blocks|key|1433649|text|你能做到，就像这样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1433650|[2,3].forEach(key+=>+{
++++delete+object[key];
})|code-block|syntax|javascript|1433651|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<p>You can do it, like this:</p>

<pre class="lang-js prettyprint-override"><code>[2,3].forEach(key =&gt; {
    delete object[key];
})
</code></pre>


blocks|key|1433672|text|为此，可以使用筛选方法。例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1433673|let+id+=+2;
let+list+=+[{
++Id:+1,
++Name:+'a'
},+{
++Id:+2,
++Name:+'b'
},+{
++Id:+3,
++Name:+'c'
}];
let+lists+=+list.filter(x+=>+{
++return+x.Id+!=+id;
})
console.log(lists);|code-block|syntax|javascript|1433674|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<p>You can use filter method for this.
Ex:</p>

<pre class="lang-js prettyprint-override"><code>let id = 2;
let list = [{
  Id: 1,
  Name: 'a'
}, {
  Id: 2,
  Name: 'b'
}, {
  Id: 3,
  Name: 'c'
}];
let lists = list.filter(x =&gt; {
  return x.Id != id;
})
console.log(lists);
</code></pre>


blocks|key|484622|text|您可以使用filter。在过滤器回调函数中，通过使用id检查id数组中是否也存在includes|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|484623|​|484624|let+idArr+=+[1,+2]
let+obj+=+[{
++++id:+1,
++++name:+'abc'
++},
++{
++++id:+2,
++++name:+'abc'
++},
++{
++++id:+3,
++++name:+'abc'
++},
++{
++++id:+4,
++++name:+'abc'
++}
];

let+data+=+obj.filter(item+=>+!idArr.includes(item.id));
console.log(data);
console.log(obj)|code-block|syntax|javascript|484625|484626|entityMap^0|5|6|Q|2|U|2|14|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Z|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|10|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|O|$]]

<p>You can use <code>filter</code>. In the filter callback function check if the <code>id</code> is also there in <code>id</code> array by using <code>includes</code></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 idArr = [1, 2]
let obj = [{
    id: 1,
    name: 'abc'
  },
  {
    id: 2,
    name: 'abc'
  },
  {
    id: 3,
    name: 'abc'
  },
  {
    id: 4,
    name: 'abc'
  }
];

let data = obj.filter(item =&gt; !idArr.includes(item.id));
console.log(data);
console.log(obj)</code></pre>
</div>
</div>
</p>


blocks|key|484672|text|在这里使用过滤器可能会很好。你可以写这样的东西：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|484673|​|484674|var+newArray+=+oldArray.filter(object+=>+!ids.includes(object.id))|code-block|syntax|javascript|484675|484676|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter^0|5|3|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|D|3|E|5|6|7|X|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|Y|8|@]|9|@]|C|$I|J]]|$1|K|3|E|5|6|7|Z|8|@]|9|@]|C|$]]|$1|L|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|M|$N|$5|O|P|Q|C|$R|S]]]]

<p>using <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter" rel="nofollow noreferrer">filter</a> might work well here. you could write something like:</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 newArray = oldArray.filter(object =&gt; !ids.includes(object.id))</code></pre>
</div>
</div>
</p>


blocks|key|484715|text|假设您希望通过从数组中完全删除元素来从原始数组中删除项(并且您不希望获得新的数组)，则可以利用Array.splice|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|484716|let+idArr+=+[1,+2];
let+obj+=+[{
++++id:+1
++},
++{
++++id:+2
++},
++{
++++id:+3
++},
++{
++++id:+4
++}
];

for+(let+id+of+idArr)+{
++//+look+for+the+element+by+its+id.
++const+objIdRef+=+obj.find(i+=>+i.id+===+id);
++//+if+it+actually+exists,+splice+it.
++objIdRef+&&+obj.splice(obj.indexOf(objIdRef),+1);
}
console.log(obj);|code-block|syntax|javascript|484717|​|484718|如果obj数组很大，您可能需要在处理id数组之前从它生成一个映射，以便在删除进程开始时将复合减少到O(1)。|484719|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/splice^0|10|4|1B|C|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|14|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

<p>Assuming you want to <strong>delete</strong> items from the original array by entirely removing the element from the array (and you don't want to get a <strong>new array</strong>), you can take advantage of 
<a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/splice" rel="nofollow noreferrer">Array.splice</a>
<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 idArr = [1, 2];
let obj = [{
    id: 1
  },
  {
    id: 2
  },
  {
    id: 3
  },
  {
    id: 4
  }
];

for (let id of idArr) {
  // look for the element by its id.
  const objIdRef = obj.find(i =&gt; i.id === id);
  // if it actually exists, splice it.
  objIdRef &amp;&amp; obj.splice(obj.indexOf(objIdRef), 1);
}
console.log(obj);</code></pre>
</div>
</div>
</p>

<p>If the obj array is big, you might want to make a map from it before processing the id array, so that the complexing is reduced to O(1) when the delete process begins.</p>


blocks|key|1433749|text|也许这就是你想要的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1433750|​|1433751|var+arr=+[{id:1,+name:+"foo"},+{id:2,+name:+"bar"},+{id:3,+name:"not+to+be+deleted"}];

var+idsToDelete+=+[1,+2];

var+res+=+arr.map((i,+idx)=>{
+++return+arr[idx]+=+idsToDelete.includes(i.id)?+undefined+:+arr[idx]
}).filter(i=>i)

console.log(res)|code-block|syntax|javascript|1433752|1433753|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

<p>Perhaps This is what you want:</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 arr= [{id:1, name: "foo"}, {id:2, name: "bar"}, {id:3, name:"not to be deleted"}];

var idsToDelete = [1, 2];

var res = arr.map((i, idx)=&gt;{
   return arr[idx] = idsToDelete.includes(i.id)? undefined : arr[idx]
}).filter(i=&gt;i)

console.log(res)</code></pre>
</div>
</div>
</p>


blocks|key|472249|text|您可以尝试Lodash.js函数_.forEach()和_.remove()|type|unstyled|depth|inlineStyleRanges|entityRanges|data|472250|let+valuesArr+=+[
++++{id:+1,+name:+"dog"},+
++++{id:+2,+name:+"cat"},+
++++{id:+3,+name:+"rat"},+
++++{id:+4,+name:+"bat"},
++++{id:+5,+name:+"pig"},
];+
let+removeValFromIndex+=+[
++++{id:+2,+name:+"cat"},+
++++{id:+5,+name:+"pig"},
];+
_.forEach(removeValFromIndex,+(indi)+=>+{
++++_.remove(valuesArr,+(item)+=>+{
++++++++return+item.id+===+indi.id;
++++});
})

console.log(valuesArr)
/*[
++++{id:+1,+name:+"dog"},++
++++{id:+3,+name:+"rat"},+
++++{id:+4,+name:+"bat"},
];+*/|code-block|syntax|javascript|472251|在改变数组之前，不要忘记克隆(_.clone(valuesArr)或...valuesArr)|472252|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

<p>You can try Lodash.js functions _.forEach() and _.remove()</p>
<pre class="lang-js prettyprint-override"><code>let valuesArr = [
    {id: 1, name: &quot;dog&quot;}, 
    {id: 2, name: &quot;cat&quot;}, 
    {id: 3, name: &quot;rat&quot;}, 
    {id: 4, name: &quot;bat&quot;},
    {id: 5, name: &quot;pig&quot;},
]; 
let removeValFromIndex = [
    {id: 2, name: &quot;cat&quot;}, 
    {id: 5, name: &quot;pig&quot;},
]; 
_.forEach(removeValFromIndex, (indi) =&gt; {
    _.remove(valuesArr, (item) =&gt; {
        return item.id === indi.id;
    });
})

console.log(valuesArr)
/*[
    {id: 1, name: &quot;dog&quot;},  
    {id: 3, name: &quot;rat&quot;}, 
    {id: 4, name: &quot;bat&quot;},
]; */
</code></pre>
<p>Don't forget to clone (_.clone(valuesArr) or [...valuesArr]) before mutate your array</p>


<p>I have a main array of objects with each object having some key/values as well as a "id" key with 1,2,3,4,5, etc
Now I have another array representing just id's (like [2,3])</p>

<p>I want to use this array to delete objects from the main array...so in this case, objects from the main array having id's 2 &amp; 3 should be deleted</p>

<p>While I am aware of findBy(id), I am not sure if that can be used to delete multiple objects at once.</p>


Delete multiple objects in an array by id

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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我有一个主要的对象数组，每个对象都有一些键/值，还有一个带有1,2,3,4,5等的"id“键，现在我有了另一个数组，表示的只是id(类似于2,3)。我想使用这个数组从主array...so中删除对象在本例中，应该删除具有id的2&3的主数组中的对象虽然我知道findBy(id)，但我不确定是否可以使用它同时删除多个对象...

问按id删除数组中的多个对象
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问按id删除数组中的多个对象
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