Question

假设我有两个列表

a= ['apple', 'orange', 'banana']
b= ['red', 'orange', 'yellow']

如何使用第二个列表作为属性名称的指南将其转换为JSON对象？

例如，我会定义attributes = ['fruit', 'color']

以获得

result = [
  {fruit: 'apple', color: 'red'}, 
  {fruit: 'orange', color: 'orange'}, 
  {fruit: 'banana', color: 'yellow'}]

Answer 1

我创建了一个接受2个参数的参数，第一个是array作为属性，第二个是array array作为列表项，并且它将处理属性编号多于给定的属性项：

＆＃13;

 var create = function(attrList, propertyLists) {
  var result = [];
  var aLen = attrList.length;
  var pLen = propertyLists.length;
  
  if (pLen === 0) {
    return result;
  }

  var itemLength = propertyLists[0].length;
  if (itemLength === 0) {
    return result;
  }

  var i, j, obj, key;
  for (i = 0; i < itemLength; ++i) {
    obj = {};
    for(j = 0; j < aLen; ++j) {
      key = attrList[j];
      if (typeof propertyLists[j] === 'undefined') {
        //
        continue;
      }
      obj[key] = propertyLists[j][i];
    }
    result.push(obj);
  }

  return result;
 };

var a = ['apple', 'orange', 'banana'];
var b= ['red', 'orange', 'yellow'];
var attrs = ['fruit', 'color'];
var jsonObj = create(attrs, [a, b]);
console.log(jsonObj);

＆＃13;

Answer 2

假设两个列表大小相同且一切都匹配，这应该可行。但是，如果它们的大小不同，则会破坏。你的数据是什么样的？

\\given
a= ['apple', 'orange', 'banana']
b= ['red', 'orange', 'yellow']
attributes = ['fruit', 'color']

\\insert this code
var result = [];
for(var i = 0; i<a.length; i++){
    result.push({
        attributes[0]:a[i], 
        attributes[1]:b[i]
    });
}

console.log(result);
\\  result = [
\\      {fruit: 'apple', color: 'red'}, 
\\      {fruit: 'orange', color: 'orange'}, 
\\      {fruit: 'banana', color: 'yellow'}]

Answer 3

如果您可以使用像下划线或lodash这样的库（或重新创建此处使用的方法），可以这样做：

var attributes = ['fruit', 'color'];
var fruits = ['apple', 'orange', 'banana'];
var colors = ['red', 'orange', 'yellow'];

//Combine arrays in to list of pairs 
//(this would be expanded with each new array of attribute values)
//ORDER IS IMPORTANT
var zipped = _.zip(fruits, colors);

//Map the zipped list, returning an object based on the keys. 
//Remember the order of the arrays in the zip operation 
//must match the order of the attributes in the attributes list
var result = _.map(zipped, function(item, index) {
    return _.object(attributes, item);
});

console.log(result);

列出到json转换javascript

3 个答案: