blocks|key|1779252|text|因为.remove并不是一个真正的数组函数。您可以只使用filter|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1779253|​|1779254|const+names+=+['Milan',+'Poudel',+'Soudel',+'Academia']
const+filteredNames+=+names.filter((name)+=>+!name.includes('ou'))

console.log(filteredNames)|code-block|syntax|javascript|1779255|1779256|entityMap^0|2|7|S|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|V|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|W|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|O|$]]

<p>Cause <code>.remove</code> is not really an Array function.
You can use just <code>filter</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>const names = ['Milan', 'Poudel', 'Soudel', 'Academia']
const filteredNames = names.filter((name) =&gt; !name.includes('ou'))

console.log(filteredNames)</code></pre>
</div>
</div>
</p>


blocks|key|1763259|text|如果你想用'ou‘删除所有的元素，可以这样做。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1763260|let+names+=+['Milan',+'Poudel',+'Soudel',+'Academia']++++
names+=+names.filter(function(name)+{
++return+!(name.includes('ou'));
})
console.log(names);|code-block|syntax|javascript|1763261|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>If you want to remove all elements with 'ou' do this. </p>

<pre><code>let names = ['Milan', 'Poudel', 'Soudel', 'Academia']    
names = names.filter(function(name) {
  return !(name.includes('ou'));
})
console.log(names);
</code></pre>


blocks|key|1779258|text|.remove()的确切来源是什么？你在使用某种类库吗？这是提供准确答案的当务之急。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1779259|为了解释你的过程，以及它的缺陷，这里有一个演练：|1779260|​|1779261|//+Array+of+strings+is+initialized
const+names+=+['Milan',+'Poudel',+'Soudel',+'Academia']

/**
*+Filter+the+names+array+and+return+
*+a+new+array+of+strings+that+INCLUDE
*+strings+with+'ou'+which+will
*+be+stored+in+a+global+variable+'found'
*/
found+=+names.filter(function(name)+{
++return+name.includes('ou')
})

//+Result+['Poudel',+'Soudel']
console.log(found);


//+Exception+thrown+because+Array.remove+is+not+a+function
found.remove()

/**
*+names+is+still+in-tact+and+in+it's+original
*+state+since+you+returned+the+result+of
*+your+names.filter+function+into+a+global
*+variable+called+'found'
*/
console.log(names)|code-block|syntax|javascript|1779262|1779263|您是否正在尝试创建一个Array.remove函数？也许它看起来像这样：|1779264|1779265|Array.prototype.remove+=+function+(name)+{
++return+this.filter(+e+=>+!e.includes(name));
}

let+names+=+['Milan',+'Poudel',+'Soudel',+'Academia']

names+=+names.remove('ou')
console.log(names)|1779266|1779267|entityMap^0|0|9|0|0|0|0|0|B|C|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|10|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|11|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|12|8|@]|D|@]|E|$M|N]]|$1|O|3|I|5|6|7|13|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|R|3|I|5|6|7|17|8|@]|D|@]|E|$]]|$1|S|3|T|5|L|7|18|8|@]|D|@]|E|$M|N]]|$1|U|3|I|5|6|7|19|8|@]|D|@]|E|$]]|$1|V|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|W|$]]

<p>Where is <code>.remove()</code> coming from exactly? Are you using an library of some sort? That's is imperative to providing an accurate answer.</p>

<p>To explain your process, and where it's flawed, here is a walkthrough:</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>// Array of strings is initialized
const names = ['Milan', 'Poudel', 'Soudel', 'Academia']

/**
* Filter the names array and return 
* a new array of strings that INCLUDE
* strings with 'ou' which will
* be stored in a global variable 'found'
*/
found = names.filter(function(name) {
  return name.includes('ou')
})

// Result ['Poudel', 'Soudel']
console.log(found);


// Exception thrown because Array.remove is not a function
found.remove()

/**
* names is still in-tact and in it's original
* state since you returned the result of
* your names.filter function into a global
* variable called 'found'
*/
console.log(names)</code></pre>
</div>
</div>
</p>

<p>Are you trying to create an <code>Array.remove</code> function? Maybe that could look something like this:</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>Array.prototype.remove = function (name) {
  return this.filter( e =&gt; !e.includes(name));
}

let names = ['Milan', 'Poudel', 'Soudel', 'Academia']

names = names.remove('ou')
console.log(names)</code></pre>
</div>
</div>
</p>


<pre><code>const names = ['Milan', 'Poudel', 'Soudel', 'Academia']
found = names.filter(function(name) {
  return name.includes('ou')
})

found.remove()
console.log(names)
</code></pre>


Why is it saying that found.remove() is not a function

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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const names = ['Milan', 'Poudel', 'Soudel', 'Academia']found = names.filter(function(name) {  return name.includes('ou')})found.remove()console.log(names)

问为什么说found.remove()不是一个函数
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问为什么说found.remove()不是一个函数EN