在ES6中进行破坏和/或批量分配

Question

我有2个数据来源。其中一个来源是＆＃34;模板＆＃34;到数据可接受的范围。但是，第二个来源可能包含大量我不关心的数据（JSON中有100多个属性）。这是模式：

>unique(unlist(projectdf$ProjectSubject))

[1] "Applied Learning"                           "Applied Learning, Literacy 
& Language"     
[3] "Literacy & Language"                        "Special Needs"                             
[5] "Literacy & Language, History & Civics"      "Math & Science"                            
[7] "History & Civics, Math & Science"           "Literacy & Language, 
Special Needs"        
[9] "Applied Learning, Special Needs"            "Health & Sports, Special 
Needs"            
[11] "Math & Science, Literacy & Language"        "Literacy & Language, Math 
& Science"       
[13] "Literacy & Language, Music & The Arts"      "Math & Science, Special 
Needs"             
[15] "Health & Sports"                            "Music & The Arts"                          
[17] "Math & Science, Applied Learning"           "Literacy & Language, 
Applied Learning"     
[19] "Applied Learning, Music & The Arts"         "History & Civics, 
Literacy & Language"     
[21] "Applied Learning, Math & Science"           "Health & Sports, Math & 
Science"           
[23] "Applied Learning, Health & Sports"          "History & Civics"                          
[25] "History & Civics, Music & The Arts"         "Math & Science, History & 
Civics"          
[27] "Math & Science, Music & The Arts"           "Special Needs, Music & 
The Arts"           
[29] "History & Civics, Applied Learning"         "History & Civics, Special 
Needs"           

第二个来源将在上面的数据模式中具有4个属性（加上我不在乎的许多属性）。目前，我正在这样分配它们：

// Only store the data we care about.  Only a small subset of
// data that I need for this particular dataset.
state = {
    isDirty: false,
    data: {
        name: '',
        address: '',
        city: '',
        state: ''
    }
 }

使用ES6，也许是破坏，是否有更好的方法来批量分配这样的变量？

再次感谢！

修改

添加以便澄清循环后的分配。

Answer 1

Lodash pick可用于挑选特定键，或者辅助函数可用于相同目的：

const pick = (obj, keys) => Object.keys(obj)
    .filter((key) => keys.indexOf(key) >= 0)
    .reduce(
        (newObj, key) => Object.assign(newObj, { [key]: obj[key] }),
        {}
    );

许多相关问题已经提出了这一点。这个问题特有的是：

this.state.data = pick(someDataSource, Object.keys(this.state.data));

Answer 2

你可以做的一个技巧（技巧，因为它需要吞下错误）是使用不可扩展的对象，使用Object.preventExtensions然后使用Object.assign来填充它包含数据（try / catch块中的）。

＆＃13;

＆＃13;

// Only store the data we care about.  Only a small subset of
// data that I need for this particular dataset.
state = {
    isDirty: false,
    data: {
        name: '',
        address: '',
        city: '',
        state: ''
    }
 }

const newData = {
    name:'name',
    address:'address',
    city:'city',
    state:'state',
    phone:'phone',
    zip:'zip'
}

const updatedData = Object.preventExtensions({...state.data});

try{
   Object.assign(updatedData, newData);
} catch(throwaway){};

console.log(updatedData);

＆＃13;

＆＃13;

＆＃13;

作为重用的功能

＆＃13;

＆＃13;

function schemaMerge(schema, data) {
  const mergedData = Object.preventExtensions({...schema});

  try {
    Object.assign(mergedData, data);
  } catch (throwaway) {};
  
  return ({...mergedData}); // create a new object from the merged one so that it no longer is extensionless
}

// Only store the data we care about.  Only a small subset of
// data that I need for this particular dataset.
state = {
  isDirty: false,
  data: {
    name: '',
    address: '',
    city: '',
    state: ''
  }
}

const newData = {
  name: 'name',
  address: 'address',
  city: 'city',
  state: 'state',
  phone: 'phone',
  zip: 'zip'
}


const updatedData = schemaMerge(state.data, newData);

state.data = updatedData;

console.log(state.data);

＆＃13;

＆＃13;

＆＃13;

Answer 3

可以在JSON.parse reviver：

中排除和修改属性

var o = JSON.parse('{"a":1, "b":2}', (k, v) => k === 'a' ? void 0 : k === 'b' ? 3 : v)

console.log( o )