blocks|key|569479|text|按照您的要求，根据javascript中字符串的长度，排序如下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|569480|用冒泡排序法解决问题|569481|1：http://jsfiddle.net/sssonline2/vcme3/2/enter+code+here|offset|length|style|CODE|569482|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/sssonline2/vcme3/2/^0|0|0|15|F|2|13|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|T|8|@$F|U|G|V|H|I]]|9|@$F|W|G|X|1|Y]]|A|$]]|$1|J|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

Here is the sort, depending on the length of a string with javascript as you asked:

[the solution of the problem by bubble sort][1]

[1]: <a href="http://jsfiddle.net/sssonline2/vcme3/2/" rel="nofollow noreferrer">http://jsfiddle.net/sssonline2/vcme3/2/</a><code>enter code here</code>

blocks|key|569765|text|#created+a+sorting+function+to+sort+by+length+of+elements+of+list
def+sort_len(a):
++++num+=+len(a)
++++d+=+{}
++++i+=+0
++++while+i<num:
++++++++d[i]+=+len(a[i])
++++++++i+%2B=+1
++++b+=+list(d.values())
++++b.sort()
++++c+=+[]
++++for+i+in+b:
++++++++for+j+in+range(num):
++++++++++++if+j+in+list(d.keys()):
++++++++++++++++if+d[j]+==+i:
++++++++++++++++++++c.append(a[j])
++++++++++++++++++++d.pop(j)
++++return+c|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|569766|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>#created a sorting function to sort by length of elements of list
def sort_len(a):
 num = len(a)
 d = {}
 i = 0
 while i&lt;num:
 d[i] = len(a[i])
 i += 1
 b = list(d.values())
 b.sort()
 c = []
 for i in b:
 for j in range(num):
 if j in list(d.keys()):
 if d[j] == i:
 c.append(a[j])
 d.pop(j)
 return c
</code></pre>

blocks|key|569802|text|如果希望保留与原始数组长度相同的元素顺序，请使用冒泡排序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|569803|Input+=+["ab","cdc","abcd","de"];

Output++=+["ab","cd","cdc","abcd"]|code-block|syntax|javascript|569804|功能：|569805|function+bubbleSort(strArray){
++const+arrayLength+=+Object.keys(strArray).length;
++++var+swapp;
++++var+newLen+=+arrayLength-1;
++++var+sortedStrArrByLenght=strArray;
++++do+{
++++++++swapp+=+false;
++++++++for+(var+i=0;+i+<+newLen;+i%2B%2B)
++++++++{
++++++++++++if+(sortedStrArrByLenght[i].length+>+sortedStrArrByLenght[i%2B1].length)
++++++++++++{
+++++++++++++++var+temp+=+sortedStrArrByLenght[i];
+++++++++++++++sortedStrArrByLenght[i]+=+sortedStrArrByLenght[i%2B1];
+++++++++++++++sortedStrArrByLenght[i%2B1]+=+temp;
+++++++++++++++swapp+=+true;
++++++++++++}
++++++++}
++++++++newLen--;
++++}+while+(swap);
++return+sortedStrArrByLenght;
}|569806|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|$]]

If you want to preserve the order of the element with the same length as the original array, use bubble sort.
<pre><code>Input = [&quot;ab&quot;,&quot;cdc&quot;,&quot;abcd&quot;,&quot;de&quot;];

Output = [&quot;ab&quot;,&quot;cd&quot;,&quot;cdc&quot;,&quot;abcd&quot;]
</code></pre>
Function:
<pre><code>function bubbleSort(strArray){
 const arrayLength = Object.keys(strArray).length;
 var swapp;
 var newLen = arrayLength-1;
 var sortedStrArrByLenght=strArray;
 do {
 swapp = false;
 for (var i=0; i &lt; newLen; i++)
 {
 if (sortedStrArrByLenght[i].length &gt; sortedStrArrByLenght[i+1].length)
 {
 var temp = sortedStrArrByLenght[i];
 sortedStrArrByLenght[i] = sortedStrArrByLenght[i+1];
 sortedStrArrByLenght[i+1] = temp;
 swapp = true;
 }
 }
 newLen--;
 } while (swap);
 return sortedStrArrByLenght;
}
</code></pre>

blocks|key|353945|text|根据Salman的回答，我编写了一个小函数来封装它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|353946|function+sortArrayByLength(arr,+ascYN)+{
++++++++arr.sort(function+(a,+b)+{+++++++++++//+sort+array+by+length+of+text
++++++++++++if+(ascYN)+return+a.length+-+b.length;++++++++++++++//+ASC+->+a+-+b
++++++++++++else+return+b.length+-+a.length;++++++++++++++++++++//+DESC+->+b+-+a
++++++++});
++++}|code-block|syntax|javascript|353947|然后使用以下命令调用它|353948|sortArrayByLength(+myArray,+true+);|353949|注意，不幸的是，函数可以/不应该添加到数组原型中，正如this+page上所解释的那样。|offset|length|353950|此外，它还修改了作为参数传递的数组，并且不返回任何内容。这将强制复制数组，并且对于大型数组来说不是很好。如果有人有更好的想法，请发表评论！|353951|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/948358/adding-custom-functions-into-array-prototype^0|0|0|0|0|R|9|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@$M|13|N|14|1|15]]|A|$]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

Based on Salman's answer, I've written a small function to encapsulate it: 

<pre><code>function sortArrayByLength(arr, ascYN) {
 arr.sort(function (a, b) { // sort array by length of text
 if (ascYN) return a.length - b.length; // ASC -&gt; a - b
 else return b.length - a.length; // DESC -&gt; b - a
 });
 }
</code></pre>

then just call it with

<pre><code>sortArrayByLength( myArray, true );
</code></pre>

Note that unfortunately, functions can/should not be added to the Array prototype, as explained on <a href="https://stackoverflow.com/questions/948358/adding-custom-functions-into-array-prototype">this page</a>.

Also, it modified the array passed as a parameter and doesn't return anything. This would force the duplication of the array and wouldn't be great for large arrays. If someone has a better idea, please do comment!

blocks|key|353997|text|我修改了@shareef的回答，使其简明扼要。我使用，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|353998|.sort(function(arg1,+arg2)+{+return+arg1.length+-+arg2.length+})|offset|length|style|CODE|353999|entityMap^0|0|0|1S|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

I adapted @shareef's answer to make it concise. I use,

<code>.sort(function(arg1, arg2) { return arg1.length - arg2.length })</code>

blocks|key|354035|text|下面的代码应该可以做到这一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|354036|var+array+=+["ab",+"abcdefgh",+"abcd"];

array.sort(function(a,+b){return+b.length+-+a.length});

console.log(JSON.stringify(array,+null,+'\t'));|code-block|syntax|javascript|354037|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 code should do the trick:

<pre class="lang-js prettyprint-override"><code>var array = ["ab", "abcdefgh", "abcd"];

array.sort(function(a, b){return b.length - a.length});

console.log(JSON.stringify(array, null, '\t'));
</code></pre>

blocks|key|569832|text|let+array+=+[`ab`,+`abcdefgh`,+`abcd`];
let+newArray+=+array.sort((a,b)+=>+{
++++return+b.length+-+a.length
})
console.log(newArray);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|569833|请输入以下代码|unstyled|569834|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>let array = [`ab`, `abcdefgh`, `abcd`];
let newArray = array.sort((a,b) =&gt; {
 return b.length - a.length
})
console.log(newArray);
</code></pre>
Please the following code

blocks|key|354185|text|let+arr++=+[5,2,100,1,20,3];
arr.sort((a,b)=>{
++return+a-b+
})

console.log(arr)+//[1,+2,+3,+5,+20,+100]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|354186|在返回值上，排序方法将执行交换元素的功能|unstyled|354187|return+<+0++{+i.e+-ve+number+then++a+comes+before+b}
return+>+0++{+i.e+%2Bve+number+then++b+comes+before+a}
return+==+0+{+order+of+a+and+b+remains+same+}|354188|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>let arr = [5,2,100,1,20,3];
arr.sort((a,b)=&gt;{
 return a-b 
})

console.log(arr) //[1, 2, 3, 5, 20, 100]
</code></pre>
on the return value, the sort method will perform the functionality of swapping of an elements
<pre><code>return &lt; 0 { i.e -ve number then a comes before b}
return &gt; 0 { i.e +ve number then b comes before a}
return == 0 { order of a and b remains same }
</code></pre>

blocks|key|569587|text|<script>
+++++++++arr+=+[]
+++++++++arr[0]+=+"ab"
+++++++++arr[1]+=+"abcdefgh"
+++++++++arr[2]+=+"sdfds"
+++++++++arr.sort(function(a,b){
++++++++++++return+a.length<b.length
+++++++++})
+++++++++document.write(arr)

</script>|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|569588|传递给array.hope的匿名函数告诉它如何对给定的排序进行排序，这很有帮助。我知道这很让人困惑，但您可以通过将一个函数作为参数传递给排序函数来告诉它如何对数组的元素进行排序|unstyled|569589|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>&lt;script&gt;
 arr = []
 arr[0] = "ab"
 arr[1] = "abcdefgh"
 arr[2] = "sdfds"
 arr.sort(function(a,b){
 return a.length&lt;b.length
 })
 document.write(arr)

&lt;/script&gt;
</code></pre>

The anonymous function that you pass to sort tells it how to sort the given array.hope this helps.I know this is confusing but you can tell the sort function how to sort the elements of the array by passing it a function as a parameter telling it what to do

I have an array like this:
<pre><code>arr = []
arr[0] = &quot;ab&quot;
arr[1] = &quot;abcdefgh&quot;
arr[2] = &quot;abcd&quot;
</code></pre>
After sorting, the output array should be:
<pre><code>arr[0] = &quot;abcdefgh&quot;
arr[1] = &quot;abcd&quot;
arr[2] = &quot;ab&quot; 
</code></pre>
I want in the <code>descending</code> order of the length of each element.

How to sort an array based on the length of each element?

JSON

我有一个这样的数组：arr = []arr[0] = "ab"arr[1] = "abcdefgh"arr[2] = "abcd"排序后，输出数组应为：arr[0] = "abcdefgh"arr[1] = "abcd"arr[2] = "ab"  我希望每个元素的长度按descending顺序排列。

问如何根据每个元素的长度对数组进行排序？
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回答 9

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问如何根据每个元素的长度对数组进行排序？EN

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问如何根据每个元素的长度对数组进行排序？
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