基于一次性密钥展平

Question

有

{ _id: "123", super: { super: { super: { someString: "here", someOtherString: "here", someObject: { withSomeContents: true }}}}}

带有超级键的关卡可以任意深，我希望每个关卡及其内容都被压平，以便得到

{ _id: "123", someString: "here", someOtherString: "here", someObject: { withSomeContents: true }}

这样做的好方法是什么？最好使用下划线。

Answer 1

您需要递归合并对象，同时检查所需的键：

var data = {
  _id: "123",
  super: {
    super: {
      super: {
        someString: "here",
        someOtherString: "here",
        someObject: {
          withSomeContents: true
        }
      }
    }
  }
};

function merge(dest, src) {
  Object.keys(src).forEach(function(key) {
    dest[key] = src[key];
  });
}

function flattenKey(data, flatKey) {
  return Object.keys(data).reduce(function(obj, key) {
    if (key === flatKey) {
      merge(obj, flattenKey(data[key], flatKey));
    } else {
      obj[key] = data[key];
    }
    return obj;
  }, {});
}

document.getElementById('r').textContent = JSON.stringify(flattenKey(data, 'super'));

<pre id=r></pre>

基于一次性密钥展平

Flatten based on disposable key

javascript

json

underscore.js