blocks|key|98635|text|如果要将对象映射到某个对象(在本例中为属性)。我认为如果你想编写函数式代码，Array.prototype.map就是你要找的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|98636|​|98637|console.log([
++++++{name:+"Joe",+age:+22},
++++++{name:+"Kevin",+age:+24},
++++++{name:+"Peter",+age:+21}
++++].map(function(elem){
++++++++return+elem.name;
++++}).join(","));|code-block|syntax|javascript|98638|98639|在现代JavaScript中：|98640|98641|console.log([
++++++{name:+"Joe",+age:+22},
++++++{name:+"Kevin",+age:+24},
++++++{name:+"Peter",+age:+21}
++++].map(e+=>+e.name).join(","));|98642|98643|(fiddle)|98644|如果你想支持不是ES5+compliant的旧浏览器，你可以填充它(上面的MDN页面上有一个polyfill+)。另一种选择是使用underscorejs的pluck方法：|98645|var+users+=+[
++++++{name:+"Joe",+age:+22},
++++++{name:+"Kevin",+age:+24},
++++++{name:+"Peter",+age:+21}
++++];
var+result+=+_.pluck(users,'name').join(",")|98646|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/Array/map|1|http://codepen.io/anon/pen/HohnI|2|http://es5.github.io/#x15.4.4.19|3|http://underscorejs.org/#pluck^0|12|J|12|J|0|0|0|0|0|0|0|0|0|0|8|1|0|26|5|8|D|2|26|5|3|0|0^^$0|@$1|2|3|4|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@$9|1G|A|1H|1|1I]]|E|$]]|$1|F|3|G|5|6|7|1J|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1K|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|1L|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|1M|8|@]|D|@]|E|$]]|$1|P|3|G|5|6|7|1N|8|@]|D|@]|E|$]]|$1|Q|3|R|5|J|7|1O|8|@]|D|@]|E|$K|L]]|$1|S|3|G|5|6|7|1P|8|@]|D|@]|E|$]]|$1|T|3|U|5|6|7|1Q|8|@]|D|@$9|1R|A|1S|1|1T]]|E|$]]|$1|V|3|W|5|6|7|1U|8|@$9|1V|A|1W|B|C]]|D|@$9|1X|A|1Y|1|1Z]|$9|20|A|21|1|22]]|E|$]]|$1|X|3|Y|5|J|7|23|8|@]|D|@]|E|$K|L]]|$1|Z|3|-4|5|6|7|24|8|@]|D|@]|E|$]]]|10|$11|$5|12|13|14|E|$15|16]]|17|$5|12|13|14|E|$15|18]]|19|$5|12|13|14|E|$15|1A]]|1B|$5|12|13|14|E|$15|1C]]]]

<p>If you want to map objects to something (in this case a property). I think <a href="https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/Array/map" rel="noreferrer"><code>Array.prototype.map</code></a> is what you're looking for if you want to code functionally.</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>console.log([
      {name: "Joe", age: 22},
      {name: "Kevin", age: 24},
      {name: "Peter", age: 21}
    ].map(function(elem){
        return elem.name;
    }).join(","));</code></pre>
</div>
</div>
</p>
<p>In modern JavaScript:</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>console.log([
      {name: "Joe", age: 22},
      {name: "Kevin", age: 24},
      {name: "Peter", age: 21}
    ].map(e =&gt; e.name).join(","));</code></pre>
</div>
</div>
</p>
<p><a href="http://codepen.io/anon/pen/HohnI" rel="noreferrer">(fiddle)</a></p>
<p>If you want to support older browsers, that are not <a href="http://es5.github.io/#x15.4.4.19" rel="noreferrer">ES5 compliant</a> you can shim it (there is a polyfill on the MDN page above). Another alternative would be to use underscorejs's <a href="http://underscorejs.org/#pluck" rel="noreferrer"><code>pluck</code></a> method:</p>
<pre><code>var users = [
      {name: &quot;Joe&quot;, age: 22},
      {name: &quot;Kevin&quot;, age: 24},
      {name: &quot;Peter&quot;, age: 21}
    ];
var result = _.pluck(users,'name').join(&quot;,&quot;)
</code></pre>


blocks|key|315519|text|你可以随时覆盖你的对象的toString方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|315520|​|315521|++++var+arr+=+[
++++++++{name:+"Joe",+age:+22,+toString:+function(){return+this.name;}},
++++++++{name:+"Kevin",+age:+24,+toString:+function(){return+this.name;}},
++++++++{name:+"Peter",+age:+21,+toString:+function(){return+this.name;}}
++++];
+++++++++
++++var+result+=+arr.join(",+");
++++console.log(result);|code-block|syntax|javascript|315522|315523|entityMap^0|C|8|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>Well you can always override the <code>toString</code> method of your objects:</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 = [
        {name: "Joe", age: 22, toString: function(){return this.name;}},
        {name: "Kevin", age: 24, toString: function(){return this.name;}},
        {name: "Peter", age: 21, toString: function(){return this.name;}}
    ];
         
    var result = arr.join(", ");
    console.log(result);</code></pre>
</div>
</div>
</p>


blocks|key|315535|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|315536|var+x=+[
++{name:+"Joe",+age:+22},
++{name:+"Kevin",+age:+24},
++{name:+"Peter",+age:+21}
]

function+joinObj(a,+attr)+{
++var+out+=+[];+
++for+(var+i=0;+i<a.length;+i%2B%2B)+{++
++++out.push(a[i][attr]);+
++}+
+return+out.join(",+");
}

var+z+=+joinObj(x,'name');
z+>+"Joe,+Kevin,+Peter"
var+y+=+joinObj(x,'age');
y+>+"22,+24,+21"|code-block|syntax|javascript|315537|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>try this</p>

<pre><code>var x= [
  {name: "Joe", age: 22},
  {name: "Kevin", age: 24},
  {name: "Peter", age: 21}
]

function joinObj(a, attr) {
  var out = []; 
  for (var i=0; i&lt;a.length; i++) {  
    out.push(a[i][attr]); 
  } 
 return out.join(", ");
}

var z = joinObj(x,'name');
z &gt; "Joe, Kevin, Peter"
var y = joinObj(x,'age');
y &gt; "22, 24, 21"
</code></pre>


blocks|key|98775|text|我也遇到过使用reduce方法，它看起来是这样的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|98776|​|98777|console.log(
++++[
++++++++{name:+"Joe",+age:+22},
++++++++{name:+"Kevin",+age:+24},
++++++++{name:+"Peter",+age:+21}
++++]
++++.reduce(function+(a,+b)+{
++++++++return+(a.name+%7C%7C+a)+%2B+",+"+%2B+b.name}
++++)
)|code-block|syntax|javascript|98778|98779|(a.name+%7C%7C+a)是正确的，所以第一个元素被正确地处理，但是其余的(其中a是一个字符串，所以a.name是未定义的)不被当作对象来处理。|98780|编辑:我现在将其进一步重构为：|98781|x.reduce(function(a,+b)+{return+a+%2B+["",+",+"][%2B!!a.length]+%2B+b.name;},+"");|98782|我认为这样更简洁，因为a总是一个字符串，b总是一个对象(因为在reduce中使用了optional+initialValue+parameter+)|98783|6个月后编辑:哦，我到底在想什么。“清洁工”我激怒了代码之神。|98784|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce#Syntax^0|7|6|0|0|0|0|0|D|14|1|1E|6|0|0|0|B|1|K|1|V|6|15|V|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|18|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|19|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|1A|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|1B|8|@$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1I|8|@]|D|@]|E|$]]|$1|R|3|S|5|J|7|1J|8|@]|D|@]|E|$K|L]]|$1|T|3|U|5|6|7|1K|8|@$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]]|D|@$9|1R|A|1S|1|1T]]|E|$]]|$1|V|3|W|5|6|7|1U|8|@]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1V|8|@]|D|@]|E|$]]]|Y|$Z|$5|10|11|12|E|$13|14]]]]

<p>I've also come across using the <code>reduce</code> method, this is what it looks 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>console.log(
    [
        {name: "Joe", age: 22},
        {name: "Kevin", age: 24},
        {name: "Peter", age: 21}
    ]
    .reduce(function (a, b) {
        return (a.name || a) + ", " + b.name}
    )
)</code></pre>
</div>
</div>
</p>
<p>The <code>(a.name || a)</code> is so the first element is treated correctly, but the rest (where <code>a</code> is a string, and so <code>a.name</code> is undefined) isn't treated as an object.</p>
<p>Edit: I've now refactored it further to this:</p>
<pre><code>x.reduce(function(a, b) {return a + [&quot;&quot;, &quot;, &quot;][+!!a.length] + b.name;}, &quot;&quot;);
</code></pre>
<p>which I believe is cleaner as <code>a</code> is always a string, <code>b</code> is always an object (due to the use of the <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce#Syntax" rel="nofollow noreferrer">optional initialValue parameter</a> in <code>reduce</code>)</p>
<p>Edit 6 months later: Oh what was I thinking. &quot;cleaner&quot;. I've angered the code Gods.</p>


blocks|key|98803|text|在节点或ES6%2B上：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|98804|users.map(u+=>+u.name).join(',+')|code-block|syntax|javascript|98805|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>On node or ES6+:</p>

<pre class="lang-js prettyprint-override"><code>users.map(u =&gt; u.name).join(', ')
</code></pre>


blocks|key|98833|text|假设对象数组被变量users引用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|98834|如果可以使用ES6，那么最简单的解决方案将是：|98835|users.map(user+=>+user.name).join(',+');|code-block|syntax|javascript|98836|如果不是，则可以这样使用lodash：|98837|+_.map(users,+function(user)+{
+++++return+user.name;
+}).join(',+');|98838|entityMap|0|LINK|mutability|MUTABLE|url|https://lodash.com/docs/4.17.4^0|9|5|0|0|0|C|6|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|13|8|@]|D|@$9|14|A|15|1|16]]|E|$]]|$1|O|3|P|5|J|7|17|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

<p>lets say the objects array is referenced by the variable <code>users</code></p>

<p>If ES6 can be used then the easiest solution will be:</p>

<pre><code>users.map(user =&gt; user.name).join(', ');
</code></pre>

<p>If not, and <a href="https://lodash.com/docs/4.17.4" rel="nofollow noreferrer">lodash</a> can be used so :</p>

<pre><code> _.map(users, function(user) {
     return user.name;
 }).join(', ');
</code></pre>


blocks|key|315660|text|这是一个我知道的老帖子，但对于遇到这个问题的任何人来说，仍然是超级相关的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|315661|这里推荐了Array.map，这是我一直在使用的一个很棒的方法。还提到了Array.reduce+..。|315662|我个人会在这个用例中使用Array.reduce。为什么？尽管代码不太干净/清晰。这比通过管道将map函数传递给连接要高效得多。|315663|这样做的原因是因为Array.map必须遍历每个元素，以返回一个新的数组，其中包含该数组中对象的所有名称。然后，Array.join遍历数组的内容以执行联接。|315664|您可以通过使用模板文字将代码放在一行中来提高jackweirdys的reduce+answer的可读性。“在所有现代浏览器中也受支持”|315665|//+a+one+line+answer+to+this+question+using+modern+JavaScript
x.reduce((a,+b)+=>+`${a.name+%7C%7C+a},+${b.name}`);|code-block|syntax|javascript|315666|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|T|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|V|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

<p>An old thread I know but still super relevant to anyone coming across this.</p>

<p>Array.map has been suggested here which is an awesome method that I use all the time. 
Array.reduce was also mentioned...</p>

<p>I would personally use an Array.reduce for this use case. 
Why? Despite the code being slightly less clean/clear. It is a much more efficient than piping the map function to a join.</p>

<p>The reason for this is because Array.map has to loop over each element to return a new array with all of the names of the object in the array. Array.join then loops over the contents of array to perform the join.</p>

<p>You can improve the readability of jackweirdys reduce answer by using template literals to get the code on to a single line. "Supported in all modern browsers too" </p>

<pre><code>// a one line answer to this question using modern JavaScript
x.reduce((a, b) =&gt; `${a.name || a}, ${b.name}`);
</code></pre>


blocks|key|98889|text|If对象和动态关键点："applications\":{\"1\":\"Element1\",\"2\":\"Element2\"}|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|98890|Object.keys(myObject).map(function+(key,+index)+{
++++return+myObject[key]
}).join(',+')|code-block|syntax|javascript|98891|entityMap^0|B|1J|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

<p>If object and dynamical keys: <code>"applications\":{\"1\":\"Element1\",\"2\":\"Element2\"}</code></p>

<pre><code>Object.keys(myObject).map(function (key, index) {
    return myObject[key]
}).join(', ')
</code></pre>


blocks|key|315731|text|不确定，但所有这些都可以回答他们的工作，但不是最佳选择，因为他们正在执行两次扫描，并且您可以在一次扫描中执行此操作。即使O(2n)被认为是一个真的O(n)，O(n)总是更好的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|315732|const+Join+=+(arr,+separator,+prop)+=>+{
++++let+combined+=+'';
++++for+(var+i+=+0;+i+<+arr.length;+i%2B%2B)+{
++++++++combined+=+`${combined}${arr[i][prop]}`;
++++++++if+(i+%2B+1+<+arr.length)
++++++++++++combined+=+`${combined}${separator}+`;
++++}
++++return+combined;
}|code-block|syntax|javascript|315733|这看起来可能很老套，但它允许我这样做：|315734|skuCombined+=+Join(option.SKUs,+',',+'SkuNum');|315735|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<p>not sure, but all this answers tho they work but are not optiomal since the are performing two scans and you can perform this in a single scan. Even though O(2n) is considered O(n) is always better to have a true O(n).</p>

<pre><code>const Join = (arr, separator, prop) =&gt; {
    let combined = '';
    for (var i = 0; i &lt; arr.length; i++) {
        combined = `${combined}${arr[i][prop]}`;
        if (i + 1 &lt; arr.length)
            combined = `${combined}${separator} `;
    }
    return combined;
}
</code></pre>

<p>This might look like old school, but allows me to do thig like this:</p>

<pre><code>skuCombined = Join(option.SKUs, ',', 'SkuNum');
</code></pre>


blocks|key|315751|text|您可以转换为数组，以便获取对象名称|type|unstyled|depth|inlineStyleRanges|entityRanges|data|315752|​|315753|var+objs+=+[
++++++{name:+"Joe",+age:+22},
++++++{name:+"Kevin",+age:+24},
++++++{name:+"Peter",+age:+21}
++++];
document.body.innerHTML+=+Object.values(objs).map(function(obj){
+++return+obj.name;
});|code-block|syntax|javascript|315754|315755|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>you can convert to array so get object name</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 objs = [
      {name: "Joe", age: 22},
      {name: "Kevin", age: 24},
      {name: "Peter", age: 21}
    ];
document.body.innerHTML = Object.values(objs).map(function(obj){
   return obj.name;
});</code></pre>
</div>
</div>
</p>


blocks|key|315793|text|最简单的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|315794|​|315795|const+oldArrayOfObjects+=+[
{name:+"Bob",+age:40},+
{name:+"Andrew",+age:25},+
{name:+"Peter",+age:30}];
const+newArrayOfStringNames+=+oldArrayOfObjects.map((OA)+=>+OA.name);
const+newArrayOfAges+=+oldArrayOfObjects.map((OA)+=>+OA.age);
console.log({newArrayOfStringNames,+newArrayOfAges})|code-block|syntax|javascript|315796|315797|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>Easiest way:</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 oldArrayOfObjects = [
{name: "Bob", age:40}, 
{name: "Andrew", age:25}, 
{name: "Peter", age:30}];
const newArrayOfStringNames = oldArrayOfObjects.map((OA) =&gt; OA.name);
const newArrayOfAges = oldArrayOfObjects.map((OA) =&gt; OA.age);
console.log({newArrayOfStringNames, newArrayOfAges})</code></pre>
</div>
</div>
</p>


<p>If I have an array of strings, I can use the <code>.join()</code> method to get a single string, with each element separated by commas, like so:</p>

<pre><code>["Joe", "Kevin", "Peter"].join(", ") // =&gt; "Joe, Kevin, Peter"
</code></pre>

<p>I have an array of objects, and I’d like to perform a similar operation on a value held within it; so from</p>

<pre><code>[
  {name: "Joe", age: 22},
  {name: "Kevin", age: 24},
  {name: "Peter", age: 21}
]
</code></pre>

<p>perform the <code>join</code> method only on the <code>name</code> attribute, to achieve the same output as before.</p>

<p>Currently I have the following function:</p>

<pre><code>function joinObj(a, attr){
  var out = [];

  for (var i = 0; i &lt; a.length; i++){
    out.push(a[i][attr]);
  }

  return out.join(", ");
}
</code></pre>

<p>There’s nothing wrong with that code, it works, but all of a sudden I’ve gone from a simple, succinct line of code to a very imperative function. Is there a more succinct, ideally more functional way of writing this?</p>


Perform .join on value in 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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如果我有一个字符串数组，我可以使用.join()方法来获取单个字符串，每个元素用逗号分隔，如下所示：["Joe", "Kevin", "Peter"].join(", ") // => "Joe, Kevin, Peter"我有一个对象数组，并且我想对其中保存的值执行类似的操作；所以从[  {name: "Joe", ...

问对对象数组中的值执行.join
EN

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问对对象数组中的值执行.joinEN