Question

我有这样的JSON对象数组：

[ { id: 841,
    when: 'date',
    who: 'asd',
    what: 'what',
    key1 : 'a key',
    key2 : 'a key'
    _id: 544034ab914ae3b9270545c1,
    __v: 0 }, 
{ id: 841,
    when: 'date',
    who: 'asd',
    what: 'what',
    key1 : 'a key',
    key2 : 'a key'
    _id: 544034ab914ae3b9270545c1,
    __v: 0 } ]

我想从这些对象中剪切key1和key2，并希望看到这一点：

[ { id: 841,
    when: 'date',
    who: 'asd',
    what: 'what',
    _id: 544034ab914ae3b9270545c1,
    __v: 0 }, 
{ id: 841,
    when: 'date',
    who: 'asd',
    what: 'what',
    _id: 544034ab914ae3b9270545c1,
    __v: 0 } ]

如何剪切键和值？

我的方法不起作用。（伪）：

var new_array
for i  old_array.length
   delete old_array[i].key1  
   delete old_array[i].key2
new_array.push(old_array[i])

Answer 1

yourArray = yourArray.map(function(current, index, arr){
  delete current.key1;
  delete current.key2;
  return current;
});

那应该按你的意愿行事; - ）

希望有所帮助，扬

Answer 2

您的伪代码基本上是您所需要的，但您可以只删除属性：

var arr = [...]; // your array
for (var i = 0; i < arr.length; i++) {
    delete arr[i].key1;
    delete arr[i].key2;
}

Answer 3

像这样：

function deleteProps(){
  var a = Array.slice(arguments) || Array.prototype.slice.call(arguments);
  var oa = a[0], dp = a.slice(1);
  for(var i=0,l=oa.length; i<l; i++){
    for(var n=0,c=dp.length; n<c; n++){
      delete oa[i][dp[n]];
    }
  }
  return oa;
}
var newObjectsArray = deleteProps(objectsArray, 'key1', 'key2');

Answer 4

我使用了underscore.js

var cutted_object =  _.omit(json_object, "key1", "key2" 
                                         "key2", "date",
                                         "id33", "etc");

从JSON对象数组中剪切数据

4 个答案: