Question

我必须删除与我的模型不匹配的不需要的对象属性。我怎样才能用lodash实现它。

我的模特是

var model = {
   fname:null,
   lname:null
}

发送到服务器之前我的控制器输出将是

var credentials = {
    fname:"xyz",
    lname:"abc",
    age:23
}

如果我使用

 _.extend (model, credentials)

我也正在获得年龄。我知道我可以使用

delete credentials.age

但如果我有超过10个不需要的对象怎么办？我可以用lodash实现吗。

Answer 1

您可以从白名单或黑名单方式接近它：

// Black list
// Remove the values you don't want
var result = _.omit(credentials, ['age']);

// White list
// Only allow certain values
var result = _.pick(credentials, ['fname', 'lname']);

如果它是可重用的业务逻辑，您也可以将其分开：

// Partial out a black list version
var clean = _.partial(_.omit, _, ['age']); 

// and later
var result = clean(credentials);

Answer 2

使用_.keys()从model获取属性列表，并使用_.pick()将属性从credentials提取到新对象：

var model = {
   fname:null,
   lname:null
};

var credentials = {
    fname:"xyz",
    lname:"abc",
    age:23
};

var result = _.pick(credentials, _.keys(model));

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.16.4/lodash.min.js"></script>

如果您不想使用lodash，可以使用Object.keys()和Array.prototype.reduce()：

var model = {
   fname:null,
   lname:null
};

var credentials = {
    fname:"xyz",
    lname:"abc",
    age:23
};

var result = Object.keys(model).reduce(function(obj, key) {
  obj[key] = credentials[key];
  return obj;
}, {});

console.log(result);

Answer 3

您可以使用_.pick：

轻松完成此操作

var model = {
 fname:null,
 lname:null
};

var credentials = {
  fname: 'abc',
  lname: 'xyz',
  age: 2
};

var result = _.pick(credentials, _.keys(model));


console.log('result =', result);

<script src="https://cdn.jsdelivr.net/lodash/4.16.4/lodash.min.js"></script>

但你可以简单地使用纯JavaScript（特别是如果你使用ES6），就像那样：

const model = {
  fname: null,
  lname: null
};

const credentials = {
  fname: 'abc',
  lname: 'xyz',
  age: 2
};

const newModel = {};

Object.keys(model).forEach(key => newModel[key] = credentials[key]);

console.log('newModel =', newModel);

Answer 4

在这里，我通过使用lodash库对要删除的各个“键”使用了省略：

private static String getAccessToken() throws IOException {
  GoogleCredential googleCredential = GoogleCredential
      .fromStream(new FileInputStream("service-account.json"))
      .createScoped(Arrays.asList(SCOPES));
  googleCredential.refreshToken();
  return googleCredential.getAccessToken();
}

Answer 5

如果对象较少，则可以使用_.omit()从json数组中发出密钥

   _.forEach(data,(d)=>{
      _.omit(d,['keyToEmit1','keyToEmit2'])
   });

如果有更多对象，则可以使用其相反的_.pick()

   _.forEach(data,(d)=>{
      _.pick(d,['keyToPick1','keyToPick2'])
   });

Answer 6

要深入选择（或删除）满足给定条件的对象属性，可以使用类似以下内容的

：

function pickByDeep(object, condition, arraysToo=false) {
  return _.transform(object, (acc, val, key) => {
    if (_.isPlainObject(val) || arraysToo && _.isArray(val)) {
      acc[key] = pickByDeep(val, condition, arraysToo);
    } else if (condition(val, key, object)) {
      acc[key] = val;
    }
  });
}

https://codepen.io/aercolino/pen/MWgjyjm

Answer 7

这是我用lodash深度删除空属性的解决方案：

const compactDeep = obj => {
    const emptyFields = [];

    function calculateEmpty(prefix, source) {
        _.each(source, (val, key) => {
           if (_.isObject(val) && !_.isEmpty(val)) {
                calculateEmpty(`${prefix}${key}.`, val);
            } else if ((!_.isBoolean(val) && !_.isNumber(val) && !val) || (_.isObject(val) && _.isEmpty(val))) {
                emptyFields.push(`${prefix}${key}`);
            }
        });
    }

    calculateEmpty('', obj);

    return _.omit(obj, emptyFields);
};

Answer 8

Lodash unset适合删除一些不需要的键。

const myObj = {
    keyOne: "hello",
    keyTwo: "world"
}

unset(myObj, "keyTwo");

console.log(myObj); /// myObj = { keyOne: "hello" }

使用lodash删除对象属性

8 个答案: