blocks|key|577689|text|不要使用jquery...use纯javascript|type|unstyled|depth|inlineStyleRanges|entityRanges|data|577690|var+a+=+[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];

var+b+=+a.splice(0,10);

//a+is+now+[11,12,13,14,15];
//b+is+now+[1,2,3,4,5,6,7,8,9,10];|code-block|syntax|javascript|577691|你可以循环这段代码来获得你想要的行为。|577692|var+a+=+YOUR_ARRAY;
while(a.length)+{
++++console.log(a.splice(0,10));
}|577693|这会给你一个time...if的10个元素，比如说15个元素，你会得到1-10，11-15，这是你想要的。|577694|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Don't use jquery...use plain javascript

<pre><code>var a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];

var b = a.splice(0,10);

//a is now [11,12,13,14,15];
//b is now [1,2,3,4,5,6,7,8,9,10];
</code></pre>

You could loop this to get the behavior you want.

<pre><code>var a = YOUR_ARRAY;
while(a.length) {
 console.log(a.splice(0,10));
}
</code></pre>

This would give you 10 elements at a time...if you have say 15 elements, you would get 1-10, the 11-15 as you wanted.

blocks|key|793666|text|var+size+=+10;+var+arrayOfArrays+=+[];
for+(var+i=0;+i<bigarray.length;+i%2B=size)+{
+++++arrayOfArrays.push(bigarray.slice(i,i%2Bsize));
}
console.log(arrayOfArrays);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|793667|与splice()不同，slice()对原始数组是非破坏性的。|unstyled|offset|length|style|CODE|793668|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/Array/splice|1|https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/Array/slice^0|0|1|8|C|7|1|8|0|C|7|1|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@$G|W|H|X|I|J]|$G|Y|H|Z|I|J]]|9|@$G|10|H|11|1|12]|$G|13|H|14|1|15]]|A|$]]|$1|K|3|-4|5|F|7|16|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]|S|$5|N|O|P|A|$Q|T]]]]

<pre><code>var size = 10; var arrayOfArrays = [];
for (var i=0; i&lt;bigarray.length; i+=size) {
 arrayOfArrays.push(bigarray.slice(i,i+size));
}
console.log(arrayOfArrays);
</code></pre>

Unlike <a href="https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/Array/splice" rel="noreferrer"><code>splice()</code></a>, <a href="https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/Array/slice" rel="noreferrer"><code>slice()</code></a> is non-destructive to the original array.

blocks|key|577517|text|只需在数组上循环，拼接它，直到它全部用完。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|577518|var+a+=+['a','b','c','d','e','f','g']
++,+chunk

while+(a.length+>+0)+{

++chunk+=+a.splice(0,3)

++console.log(chunk)

}|code-block|syntax|javascript|577519|输出|577520|[+'a',+'b',+'c'+]
[+'d',+'e',+'f'+]
[+'g'+]|577521|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|$]]

Just loop over the array, splicing it until it's all consumed.

<pre><code>

var a = ['a','b','c','d','e','f','g']
 , chunk

while (a.length > 0) {

 chunk = a.splice(0,3)

 console.log(chunk)

}

</code></pre>

output

<pre><code>
[ 'a', 'b', 'c' ]
[ 'd', 'e', 'f' ]
[ 'g' ]

</code></pre>

blocks|key|577762|text|您可以使用lodash：https://lodash.com/docs|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|577763|_.chunk(['a',+'b',+'c',+'d'],+2);
//+→+[['a',+'b'],+['c',+'d']]|code-block|syntax|javascript|577764|entityMap|0|LINK|mutability|MUTABLE|url|https://lodash.com/docs^0|C|N|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

You can use lodash:
<a href="https://lodash.com/docs">https://lodash.com/docs</a>

<pre><code>_.chunk(['a', 'b', 'c', 'd'], 2);
// → [['a', 'b'], ['c', 'd']]
</code></pre>

blocks|key|793738|text|假设您不想销毁原始数组，您可以使用如下代码将长数组分解为较小的数组，然后可以迭代这些数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|793739|var+longArray+=+[];+++//+assume+this+has+100+or+more+email+addresses+in+it
var+shortArrays+=+[],+i,+len;

for+(i+=+0,+len+=+longArray.length;+i+<+len;+i+%2B=+10)+{
++++shortArrays.push(longArray.slice(i,+i+%2B+10));
}

//+now+you+can+iterate+over+shortArrays+which+is+an+
//+array+of+arrays+where+each+array+has+10+or+fewer+
//+of+the+original+email+addresses+in+it

for+(i+=+0,+len+=+shortArrays.length;+i+<+len;+i%2B%2B)+{
++++//+shortArrays[i]+is+an+array+of+email+addresss+of+10+or+less
}|code-block|syntax|javascript|793740|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|$]]

Assuming you don't want to destroy the original array, you can use code like this to break up the long array into smaller arrays which you can then iterate over:

<pre><code>var longArray = []; // assume this has 100 or more email addresses in it
var shortArrays = [], i, len;

for (i = 0, len = longArray.length; i &lt; len; i += 10) {
 shortArrays.push(longArray.slice(i, i + 10));
}

// now you can iterate over shortArrays which is an 
// array of arrays where each array has 10 or fewer 
// of the original email addresses in it

for (i = 0, len = shortArrays.length; i &lt; len; i++) {
 // shortArrays[i] is an array of email addresss of 10 or less
}
</code></pre>

blocks|key|794571|text|对于大型数组，尤其是使用mod运算符时，Array.reduce的效率可能很低。我认为一个更简洁(也可能更容易阅读)的函数解决方案应该是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|794572|const+chunkArray+=+(arr,+size)+=>
++arr.length+>+size
++++?+[arr.slice(0,+size),+...chunkArray(arr.slice(size),+size)]
++++:+[arr];|code-block|syntax|javascript|794573|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|$]]

Array.reduce could be inefficient for large arrays, especially with the mod operator. I think a cleaner (and possibly easier to read) functional solution would be this:

<pre><code>const chunkArray = (arr, size) =&gt;
 arr.length &gt; size
 ? [arr.slice(0, size), ...chunkArray(arr.slice(size), size)]
 : [arr];
</code></pre>

blocks|key|577737|text|作为对@jyore的answer的补充，以防您仍然想保留原始数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|577738|var+originalArray+=+[1,2,3,4,5,6,7,8];

var+splitArray+=+function+(arr,+size)+{

++var+arr2+=+arr.slice(0),
++++++arrays+=+[];

++while+(arr2.length+>+0)+{
++++++arrays.push(arr2.splice(0,+size));
++}

++return+arrays;
}

splitArray(originalArray,+2);
//+originalArray+is+still+=+[1,2,3,4,5,6,7,8];|code-block|syntax|javascript|577739|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/7273794/1778898^0|A|6|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

As a supplement to @jyore's <a href="https://stackoverflow.com/a/7273794/1778898">answer</a>, and in case you still want to keep the original array: 

<pre><code>var originalArray = [1,2,3,4,5,6,7,8];

var splitArray = function (arr, size) {

 var arr2 = arr.slice(0),
 arrays = [];

 while (arr2.length &gt; 0) {
 arrays.push(arr2.splice(0, size));
 }

 return arrays;
}

splitArray(originalArray, 2);
// originalArray is still = [1,2,3,4,5,6,7,8];
</code></pre>

blocks|key|794099|text|另一种实现方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|794100|const+arr+=+["H",+"o",+"w",+"+",+"t",+"o",+"+",+"s",+"p",+"l",+"i",+"t",+"+",+"a",+"+",+"l",+"o",+"n",+"g",+"+",+"a",+"r",+"r",+"a",+"y",+"+",+"i",+"n",+"t",+"o",+"+",+"s",+"m",+"a",+"l",+"l",+"e",+"r",+"+",+"a",+"r",+"r",+"a",+"y",+"s",+",",+"+",+"w",+"i",+"t",+"h",+"+",+"J",+"a",+"v",+"a",+"S",+"c",+"r",+"i",+"p",+"t"];

const+size+=+3;+
const+res+=+arr.reduce((acc,+curr,+i)+=>+{
++if+(+!(i+%25+size)++)+{++++//+if+index+is+0+or+can+be+divided+by+the+`size`...
++++acc.push(arr.slice(i,+i+%2B+size));+++//+..push+a+chunk+of+the+original+array+to+the+accumulator
++}
++return+acc;
},+[]);

//+=>+[["H",+"o",+"w"],+["+",+"t",+"o"],+["+",+"s",+"p"],+["l",+"i",+"t"],+["+",+"a",+"+"],+["l",+"o",+"n"],+["g",+"+",+"a"],+["r",+"r",+"a"],+["y",+"+",+"i"],+["n",+"t",+"o"],+["+",+"s",+"m"],+["a",+"l",+"l"],+["e",+"r",+"+"],+["a",+"r",+"r"],+["a",+"y",+"s"],+[",",+"+",+"w"],+["i",+"t",+"h"],+["+",+"J",+"a"],+["v",+"a",+"S"],+["c",+"r",+"i"],+["p",+"t"]]|code-block|syntax|javascript|794101|注意-这不会修改原始数组。|794102|或者，如果你更喜欢函数式的，100%25不可变的(尽管像上面那样就地变异没有什么不好的)和自包含的方法：|794103|function+splitBy(size,+list)+{
++return+list.reduce((acc,+curr,+i,+self)+=>+{
++++if+(+!(i+%25+size)++)+{++
++++++return+[
++++++++++...acc,
++++++++++self.slice(i,+i+%2B+size),
++++++++];
++++}
++++return+acc;
++},+[]);
}|794104|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Another implementation:

<pre class="lang-js prettyprint-override"><code>const arr = ["H", "o", "w", " ", "t", "o", " ", "s", "p", "l", "i", "t", " ", "a", " ", "l", "o", "n", "g", " ", "a", "r", "r", "a", "y", " ", "i", "n", "t", "o", " ", "s", "m", "a", "l", "l", "e", "r", " ", "a", "r", "r", "a", "y", "s", ",", " ", "w", "i", "t", "h", " ", "J", "a", "v", "a", "S", "c", "r", "i", "p", "t"];

const size = 3; 
const res = arr.reduce((acc, curr, i) =&gt; {
 if ( !(i % size) ) { // if index is 0 or can be divided by the `size`...
 acc.push(arr.slice(i, i + size)); // ..push a chunk of the original array to the accumulator
 }
 return acc;
}, []);

// =&gt; [["H", "o", "w"], [" ", "t", "o"], [" ", "s", "p"], ["l", "i", "t"], [" ", "a", " "], ["l", "o", "n"], ["g", " ", "a"], ["r", "r", "a"], ["y", " ", "i"], ["n", "t", "o"], [" ", "s", "m"], ["a", "l", "l"], ["e", "r", " "], ["a", "r", "r"], ["a", "y", "s"], [",", " ", "w"], ["i", "t", "h"], [" ", "J", "a"], ["v", "a", "S"], ["c", "r", "i"], ["p", "t"]]
</code></pre>

NB - This does not modify the original array.

Or, if you prefer a functional, 100% immutable (although there's really nothing bad in mutating in place like done above) and self-contained method:

<pre><code>function splitBy(size, list) {
 return list.reduce((acc, curr, i, self) =&gt; {
 if ( !(i % size) ) { 
 return [
 ...acc,
 self.slice(i, i + size),
 ];
 }
 return acc;
 }, []);
}
</code></pre>

blocks|key|794423|text|我也想分享我的解决方案。它有点冗长，但也很好用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|794424|var+data+=+[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];

var+chunksize+=+4;


var+chunks+=+[];

data.forEach((item)=>{
++if(!chunks.length+%7C%7C+chunks[chunks.length-1].length+==+chunksize)
++chunks.push([]);

++chunks[chunks.length-1].push(item);
});

console.log(chunks);|code-block|syntax|javascript|794425|输出(格式化)：|794426|[+[+1,++2,++3,++4],
++[+5,++6,++7,++8],
++[+9,+10,+11,+12],
++[13,+14,+15++++]+]|794427|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|$]]

I would like to share my solution as well. It's a little bit more verbose but works as well.

<pre><code>var data = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];

var chunksize = 4;


var chunks = [];

data.forEach((item)=&gt;{
 if(!chunks.length || chunks[chunks.length-1].length == chunksize)
 chunks.push([]);

 chunks[chunks.length-1].push(item);
});

console.log(chunks);
</code></pre>

Output (formatted):

<pre><code>[ [ 1, 2, 3, 4],
 [ 5, 6, 7, 8],
 [ 9, 10, 11, 12],
 [13, 14, 15 ] ]
</code></pre>

blocks|key|578035|text|另一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|578036|var+longArray+=+[1,+2,+3,+4,+5,+6,+7,+8,+9,+10];
var+size+=+2;

var+newArray+=+new+Array(Math.ceil(longArray.length+/+size)).fill("")
++++.map(function()+{+return+this.splice(0,+size)+},+longArray.slice());

//+newArray+=+[[1,+2],+[3,+4],+[5,+6],+[7,+8],+[9,+10]];|code-block|syntax|javascript|578037|这不会影响原始数组，因为使用slice制作的副本被传递到map的'+This‘参数中。|578038|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Another method:

<pre><code>var longArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
var size = 2;

var newArray = new Array(Math.ceil(longArray.length / size)).fill("")
 .map(function() { return this.splice(0, size) }, longArray.slice());

// newArray = [[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]];
</code></pre>

This doesn't affect the original array as a copy, made using slice, is passed into the 'this' argument of map.

blocks|key|578185|text|另一个实现，使用Array.reduce+(我认为它是唯一缺少的！)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|578186|const+splitArray+=+(arr,+size)+=>
{
++++if+(size+===+0)+{
++++++++return+[];
++++}

++++return+arr.reduce((split,+element,+index)+=>+{
++++++++index+%25+size+===+0+?+split.push([element])+:+split[Math.floor(index+/+size)].push(element);
++++++++return+split;
++++},+[]);
};|code-block|syntax|javascript|578187|正如上面的许多解决方案一样，这个解决方案是非破坏性的。当大小为0时返回一个空数组只是一种惯例。如果省略了if块，您会得到一个错误，这可能是您想要的结果。|offset|length|style|CODE|578188|entityMap^0|0|0|1G|2|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Another implementation, using Array.reduce (I think it’s the only one missing!):

<pre><code>const splitArray = (arr, size) =&gt;
{
 if (size === 0) {
 return [];
 }

 return arr.reduce((split, element, index) =&gt; {
 index % size === 0 ? split.push([element]) : split[Math.floor(index / size)].push(element);
 return split;
 }, []);
};
</code></pre>

As many solutions above, this one’s non-destructive. Returning an empty array when the size is 0 is just a convention. If the <code>if</code> block is omitted you get an error, which might be what you want.

blocks|key|794593|text|更紧凑：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|794594|​|794595|const+chunk+=+(xs,+size)+=>
++xs.map((_,+i)+=>
++++(i+%25+size+===+0+?+xs.slice(i,+i+%2B+size)+:+null)).filter(Boolean);
++++
//+Usage:
const+sampleArray+=+new+Array(33).fill(undefined).map((_,+i)+=>+i);

console.log(chunk(sampleArray,+5));|code-block|syntax|javascript|794596|794597|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|$]]

More compact:

<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="true">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const chunk = (xs, size) =&gt;
 xs.map((_, i) =&gt;
 (i % size === 0 ? xs.slice(i, i + size) : null)).filter(Boolean);
 
// Usage:
const sampleArray = new Array(33).fill(undefined).map((_, i) =&gt; i);

console.log(chunk(sampleArray, 5));</code></pre>
</div>
</div>

blocks|key|578917|text|const+originalArr+=+[1,2,3,4,5,6,7,8,9,10,11];
const+splittedArray+=+[];
++while+(originalArr.length+>+0)+{
++++splittedArray.push(originalArr.splice(0,range));++
++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|578918|范围3的输出|unstyled|578919|splittedArray+===+[[1,2,3][4,5,6][7,8,9][10,11]]|578920|范围4的输出|578921|splittedArray+===+[[1,2,3,4][5,6,7,8][9,10,11]]|578922|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>const originalArr = [1,2,3,4,5,6,7,8,9,10,11];
const splittedArray = [];
 while (originalArr.length &gt; 0) {
 splittedArray.push(originalArr.splice(0,range)); 
 }
</code></pre>
output for range 3
<pre><code>splittedArray === [[1,2,3][4,5,6][7,8,9][10,11]]
</code></pre>
output for range 4
<pre><code>splittedArray === [[1,2,3,4][5,6,7,8][9,10,11]]
</code></pre>

blocks|key|577807|text|你可以看看这段代码。简单而有效。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|577808|function+chunkArrayInGroups(array,+unit)+{
var+results+=+[],
length+=+Math.ceil(array.length+/+unit);

for+(var+i+=+0;+i+<+length;+i%2B%2B)+{
++++results.push(array.slice(i+*+unit,+(i+%2B+1)+*+unit));
}
+return+results;
}

chunkArrayInGroups(["a",+"b",+"c",+"d"],+2);|code-block|syntax|javascript|577809|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|$]]

You can take a look at this code . Simple and Effective . 

<pre><code>function chunkArrayInGroups(array, unit) {
var results = [],
length = Math.ceil(array.length / unit);

for (var i = 0; i &lt; length; i++) {
 results.push(array.slice(i * unit, (i + 1) * unit));
}
 return results;
}

chunkArrayInGroups(["a", "b", "c", "d"], 2);
</code></pre>

blocks|key|794383|text|下面是一个简单的线条|type|unstyled|depth|inlineStyleRanges|entityRanges|data|794384|​|794385|var+segment+=+(arr,+n)+=>+arr.reduce((r,e,i)+=>+i%25n+?+(r[r.length-1].push(e),+r)
++++++++++++++++++++++++++++++++++++++++++++++++++++:+(r.push([e]),+r),+[]),
++++++++arr+=+Array.from({length:+31}).map((_,i)+=>+i%2B1);
console.log(segment(arr,7));|code-block|syntax|javascript|794386|794387|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|$]]

Here is a simple one liner

<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 segment = (arr, n) =&gt; arr.reduce((r,e,i) =&gt; i%n ? (r[r.length-1].push(e), r)
 : (r.push([e]), r), []),
 arr = Array.from({length: 31}).map((_,i) =&gt; i+1);
console.log(segment(arr,7));</code></pre>
</div>
</div>

blocks|key|578650|text|作为函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|578651|var+arrayChunk+=+function+(array,+chunkSize)+{
++++++++++++var+arrayOfArrays+=+[];

++++++++++++if+(array.length+<=+chunkSize)+{
++++++++++++++++arrayOfArrays.push(array);
++++++++++++}+else+{
++++++++++++++++for+(var+i=0;+i<array.length;+i%2B=chunkSize)+{
++++++++++++++++++++arrayOfArrays.push(array.slice(i,i%2BchunkSize));
++++++++++++++++}
++++++++++++}
++++++++++++return+arrayOfArrays;
++++++++}|code-block|syntax|javascript|578652|要使用|578653|arrayChunk(originalArray,+10)+//10+being+the+chunk+size.|578654|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|$]]

as a function

<pre><code>var arrayChunk = function (array, chunkSize) {
 var arrayOfArrays = [];

 if (array.length &lt;= chunkSize) {
 arrayOfArrays.push(array);
 } else {
 for (var i=0; i&lt;array.length; i+=chunkSize) {
 arrayOfArrays.push(array.slice(i,i+chunkSize));
 }
 }
 return arrayOfArrays;
 }
</code></pre>

to use

<pre><code>arrayChunk(originalArray, 10) //10 being the chunk size.
</code></pre>

blocks|key|794797|text|使用递归|type|unstyled|depth|inlineStyleRanges|entityRanges|data|794798|let+myArr+=+[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16];
let+size+=+4;+//Math.sqrt(myArr.length);+-->+For+a+n+x+n+matrix
let+tempArr+=+[];
function+createMatrix(arr,+i)+{
++++if+(arr.length+!==+0)+{
++++++if(i+%25+size+==+0)+{
++++++++tempArr.push(arr.splice(0,size))
++++++}+
++++++createMatrix(arr,+i+-+1)
++++}
}
createMatrix(myArr,+myArr.length);
console.log(tempArr);|code-block|syntax|javascript|794799|注意:将修改已有的数组，即myArr。|794800|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

using recursion

<pre><code>let myArr = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16];
let size = 4; //Math.sqrt(myArr.length); --&gt; For a n x n matrix
let tempArr = [];
function createMatrix(arr, i) {
 if (arr.length !== 0) {
 if(i % size == 0) {
 tempArr.push(arr.splice(0,size))
 } 
 createMatrix(arr, i - 1)
 }
}
createMatrix(myArr, myArr.length);
console.log(tempArr);
</code></pre>

Note: The existing array i.e. myArr will be modified.

blocks|key|794824|text|使用prototype，我们可以直接设置数组类|type|unstyled|depth|inlineStyleRanges|entityRanges|data|794825|​|794826|Array.prototype.chunk+=+function(n)+{
++if+(!this.length)+{
++++return+[];
++}
++return+[this.slice(0,+n)].concat(this.slice(n).chunk(n));
};
console.log([1,+2,+3,+4,+5,+6,+7,+8,+9,+10,+11,+12,+13,+14,+15].chunk(5));|code-block|syntax|javascript|794827|794828|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|$]]

using prototype we can set directly to array class 

<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.chunk = function(n) {
 if (!this.length) {
 return [];
 }
 return [this.slice(0, n)].concat(this.slice(n).chunk(n));
};
console.log([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15].chunk(5));</code></pre>
</div>
</div>

blocks|key|794050|text|如果你想要一个不修改现有数组的方法，试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|794051|let+oldArray+=+[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];
let+newArray+=+[];
let+size+=+3;+//+Size+of+chunks+you+are+after
let+j+=+0;+//+This+helps+us+keep+track+of+the+child+arrays

for+(var+i+=+0;+i+<+oldArray.length;+i%2B%2B)+{
++if+(i+%25+size+===+0)+{
++++j%2B%2B
++}
++if(!newArray[j])+newArray[j]+=+[];
++newArray[j].push(oldArray[i])
}|code-block|syntax|javascript|794052|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|$]]

If you want a method that doesn't modify the existing array, try this:

<pre><code>let oldArray = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15];
let newArray = [];
let size = 3; // Size of chunks you are after
let j = 0; // This helps us keep track of the child arrays

for (var i = 0; i &lt; oldArray.length; i++) {
 if (i % size === 0) {
 j++
 }
 if(!newArray[j]) newArray[j] = [];
 newArray[j].push(oldArray[i])
}
</code></pre>

blocks|key|794449|text|function+chunkArrayInGroups(arr,+size)+{
++++var+newArr=[];

++++for+(var+i=0;+arr.length>size;+i%2B%2B){
++++newArr.push(arr.splice(0,size));
++++}
++++newArr.push(arr.slice(0));
++++return+newArr;

}

chunkArrayInGroups([0,+1,+2,+3,+4,+5,+6],+3);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|794450|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>function chunkArrayInGroups(arr, size) {
 var newArr=[];

 for (var i=0; arr.length&gt;size; i++){
 newArr.push(arr.splice(0,size));
 }
 newArr.push(arr.slice(0));
 return newArr;

}

chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6], 3);
</code></pre>

blocks|key|578497|text|function+chunkArrayInGroups(arr,+size)+{
++++var+newArr=[];

++++for+(var+i=0;+i+<+arr.length;+i%2B=+size){
++++newArr.push(arr.slice(i,i%2Bsize));
++++}
++++return+newArr;

}

chunkArrayInGroups([0,+1,+2,+3,+4,+5,+6],+3);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|578498|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>function chunkArrayInGroups(arr, size) {
 var newArr=[];

 for (var i=0; i &lt; arr.length; i+= size){
 newArr.push(arr.slice(i,i+size));
 }
 return newArr;

}

chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6], 3);
</code></pre>

blocks|key|794879|text|let+original+=+[1,+2,+3,+4,+5,+6,+7,+8,+9,+10,+11,+12,+13,+14];
let+size+=+5;
let+fragments+=+Array.from(Array(Math.ceil(a.length+/+size))).map((_,index)+=>+a.slice(index+*+size,(index+%2B+1)+*+size))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|794880|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>let original = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14];
let size = 5;
let fragments = Array.from(Array(Math.ceil(a.length / size))).map((_,index) =&gt; a.slice(index * size,(index + 1) * size))
</code></pre>

I have an array of e-mails (it can be just 1 email, or 100 emails), and I need to send the array with an ajax request (that I know how to do), but I can only send an array that has 10 or less e-mails in it. So if there is an original array of 20 e-mails I will need to split them up into 2 arrays of 10 each. or if there are 15 e-mails in the original array, then 1 array of 10, and another array of 5. I'm using jQuery, what would be the best way to do this?

How to split a long array into smaller arrays, with JavaScript

jQuery

我有一个电子邮件数组(可以只是1个电子邮件，也可以是100个电子邮件)，我需要用一个ajax请求发送这个数组(我知道怎么做)，但是我只能发送一个包含10个或更少电子邮件的数组。因此，如果有一个包含20封电子邮件的原始数组，我需要将它们分成2个数组，每个数组10个。或者，如果原始数组中有15封电子邮件，那么一个包含10封邮件的数组和另一个包含5封邮件的数组。我使用的是jQuery，那么最好的方法是什

问如何使用JavaScript将长数组拆分为较小的数组
EN

回答 22

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问如何使用JavaScript将长数组拆分为较小的数组EN

回答 22

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问如何使用JavaScript将长数组拆分为较小的数组
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