blocks|key|2142431|text|您可以使用Array#filter达到所需的结果。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2142432|​|2142433|const+oneAndOnlyOne+=+arr+=>+arr.filter(v+=>+v+>=+10).length+==+1;

console.log(oneAndOnlyOne([12,+5,+12,+13,+4]));
console.log(oneAndOnlyOne([2,+11,+6,+1,+4]));|code-block|syntax|javascript|2142434|2142435|entityMap^0|5|C|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|T|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|O|$]]

<p>You could reach desired result using <code>Array#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 oneAndOnlyOne = arr =&gt; arr.filter(v =&gt; v &gt;= 10).length == 1;

console.log(oneAndOnlyOne([12, 5, 12, 13, 4]));
console.log(oneAndOnlyOne([2, 11, 6, 1, 4]));</code></pre>
</div>
</div>
</p>


blocks|key|2142454|text|尝试使用filter和数组长度匹配1+.And使用与Array.prototype.oneAndOnlyOne添加的回调方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2142455|​|2142456|Array.prototype.oneAndOnlyOne=function(callback){
++return+this.filter(callback).length+==+1
}

console.log([12,+11,+6,+1,+4].oneAndOnlyOne(x=>+x+>=+10))+//+false
console.log([2,+11,+6,+1,+4].oneAndOnlyOne(x=>+x+>=+10))+//+true|code-block|syntax|javascript|2142457|2142458|entityMap^0|4|6|H|1|Q|T|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]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|X|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|Y|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|O|$]]

<p>Try with <code>filter</code> and array length match with <code>1</code> .And use callback method added with <code>Array.prototype.oneAndOnlyOne</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>Array.prototype.oneAndOnlyOne=function(callback){
  return this.filter(callback).length == 1
}

console.log([12, 11, 6, 1, 4].oneAndOnlyOne(x=&gt; x &gt;= 10)) // false
console.log([2, 11, 6, 1, 4].oneAndOnlyOne(x=&gt; x &gt;= 10)) // true</code></pre>
</div>
</div>
</p>


blocks|key|2142480|text|您可以向数组中添加一个函数(毕竟它是一个对象)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2142481|var+arr+=+[30];
arr.oneAndOnlyOne+=+function(numToCheck)+{+
++++return+this.filter(function(x)+{
++++++++return+x+>=+numToCheck
++++}).length+===+1
}

arr.oneAndOnlyOne(30);+//true
arr.push(30);
arr.oneAndOnlyOne(30);+//false|code-block|syntax|javascript|2142482|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 add a function to your array (since it is an object after all):</p>

<pre><code>var arr = [30];
arr.oneAndOnlyOne = function(numToCheck) { 
    return this.filter(function(x) {
        return x &gt;= numToCheck
    }).length === 1
}

arr.oneAndOnlyOne(30); //true
arr.push(30);
arr.oneAndOnlyOne(30); //false
</code></pre>


blocks|key|2142491|text|最有效的解决方案可能是简单的循环：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2142492|function+findOne(arr)+{
++var+foundOne+=+false;
++for(var+i+=+0;+i+<+arr.length;+%2B%2Bi)+{
++++if(arr[i]+>+10)
++++++if(foundOne)+return+false;
++++++else+foundOne+=+true;
++}
++return+foundOne;
}|code-block|syntax|javascript|2142493|已经有一些工具可以做到这一点。比如兰达。你可以找到一个适合你的。|offset|length|2142494|entityMap|0|LINK|mutability|MUTABLE|url|http://ramdajs.com/docs/#countBy^0|0|0|H|2|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

<p>Probably the most efficient solution is the simple loop:</p>

<pre><code>function findOne(arr) {
  var foundOne = false;
  for(var i = 0; i &lt; arr.length; ++i) {
    if(arr[i] &gt; 10)
      if(foundOne) return false;
      else foundOne = true;
  }
  return foundOne;
}
</code></pre>

<p>There are some tools out there that already do this. Like <a href="http://ramdajs.com/docs/#countBy" rel="nofollow noreferrer">ramda</a> for instance. You can just find one that suits you.</p>


blocks|key|844500|text|在找到第一个匹配点时将数组切片，然后在数组的其余部分进行搜索，如|type|unstyled|depth|inlineStyleRanges|entityRanges|data|844501|​|844502|function+oneAndOnlyOne(arr,+fn)+{
++const+pos+=+arr.findIndex(fn);
++return+pos+>=+0+&&+!arr.slice(pos+%2B+1).find(fn);
}

const+tests+=+[[12,+5,+12,+13,+4],+[2,+11,+6,+1,+4],+[1,+2,+3]];
++
for+(const+test+of+tests)+
++console.log(String(test),+oneAndOnlyOne(test,+x+=>+x+>=+10));|code-block|syntax|javascript|844503|844504|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>Slice the array at the point the first match is found, then search in the remaining part of the array, as in</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>function oneAndOnlyOne(arr, fn) {
  const pos = arr.findIndex(fn);
  return pos &gt;= 0 &amp;&amp; !arr.slice(pos + 1).find(fn);
}

const tests = [[12, 5, 12, 13, 4], [2, 11, 6, 1, 4], [1, 2, 3]];
  
for (const test of tests) 
  console.log(String(test), oneAndOnlyOne(test, x =&gt; x &gt;= 10));</code></pre>
</div>
</div>
</p>


<p>I know that in Javascript I can use <code>.some</code> or <code>.every</code> on arrays to check if they have atleast one item or every item that passes the test implemented by the provided function.</p>

<p>Here is some examples:</p>

<pre><code>[2, 5, 8, 1, 4].some(x =&gt; x &gt; 10);  // false
[12, 5, 8, 1, 4].some(x =&gt; x &gt; 10); // true

[12, 5, 8, 130, 44].every(x =&gt; x &gt;= 10); // false
[12, 54, 18, 130, 44].every(x =&gt; x &gt;= 10); // true
</code></pre>

<p>I'm looking for checking if an array has "one and only one" item that passes the given function.
I would like to have some method like the following:</p>

<pre><code>[12, 5, 12, 13, 4].oneAndOnlyOne(x =&gt; x &gt;= 10); // false
[2, 11, 6, 1, 4].oneAndOnlyOne(x =&gt; x &gt;= 10); // true
</code></pre>

<p>Do you know any new ECMA Script 6 way or any easy/quick way, even using <code>lodash</code>, to check if in array there is one and one only occurrence of item that has certain values?</p>


Javascript: How to check if array of object has one and only one item with given value?

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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我知道，在Javascript中，我可以在数组上使用.some或.every来检查它们是否至少有一项或每一项通过了所提供的函数实现的测试。以下是一些例子：[2, 5, 8, 1, 4].some(x => x > 10);  // false[12, 5, 8, 1, 4].some(x => x > 10); // ...

问Javascript:如何检查对象数组是否只有一个具有给定值的项？
EN

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问Javascript:如何检查对象数组是否只有一个具有给定值的项？EN