blocks|key|665261|text|如何访问数组的最后一个元素|type|unstyled|depth|inlineStyleRanges|entityRanges|data|665262|看起来是这样的：|665263|var+my_array+=+/*+some+array+here+*/;
var+last_element+=+my_array[my_array.length+-+1];|code-block|syntax|javascript|665264|在你的例子中是这样的：|665265|var+array1+=+loc['f096012e-2497-485d-8adb-7ec0b9352c52'];
var+last_element+=+array1[array1.length+-+1];|665266|或者，在更长的版本中，不创建新的变量：|665267|loc['f096012e-2497-485d-8adb-7ec0b9352c52'][loc['f096012e-2497-485d-8adb-7ec0b9352c52'].length+-+1];|665268|如何添加方法以使其更简单|665269|如果您热衷于创建函数/快捷方式来完成这些任务，请使用以下代码：|665270|if+(!Array.prototype.last){
++++Array.prototype.last+=+function(){
++++++++return+this[this.length+-+1];
++++};
};|665271|将允许您通过调用数组的last()方法来获取数组的最后一个元素，例如：|offset|length|style|CODE|665272|loc['f096012e-2497-485d-8adb-7ec0b9352c52'].last();|665273|你可以在这里检查它是否正常工作：http://jsfiddle.net/D4NRN/|665274|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/D4NRN/^0|0|0|0|0|0|0|0|0|0|0|B|6|0|0|G|Q|0|0^^$0|@$1|2|3|4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1F|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1G|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|1H|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|1I|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|1J|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|1K|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|1L|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1M|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|1N|8|@]|9|@]|A|$G|H]]|$1|W|3|X|5|6|7|1O|8|@$Y|1P|Z|1Q|10|11]]|9|@]|A|$]]|$1|12|3|13|5|F|7|1R|8|@]|9|@]|A|$G|H]]|$1|14|3|15|5|6|7|1S|8|@]|9|@$Y|1T|Z|1U|1|1V]]|A|$]]|$1|16|3|-4|5|6|7|1W|8|@]|9|@]|A|$]]]|17|$18|$5|19|1A|1B|A|$1C|1D]]]]

<h2>How to access last element of an array</h2>
It looks like that:
<pre class="lang-js prettyprint-override"><code>var my_array = /* some array here */;
var last_element = my_array[my_array.length - 1];
</code></pre>
Which in your case looks like this:
<pre class="lang-js prettyprint-override"><code>var array1 = loc['f096012e-2497-485d-8adb-7ec0b9352c52'];
var last_element = array1[array1.length - 1];
</code></pre>
or, in longer version, without creating new variables:
<pre class="lang-js prettyprint-override"><code>loc['f096012e-2497-485d-8adb-7ec0b9352c52'][loc['f096012e-2497-485d-8adb-7ec0b9352c52'].length - 1];
</code></pre>
<h2>How to add a method for getting it simpler</h2>
If you are a fan for creating functions/shortcuts to fulfill such tasks, the following code:
<pre><code>if (!Array.prototype.last){
 Array.prototype.last = function(){
 return this[this.length - 1];
 };
};
</code></pre>
will allow you to get the last element of an array by invoking array's <code>last()</code> method, in your case eg.:
<pre><code>loc['f096012e-2497-485d-8adb-7ec0b9352c52'].last();
</code></pre>
You can check that it works here: <a href="http://jsfiddle.net/D4NRN/" rel="noreferrer">http://jsfiddle.net/D4NRN/</a>

blocks|key|451125|text|var+last+=+function(+obj,+key+)+{+
++++var+a+=+obj[key];
++++return+a[a.length+-+1];
};

last(loc,+'f096012e-2497-485d-8adb-7ec0b9352c52');|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|451126|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>var last = function( obj, key ) { 
 var a = obj[key];
 return a[a.length - 1];
};

last(loc, 'f096012e-2497-485d-8adb-7ec0b9352c52');
</code></pre>

blocks|key|451141|text|使用slice()方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|451142|my_array.slice(-1)[0]|code-block|syntax|javascript|451143|entityMap^0|2|7|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|$]]

Use the <code>slice()</code> method:

<pre><code>my_array.slice(-1)[0]
</code></pre>

blocks|key|665341|text|var+last+=+array.slice(-1)[0];|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|665342|我发现slice+at+-1对于获取最后一个元素(尤其是长度未知的数组)很有用，而且它的性能比计算长度小于1要好得多。|unstyled|665343|Mozilla+Docs+on+Slice|offset|length|665344|Performance+of+the+various+methods+for+selecting+last+array+element|665345|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/slice|1|http://jsperf.com/get-last-item-from-array/13^0|0|0|0|L|0|0|0|1V|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|X|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|Y|8|@]|9|@$I|Z|J|10|1|11]]|A|$]]|$1|K|3|L|5|F|7|12|8|@]|9|@$I|13|J|14|1|15]]|A|$]]|$1|M|3|-4|5|F|7|16|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]|U|$5|P|Q|R|A|$S|V]]]]

<pre><code>var last = array.slice(-1)[0];
</code></pre>

I find slice at -1 useful for getting the last element (especially of an array of unknown length) and the performance is much better than calculating the length less 1.

<a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/slice">Mozilla Docs on Slice</a>

<a href="http://jsperf.com/get-last-item-from-array/13">Performance of the various methods for selecting last array element</a>

blocks|key|665422|text|var+arr+=+[1,+2,+3];
arr.slice(-1).pop();+//+return+3+and+arr+=+[1,+2,+3]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|665423|如果数组为空，这将返回undefined，这不会更改数组的值。|unstyled|665424|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>var arr = [1, 2, 3];
arr.slice(-1).pop(); // return 3 and arr = [1, 2, 3]
</code></pre>

This will return undefined if the array is empty and this will not change the value of the array.

blocks|key|451292|text|您可以在Array.prototype上定义getter|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|451293|​|451294|if+(!Array.prototype.hasOwnProperty("last"))+{
++Object.defineProperty(Array.prototype,+"last",+{
++++get()+{
++++++return+this[this.length+-+1];
++++}
++});
}

console.log([9,+8,+7,+6].last);+//+=>+6|code-block|syntax|javascript|451295|451296|如您所见，access看起来不像是函数调用；getter函数是在内部调用的。|451297|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/get^0|4|F|M|6|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@$9|10|A|11|1|12]]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|15|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|16|8|@]|D|@]|E|$]]|$1|P|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|Q|$R|$5|S|T|U|E|$V|W]]]]

You can define a <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/get" rel="nofollow noreferrer">getter</a> on <code>Array.prototype</code>:

<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>if (!Array.prototype.hasOwnProperty("last")) {
 Object.defineProperty(Array.prototype, "last", {
 get() {
 return this[this.length - 1];
 }
 });
}

console.log([9, 8, 7, 6].last); // =&gt; 6</code></pre>
</div>
</div>


As you can see, access doesn't look like a function call; the getter function is called internally.

blocks|key|451353|text|Underscore和Lodash具有_.last(Array)方法，该方法返回数组中的最后一个元素。它们的工作原理是一样的|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|451354|_.last([5,+4,+3,+2,+1]);
=>+1|code-block|syntax|javascript|451355|Ramda还有一个_.last函数|451356|R.last(['fi',+'fo',+'fum']);+//=>+'fum'|451357|entityMap|0|LINK|mutability|MUTABLE|url|http://underscorejs.org/#last|1|https://lodash.com/docs#last|2|http://ramdajs.com/0.19.1/docs/#last^0|J|D|0|A|0|B|6|1|0|0|9|6|0|5|2|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]|$9|16|A|17|1|18]]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|M|3|N|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]|Y|$5|R|S|T|E|$U|Z]]]]

<a href="http://underscorejs.org/#last">Underscore</a> and <a href="https://lodash.com/docs#last">Lodash</a> have the <code>_.last(Array)</code> method, that returns the last element in an Array. They both work about the same

<pre><code>_.last([5, 4, 3, 2, 1]);
=&gt; 1
</code></pre>

<a href="http://ramdajs.com/0.19.1/docs/#last">Ramda</a> also has a <code>_.last</code> function

<pre><code>R.last(['fi', 'fo', 'fum']); //=&gt; 'fum'
</code></pre>

blocks|key|665554|text|如果您使用的是ES6，您可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|665555|const+arr+=+[+1,+2,+3+];
[+...arr+].pop();+//+3
arr;+//+[+1,+2,+3+]+(wasn't+changed)|code-block|syntax|javascript|665556|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|$]]

In case you are using ES6 you can do:

<pre><code>const arr = [ 1, 2, 3 ];
[ ...arr ].pop(); // 3
arr; // [ 1, 2, 3 ] (wasn't changed)
</code></pre>

blocks|key|451411|text|这是可行的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|451412|array.reverse()[0]|code-block|syntax|javascript|451413|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|$]]

This worked:

<pre><code>array.reverse()[0]
</code></pre>

blocks|key|665699|text|所以，很多人都用pop()来回答，但他们中的大多数人似乎没有意识到这是一种破坏性的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|665700|var+a+=+[1,2,3]
a.pop()
//3
//a+is+now+[1,2]|code-block|syntax|javascript|665701|所以，对于一个非常愚蠢的，非破坏性的方法：|665702|var+a+=+[1,2,3]
a[a.push(a.pop())-1]
//3|665703|推送流行，就像90年代那样:)|665704|push将一个值追加到数组的末尾，并返回结果的长度。所以|665705|d=[]
d.push('life')+
//=>+1
d+
//=>['life']|665706|pop返回数组的最后一项的值，然后删除该索引处的值。所以|665707|c+=+[1,2,1]
c.pop()+
//=>+1
c+
//=>+[1,2]|665708|数组是0索引的，因此c.length+=>+3，cc.length+=>未定义(因为如果这样做，您将查找第四个值(此深度级别适用于任何倒霉的新手，最终在这里结束))。|665709|对于你的应用来说，可能不是最好的方法，或者甚至不是一个好的方法，比如流量、流失等等。但对于向下遍历一个数组，将其流式传输到另一个数组，只是使用低效的方法是愚蠢的，这。完全是这样的。|665710|entityMap^0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|17|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|19|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

So, a lot of people are answering with pop(), but most of them don't seem to realize that's a destructive method. 

<pre><code>var a = [1,2,3]
a.pop()
//3
//a is now [1,2]
</code></pre>

So, for a really silly, nondestructive method:

<pre><code>var a = [1,2,3]
a[a.push(a.pop())-1]
//3
</code></pre>

a push pop, like in the 90s :)

push appends a value to the end of an array, and returns the length of the result. so 

<pre><code>d=[]
d.push('life') 
//=&gt; 1
d 
//=&gt;['life']
</code></pre>

pop returns the value of the last item of an array, prior to it removing that value at that index. so 

<pre><code>c = [1,2,1]
c.pop() 
//=&gt; 1
c 
//=&gt; [1,2]
</code></pre>

arrays are 0 indexed, so c.length => 3, c[c.length] => undefined (because you're looking for the 4th value if you do that(this level of depth is for any hapless newbs that end up here)).

Probably not the best, or even a good method for your application, what with traffic, churn, blah. but for traversing down an array, streaming it onto another, just being silly with inefficient methods, this. Totally this.

I'm making an application that updates a user's location and path in real time and displays this on a Google Map. I have functionality that allows multiple users to be tracked at the same time using an object, which is updated every second. 

Right now, when a user pressed a button in the Android app, the coordinates are sent to a database and each time the location changes, a marker is updated on the map (and a polyline is formed). 

Since I have multiple users, I send a unique and randomly generated alphanumeric string so that I can display an individual path for each user. When the JS pulls this data from the database, it checks if the user exists, if it does not, it creates a new key with the value being a list. It would look something like this:

<pre><code>loc = {f096012e-2497-485d-8adb-7ec0b9352c52: [new google.maps.LatLng(39, -86),
 new google.maps.LatLng(38, -87),
 new google.maps.LatLng(37, -88)],
 44ed0662-1a9e-4c0e-9920-106258dcc3e7: [new google.maps.LatLng(40, -83),
 new google.maps.LatLng(41, -82),
 new google.maps.LatLng(42, -81)]}
</code></pre>

What I'm doing is storing a list of coordinates as the value of the key, which is the user's ID. My program keeps updating this list each time the location is changed by adding to the list (this works properly). 

What I need to do is update the marker's location each time the location changes. I would like to do this by selecting the last item in the array since that would be the last known location. Right now, each time the location is changed a new marker is added to the map (each one of the points in the example would show a marker at that location) so markers continue to be added. 

I would use a ´for (x in loc)` statement each time the location updates to grab the last location from the list and use that to update the marker. How do I select this last element from the array within the hash?

Selecting last element in JavaScript array

数据库

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我正在开发一个应用程序，可以实时更新用户的位置和路径，并将其显示在Google地图上。我有一个功能，允许使用一个对象同时跟踪多个用户，该对象每秒更新一次。现在，当用户在Android应用程序中按下一个按钮时，坐标就会被发送到数据库中，每次位置发生变化时，地图上的一个标记就会更新(并形成一条折线)。因为我有多个用户，所以...

问选择JavaScript数组中的最后一个元素
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