Question

所以我有这种来自json的数组

data : [{
    article_categories : {
        id : "xxx",
        name : "xxx"
    },
    article : "xxx",
    publisher : "xxx"
}]

我想为这些数组创建另一个多维数组，我想保留字段名称（数组中的名称"article"，"publisher"等等），但是我有值不知道获得字段名称

我还想做一些条件，如果只通过检查这个数组将一些字段包含到我的新数组中

thead: [
    { key: "article"},
    { key: "article_categories.name"},
    .....
]

所以我最后会有像这样的数组

newArray: [
 {article:"xxx",publisher: "xxx",article_categories.name:"xxx"},
 {article:"xxx",publisher: "xxx",article_categories.name:"xxx"}
 ....
]

怎么做？我试过了

thead.forEach(function(column){
  data.forEach(function(key,value){
      if(column.key == key){
         newArray[key] = value
      }
  })
})

但它只是不起作用......

Answer 1

如果你愿意使用lodash，那就太简单了。 Lodash使用.get()方法有效地评估JSON表达式，因此您不必为根据对象评估表达式而烦恼。

.chain()增加了锦上添花，使代码更容易阅读，但在幕后执行许多复杂的操作。

请尝试以下代码段：

var keys = [{
    key: "article"
  },
  {
    key: "article_categories.name"
  }
];

var data = [{
  article_categories: {
    id: "xxx",
    name: "xxx"
  },
  article: "xxx",
  publisher: "xxx"
}];

var result = _.chain(data)
  .map(function(item) {
    var object = {};

    keys.forEach(function(key) {
      object[key.key] = _.get(item, key.key);
    });

    return object;
  })
  .value();

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.min.js"></script>

Answer 2

我认为首先，你应该简化这个：

thead = [
    { key: "article"},
    { key: "article_categories.name"},
    .....
]

这样：

thead = ["article", "article_categories.name"]

我在这里：

＆＃13;

const data = [{
		article_categories : {
			id : "xxx",
			name : "xxx"
		},
		article : "xxx",
		publisher : "xxx"
	}],
	thead = ["article", "article_categories.name"]

const newArray = data.map( obj => {
	let output = {}
	thead.forEach(key => {
		if(key.includes(".")){
			let subkeys = key.split(".")
			output[key] = obj[subkeys[0]][subkeys[1]]
		} else {
			output[key] = obj[key]
		}
	})
	return output
})

console.log(newArray)

＆＃13;

Answer 3

您可以使用flatten函数以点符号展平对象。

然后map展平的对象每个item和filter仅允许key和reduce重建对象。< / p>

要回答原始问题，您可以使用Object.keys()获取对象的keys

＆＃13;

let data = [{
    article_categories : {
        id : "xxx",
        name : "xxx"
    },
    article : "xxx",
    publisher : "xxx"
},{
    article_categories : {
        id : "xxx2",
        name : "xxx2"
    },
    article : "xxx2",
    publisher : "xxx2"
}]

let thead = [
    { key: "article"},
    { key: "article_categories.name"},
];

let allowed_keys = thead.map(x=> x.key);

let flattened = data.map(item => flatten(item, '', ''));

// console.log(flattened);

let result = flattened.map(item => {
  return Object.keys(item)
    .filter(key => allowed_keys.includes(key))
    .reduce((obj, key) => {
      obj[key] = item[key];
      return obj;
    }, {})
});

console.log(result);


/**
 * Recursively flattens a JSON object using dot notation.
 *
 * NOTE: input must be an object as described by JSON spec. Arbitrary
 * JS objects (e.g. {a: () => 42}) may result in unexpected output.
 * MOREOVER, it removes keys with empty objects/arrays as value (see
 * examples bellow).
 *
 * @example
 * // returns {a:1, 'b.0.c': 2, 'b.0.d.e': 3, 'b.1': 4}
 * flatten({a: 1, b: [{c: 2, d: {e: 3}}, 4]})
 * // returns {a:1, 'b.0.c': 2, 'b.0.d.e.0': true, 'b.0.d.e.1': false, 'b.0.d.e.2.f': 1}
 * flatten({a: 1, b: [{c: 2, d: {e: [true, false, {f: 1}]}}]})
 * // return {a: 1}
 * flatten({a: 1, b: [], c: {}})
 *
 * @param obj item to be flattened
 * @param {Array.string} [prefix=[]] chain of prefix joined with a dot and prepended to key
 * @param {Object} [current={}] result of flatten during the recursion
 *
 * @see https://docs.mongodb.com/manual/core/document/#dot-notation
 */
function flatten (obj, prefix, current) {
  prefix = prefix || []
  current = current || {}
  if (typeof (obj) === 'object' && obj !== null) {
    Object.keys(obj).forEach(key => {
      flatten(obj[key], prefix.concat(key), current)
    })
  } else {
    current[prefix.join('.')] = obj
  }
  return current
}

＆＃13;

获取数组字段名称

3 个答案: