Question

我有一个包装函数，我在其中使用变量dataObject。我有一个动作来触发包装函数中的一些外部函数。

function wrapper() {
    var dataObject;
    var jsonPath = "dataObject[0]['Set1'][0]['Attribute1']";
    eval('outsideFunction(dataObject, jsonPath)');
}


function outsideFunction(dataObject, jsonPath) {
    dataObject[0]['Set1'][0]['Attribute1'] = 'asde';  //This sets the value to dataObject in the wapper
    var attrVal = '123';
    eval("jsonPath = attrVal");  //This doesn't set value to dataObject in the wrapper but in the local dataObject
}

为什么使用eval直接分配和分配的操作有区别？

Answer 1

根据您data[0]['Set1'][0]['Attribute1']的结构，可以写成data[0].Set1[0].Attribute1，这里是代码，但我认为您不太了解您要求的套数。

var wrapper, outsideFunction;

wrapper = function(){

  someOb = {};

  var data = [
    {
      Set1: [
        {
          Attribute1: null, // we will change null to 'asdf' below
        },
      ],
    },
  ];

  outsideFunction(data, someOb);

  console.log( someOb.newProp, someOb.dumb );
  // outputs 'hehehehe', undefined

};

outsideFunction = function(data, blah) {

  data[0].Set1[0].Attribute1 = 'asdf';

  //blah is a reference to someOb
  // we can change a part of blah and it will look at the reference and change someOb
  blah.newProp = 'hehehehe';

  // but if we do `blah =`, then we reference a new object
  // This won't affect  someOb, blah will just be a different object talking about something else
  blah = { dumb: false };

};

所以，就像我说的那样，你的数据对象是一个编号集（你有[0]），然后是一个命名集（Set1），然后是一个编号集（[0] ），我认为你的意思不是那么多。

numberedSet = [
  { 
    name: 'dan',
    likes: [
      'coding', 'girls', 'food',
    ],
  },
  { 
    name: 'Sreekesh',
  },
]

namedSet = {

  dan: {
    isPerson: true,
    likes: [
      'coding', 'girls', 'food',
    ],
  },

  Sreekesh: {
    isPerson: true,
    askedQuestion: function(){
      return true;
    },
  }

};

numberedSet[0].name == dan; // true
numberedSet[0].likes[1] == 'girls'; // true
namedSet.dan.isPerson == true; // true
namedSet.Sreekesh.askedQuestion(); // true

eval（）工作和分配差异

1 个答案: