entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/delete|blocks|key|1tvsg|text|您可以根据需要重新构建对象|unstyled|depth|inlineStyleRanges|entityRanges|bbkum|++{+
++++name:+rec.name,
++++surname:+rec.surname
++}|code-block|syntax|javascript|5pacp|​|2fcu7|const+results+=+[
++++{
++++++++name:+"John",
++++++++surname:+"Fool",
++++++++age:+22
++++},
++++{
++++++++name:+"Erik",
++++++++surname:+"Owl",
++++++++age:+38
++++}
]

const+result+=+results.map((rec)+=>+{
++return+{+
++++name:+rec.name,
++++surname:+rec.surname
++}
})


console.log(result)|fntbv|dn0ud|或无用的delete字段|offset|length|dnshj|5lgjo|const+results+=+[
++++{
++++++++name:+"John",
++++++++surname:+"Fool",
++++++++age:+22
++++},
++++{
++++++++name:+"Erik",
++++++++surname:+"Owl",
++++++++age:+38
++++}
]

const+result+=+results.map((rec)+=>+{
++delete+rec.age
++return+rec
})


console.log(result)|4n8vb|dm5d7^0|0|0|0|0|0|4|6|0|0|0|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|11|G|@]|H|@]|6|$]]|$A|I|C|J|2|K|F|12|G|@]|H|@]|6|$L|M]]|$A|N|C|O|2|E|F|13|G|@]|H|@]|6|$]]|$A|P|C|Q|2|K|F|14|G|@]|H|@]|6|$L|M]]|$A|R|C|O|2|E|F|15|G|@]|H|@]|6|$]]|$A|S|C|T|2|E|F|16|G|@]|H|@$U|17|V|18|A|19]]|6|$]]|$A|W|C|O|2|E|F|1A|G|@]|H|@]|6|$]]|$A|X|C|Y|2|K|F|1B|G|@]|H|@]|6|$L|M]]|$A|Z|C|O|2|E|F|1C|G|@]|H|@]|6|$]]|$A|10|C|-4|2|E|F|1D|G|@]|H|@]|6|$]]]]

<p>You can just rebuild the object as you want</p>
<pre><code>  { 
    name: rec.name,
    surname: rec.surname
  }
</code></pre>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const results = [
    {
        name: "John",
        surname: "Fool",
        age: 22
    },
    {
        name: "Erik",
        surname: "Owl",
        age: 38
    }
]

const result = results.map((rec) =&gt; {
  return { 
    name: rec.name,
    surname: rec.surname
  }
})


console.log(result)</code></pre>
</div>
</div>
</p>
<p>Or <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/delete" rel="nofollow noreferrer">delete</a> fields that is useless</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const results = [
    {
        name: "John",
        surname: "Fool",
        age: 22
    },
    {
        name: "Erik",
        surname: "Owl",
        age: 38
    }
]

const result = results.map((rec) =&gt; {
  delete rec.age
  return rec
})


console.log(result)</code></pre>
</div>
</div>
</p>


entityMap|blocks|key|a3bfr|text|您可以使用map和reduce来简化这一过程，这样就不需要创建新的数组了。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|8dknm|​|euee6|var+results+=+[+{+name:+"John",+surname:+"Fool",+age:+22+},+{+name:+"Erik",+surname:+"Owl",+age:+38+}+];
let+keys+=+['name',+'surname'];
var+filtered+=+results.map(obj=>
+++keys.reduce((acc,curr)=>(acc[curr]=obj[curr],acc),+{}));
console.log(filtered);|code-block|syntax|javascript|diidv|csoc7|您也可以使用对象解构。|8dkm3|fsc7t|var+results+=+[+{+name:+"John",+surname:+"Fool",+age:+22+},+{+name:+"Erik",+surname:+"Owl",+age:+38+}+];
var+filtered+=+results.map(({name,surname})=>({name,surname}));
console.log(filtered);|ef9nt|5ihmf^0|5|3|9|6|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|V|9|@$A|W|B|X|C|D]|$A|Y|B|Z|C|D]]|E|@]|F|$]]|$2|G|4|H|6|7|8|10|9|@]|E|@]|F|$]]|$2|I|4|J|6|K|8|11|9|@]|E|@]|F|$L|M]]|$2|N|4|H|6|7|8|12|9|@]|E|@]|F|$]]|$2|O|4|P|6|7|8|13|9|@]|E|@]|F|$]]|$2|Q|4|H|6|7|8|14|9|@]|E|@]|F|$]]|$2|R|4|S|6|K|8|15|9|@]|E|@]|F|$L|M]]|$2|T|4|H|6|7|8|16|9|@]|E|@]|F|$]]|$2|U|4|-4|6|7|8|17|9|@]|E|@]|F|$]]]]

<p>You can use <code>map</code> and <code>reduce</code> to simplify this, which obviates the need to create a new array.</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var results = [ { name: "John", surname: "Fool", age: 22 }, { name: "Erik", surname: "Owl", age: 38 } ];
let keys = ['name', 'surname'];
var filtered = results.map(obj=&gt;
   keys.reduce((acc,curr)=&gt;(acc[curr]=obj[curr],acc), {}));
console.log(filtered);</code></pre>
</div>
</div>
</p>
<p>You can also use object destructuring.</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var results = [ { name: "John", surname: "Fool", age: 22 }, { name: "Erik", surname: "Owl", age: 38 } ];
var filtered = results.map(({name,surname})=&gt;({name,surname}));
console.log(filtered);</code></pre>
</div>
</div>
</p>


entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://github.com/Level/level|blocks|key|avr3s|text|我建议您更多地了解需要对输出执行的操作，以获得有用的答案。|unstyled|depth|inlineStyleRanges|entityRanges|1velf|情况1.如果你的原始列表仍然存在，并且你接受你的“修改列表”总是跟随原始列表，你可以使用一个生成器来包装你的原始对象，总是不返回那些额外的属性。|nhsf|情况2.如果你真的想要一个查询系统，你可以尝试使用真正的数据库，比如levelDB|offset|length|32cva|情况3。如果需要显示修改后的列表，请编写一个简单的包装器来适应每个列表项的格式。|bsnp6|情况4.如果您需要将修改后的列表作为对象进行快照，那么您已经创建的方法已经是一个非常合理的方法。|escct|情况5.如果您需要将修改后的列表快照作为另一个输出，您可以尝试直接获取此类输出，而不是创建中间对象|cu9bn^0|0|0|Y|7|0|0|0|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|V|G|@]|H|@]|6|$]]|$A|I|C|J|2|E|F|W|G|@]|H|@]|6|$]]|$A|K|C|L|2|E|F|X|G|@]|H|@$M|Y|N|Z|A|10]]|6|$]]|$A|O|C|P|2|E|F|11|G|@]|H|@]|6|$]]|$A|Q|C|R|2|E|F|12|G|@]|H|@]|6|$]]|$A|S|C|T|2|E|F|13|G|@]|H|@]|6|$]]|$A|U|C|-4|2|E|F|14|G|@]|H|@]|6|$]]]]

<p>I suggest you can tell more about what you will need to perform on the output to get the answer that can help.</p>
<p>Case 1. if your original list will survive and you accept your &quot;modified list&quot; to always follow the original list, you may use a generator to wrap your original object, by always not returning those extra properties.</p>
<p>Case 2. if you really want a query system, you may try using real DB thing such as <a href="https://github.com/Level/level" rel="nofollow noreferrer">levelDB</a></p>
<p>Case 3. if you need to display the modified list, write a simple wrapper to fit the format of each list item</p>
<p>Case 4. if you need to snapshot the modified list as object, the method you already made is already a very reasonable method</p>
<p>Case 5. if you need to snapshot the modified list as another output, you can try to directly obtain such output rather than making the intermediate object</p>


entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map|1|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment|blocks|key|9dqus|text|您可以使用Array.map()来转换数组的各个元素。在回调函数中，使用Object+destructuring+assignment只使用您感兴趣的键，并返回一个只包含这些键的新对象。|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|e9qu|​|esru|let+results+=+[
++++{+name:+"John",+surname:+"Fool",+age:+22+},
++++{+name:+"Erik",+surname:+"Owl",+age:+38+}
];

let+modified+=+results.map(({name,+surname})+=>+({name,+surname}));
console.log(modified);|code-block|syntax|javascript|7dg45|chp9d^0|5|B|0|10|V|1|0|0|0|0^^$0|$1|$2|3|4|5|6|$7|8]]|9|$2|3|4|5|6|$7|A]]]|B|@$C|D|E|F|2|G|H|V|I|@]|J|@$K|W|L|X|C|Y]|$K|Z|L|10|C|11]]|6|$]]|$C|M|E|N|2|G|H|12|I|@]|J|@]|6|$]]|$C|O|E|P|2|Q|H|13|I|@]|J|@]|6|$R|S]]|$C|T|E|N|2|G|H|14|I|@]|J|@]|6|$]]|$C|U|E|-4|2|G|H|15|I|@]|J|@]|6|$]]]]

<p>You can use <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map" rel="nofollow noreferrer">Array.map()</a> to transform individual elements of the array. And in the callback function use <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment" rel="nofollow noreferrer">Object destructuring assignment</a> to use only the keys you are interested in and return a new object with only those keys.</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>let results = [
    { name: "John", surname: "Fool", age: 22 },
    { name: "Erik", surname: "Owl", age: 38 }
];

let modified = results.map(({name, surname}) =&gt; ({name, surname}));
console.log(modified);</code></pre>
</div>
</div>
</p>


entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map|blocks|key|2a93n|text|看一下Array.map.It创建了转换后的数组。|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|ar7d1|​|3a77t|let+arr+=+[
++++{
++++++++name:+"John",
++++++++surname:+"Fool",
++++++++age:+22
++++},
++++{
++++++++name:+"Erik",
++++++++surname:+"Owl",
++++++++age:+38
++++}
]

let+result+=+arr.map((elem)+=>+{
++return+{
++++name:+elem.name,
++++surname:+elem.surname
++}
});

console.log(result);|code-block|syntax|javascript|b92ka|199pt^0|3|9|0|0|0|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|T|G|@]|H|@$I|U|J|V|A|W]]|6|$]]|$A|K|C|L|2|E|F|X|G|@]|H|@]|6|$]]|$A|M|C|N|2|O|F|Y|G|@]|H|@]|6|$P|Q]]|$A|R|C|L|2|E|F|Z|G|@]|H|@]|6|$]]|$A|S|C|-4|2|E|F|10|G|@]|H|@]|6|$]]]]

<p>Take a look at <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map" rel="nofollow noreferrer">Array.map</a>.It creates the transformed array.</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>let arr = [
    {
        name: "John",
        surname: "Fool",
        age: 22
    },
    {
        name: "Erik",
        surname: "Owl",
        age: 38
    }
]

let result = arr.map((elem) =&gt; {
  return {
    name: elem.name,
    surname: elem.surname
  }
});

console.log(result);</code></pre>
</div>
</div>
</p>


<p>Let's assume I'm obtaining an array of objects from a Node Repository, for example:</p>
<pre><code>results = [
    {
        name: &quot;John&quot;,
        surname: &quot;Fool&quot;,
        age: 22
    },
    {
        name: &quot;Erik&quot;,
        surname: &quot;Owl&quot;,
        age: 38
    }
]
</code></pre>
<p>How can I filter every object taking only the keys I need, for example avoiding 'age' key?</p>
<pre><code>filteredResults = [
    {
        name: &quot;John&quot;,
        surname: &quot;Fool&quot;,
    },
    {
        name: &quot;Erik&quot;,
        surname: &quot;Owl&quot;,
    }
]
</code></pre>
<p>I've already obtained this by creating another empty array and populating it by looping on the original one, but in case of large-data this would be heavy.</p>
<pre><code>repository.retrieve((error, result) =&gt; {
  let filteredData = [];
  result.forEach(r =&gt; {
    filteredData.push({
      name: r.name,
      description: r.description,
    });
  });
});
</code></pre>
<p>In SQL, I would obtain it this way:</p>
<pre><code>SELECT `name, description` FROM results;
</code></pre>


Filtering object keys inside an array of objects

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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 假设我正在从Node Repository获取一个对象数组，例如： results = [    {        name: "John",        surname: "Fool",        age: 22    },    {        name: "Erik",        surname: ...

问过滤对象数组中的对象键
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问过滤对象数组中的对象键EN