有没有办法删除值为NULL或UNDEFINED的深属性或者对象是递归空的？

Question

示例：

let anyObject = {
  name: 'Any name',
  address: [{
    street: null, // should exclude
    city: 'any city',
    state: 'any state'
  }, {}],
  otherObject: {
    otherProperty: {
       value: {
         value1: 1,
         value2: {} // should exclude
       } 
    }
  }
}

removeAllPropertyWithoutValue(anyObject)

存在一些有效的算法吗？我尝试使用_.pick，但仅限于第一级深度对象。

Answer 1

这应该符合您的期望：

＆＃13;

＆＃13;

var anyObject = {
  name: 'Any name',
  address: [{
    street: null,
    city: 'any city',
    state: 'any state'
  }, {}],
  otherObject: {
    otherProperty: {
       value: {
         value1: 1,
         value2: {}
       } 
    }
  }
}

function cleanUp(obj) {
  Object.keys(obj).reverse().forEach(function(k) {
    if(obj[k] === null || obj[k] === undefined) {
      obj instanceof Array ? obj.splice(k, 1) : delete obj[k];
    }
    else if(typeof obj[k] == 'object') {
      cleanUp(obj[k]);

      if(Object.keys(obj[k]).length == 0) {
        obj instanceof Array ? obj.splice(k, 1) : delete obj[k];
      }
    }
  });
}

cleanUp(anyObject);
console.log(anyObject);

＆＃13;

＆＃13;

＆＃13;