": [
      {
        "Id": "7",
        "Name": "Name3"
      },
      {
        "Id": "8",
        "Name": "Name4"
      }
    ]
  }
]

var newJSON = []

$(json).each(function(index,value){

  $(value.Elements).each(function(index1,value1){
        newJSON.push({"SName":value.SName,"Id":value1.Id,"Name":value1.Name})   
})

})

alert(JSON.stringify(newJSON))

这是代码,它通过第一个JSON循环,然后遍历其元素,然后将其推送到新数组

答案 4 :(得分:0)

您可以使用$ .extend方法,该方法允许您在与另一个对象合并时创建对象的副本。

var source = [] // Replace with the initalization of your source array
var destination = [];

for (var i = 0; i < source.length; i++) {
    var node = source[i];
    for (var j = 0; j < node.Elements.length; j++) {
        var subNode = node.Elements[j];
        newNode = $.extend(subNode, node);
        delete newNode["Elements"];
        destination.push(newNode);
    }
}

您可以运行代码in this fiddle