blocks|key|926493|text|你可以使用这里的解决方案http://stereofrog.com/blok/on/070910。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|926494|以防链接中断，代码如下……|926495|class+Combinations+implements+Iterator
{
++++protected+$c+=+null;
++++protected+$s+=+null;
++++protected+$n+=+0;
++++protected+$k+=+0;
++++protected+$pos+=+0;

++++function+__construct($s,+$k)+{
++++++++if(is_array($s))+{
++++++++++++$this->s+=+array_values($s);
++++++++++++$this->n+=+count($this->s);
++++++++}+else+{
++++++++++++$this->s+=+(string)+$s;
++++++++++++$this->n+=+strlen($this->s);
++++++++}
++++++++$this->k+=+$k;
++++++++$this->rewind();
++++}
++++function+key()+{
++++++++return+$this->pos;
++++}
++++function+current()+{
++++++++$r+=+array();
++++++++for($i+=+0;+$i+<+$this->k;+$i%2B%2B)
++++++++++++$r[]+=+$this->s[$this->c[$i]];
++++++++return+is_array($this->s)+?+$r+:+implode('',+$r);
++++}
++++function+next()+{
++++++++if($this->_next())
++++++++++++$this->pos%2B%2B;
++++++++else
++++++++++++$this->pos+=+-1;
++++}
++++function+rewind()+{
++++++++$this->c+=+range(0,+$this->k);
++++++++$this->pos+=+0;
++++}
++++function+valid()+{
++++++++return+$this->pos+>=+0;
++++}

++++protected+function+_next()+{
++++++++$i+=+$this->k+-+1;
++++++++while+($i+>=+0+&&+$this->c[$i]+==+$this->n+-+$this->k+%2B+$i)
++++++++++++$i--;
++++++++if($i+<+0)
++++++++++++return+false;
++++++++$this->c[$i]%2B%2B;
++++++++while($i%2B%2B+<+$this->k+-+1)
++++++++++++$this->c[$i]+=+$this->c[$i+-+1]+%2B+1;
++++++++return+true;
++++}
}


foreach(new+Combinations("1234567",+5)+as+$substring)
++++echo+$substring,+'+';|code-block|syntax|javascript|926496|12345+12346+12347+12356+12357+12367+12456+12457+12467+12567+13456+13457+13467+13567+14567+23456+23457+23467+23567+24567+34567|926497|entityMap|0|LINK|mutability|MUTABLE|url|http://stereofrog.com/blok/on/070910^0|C|10|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|Z|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

You can use the solution found here <a href="http://stereofrog.com/blok/on/070910" rel="noreferrer">http://stereofrog.com/blok/on/070910</a>.

Incase the link goes down here's the code....

<pre><code>class Combinations implements Iterator
{
 protected $c = null;
 protected $s = null;
 protected $n = 0;
 protected $k = 0;
 protected $pos = 0;

 function __construct($s, $k) {
 if(is_array($s)) {
 $this-&gt;s = array_values($s);
 $this-&gt;n = count($this-&gt;s);
 } else {
 $this-&gt;s = (string) $s;
 $this-&gt;n = strlen($this-&gt;s);
 }
 $this-&gt;k = $k;
 $this-&gt;rewind();
 }
 function key() {
 return $this-&gt;pos;
 }
 function current() {
 $r = array();
 for($i = 0; $i &lt; $this-&gt;k; $i++)
 $r[] = $this-&gt;s[$this-&gt;c[$i]];
 return is_array($this-&gt;s) ? $r : implode('', $r);
 }
 function next() {
 if($this-&gt;_next())
 $this-&gt;pos++;
 else
 $this-&gt;pos = -1;
 }
 function rewind() {
 $this-&gt;c = range(0, $this-&gt;k);
 $this-&gt;pos = 0;
 }
 function valid() {
 return $this-&gt;pos &gt;= 0;
 }

 protected function _next() {
 $i = $this-&gt;k - 1;
 while ($i &gt;= 0 &amp;&amp; $this-&gt;c[$i] == $this-&gt;n - $this-&gt;k + $i)
 $i--;
 if($i &lt; 0)
 return false;
 $this-&gt;c[$i]++;
 while($i++ &lt; $this-&gt;k - 1)
 $this-&gt;c[$i] = $this-&gt;c[$i - 1] + 1;
 return true;
 }
}


foreach(new Combinations("1234567", 5) as $substring)
 echo $substring, ' ';
</code></pre>

12345 12346 12347 12356 12357 12367 12456 12457 12467 12567 13456 13457 13467 13567 14567 23456 23457 23467 23567 24567 34567

blocks|key|926518|text|<?php

echo+"<pre>";
$test+=+array("test_1","test_2","test_3");

//+Get+Combination
$return+=+uniqueCombination($test);

//Sort
sort($return);

//Pretty+Print
print_r(array_map(function($v){+return+implode(",",+$v);+},+$return));

function+uniqueCombination($in,+$minLength+=+1,+$max+=+2000)+{
++++$count+=+count($in);
++++$members+=+pow(2,+$count);
++++$return+=+array();
++++for($i+=+0;+$i+<+$members;+$i+%2B%2B)+{
++++++++$b+=+sprintf("%250"+.+$count+.+"b",+$i);
++++++++$out+=+array();
++++++++for($j+=+0;+$j+<+$count;+$j+%2B%2B)+{
++++++++++++$b{$j}+==+'1'+and+$out[]+=+$in[$j];
++++++++}

++++++++count($out)+>=+$minLength+&&+count($out)+<=+$max+and+$return[]+=+$out;
++++++++}
++++return+$return;
}

?>|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|926519|输出|unstyled|926520|Array
(
++++[0]+=>+test_1
++++[1]+=>+test_2
++++[2]+=>+test_3
++++[3]+=>+test_1,test_2
++++[4]+=>+test_1,test_3
++++[5]+=>+test_2,test_3
++++[6]+=>+test_1,test_2,test_3
)|926521|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>&lt;?php

echo "&lt;pre&gt;";
$test = array("test_1","test_2","test_3");

// Get Combination
$return = uniqueCombination($test);

//Sort
sort($return);

//Pretty Print
print_r(array_map(function($v){ return implode(",", $v); }, $return));

function uniqueCombination($in, $minLength = 1, $max = 2000) {
 $count = count($in);
 $members = pow(2, $count);
 $return = array();
 for($i = 0; $i &lt; $members; $i ++) {
 $b = sprintf("%0" . $count . "b", $i);
 $out = array();
 for($j = 0; $j &lt; $count; $j ++) {
 $b{$j} == '1' and $out[] = $in[$j];
 }

 count($out) &gt;= $minLength &amp;&amp; count($out) &lt;= $max and $return[] = $out;
 }
 return $return;
}

?&gt;
</code></pre>

<h2>output</h2>

<pre><code>Array
(
 [0] =&gt; test_1
 [1] =&gt; test_2
 [2] =&gt; test_3
 [3] =&gt; test_1,test_2
 [4] =&gt; test_1,test_3
 [5] =&gt; test_2,test_3
 [6] =&gt; test_1,test_2,test_3
)
</code></pre>

blocks|key|926579|text|另一种基于堆栈的解决方案。它退出速度很快，但会消耗大量内存。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|926580|希望这能帮助到一些人。|926581|详细说明：|926582|function+_combine($numbers,+$length)
{
++++$combinations+=+array();
++++$stack+=+array();

++++//+every+combinations+can+be+ordered
++++sort($numbers);

++++//+startup
++++array_push($stack,+array(
++++++++'store'+=>+array(),
++++++++'options'+=>+$numbers,
++++));

++++while+(true)+{
++++++++//+pop+a+item
++++++++$item+=+array_pop($stack);

++++++++//+end+of+stack
++++++++if+(!$item)+{
++++++++++++break;
++++++++}

++++++++//+valid+store
++++++++if+($length+<=+count($item['store']))+{
++++++++++++$combinations[]+=+$item['store'];
++++++++++++continue;
++++++++}

++++++++//+bypass+when+options+are+not+enough
++++++++if+(count($item['store'])+%2B+count($item['options'])+<+$length)+{
++++++++++++continue;
++++++++}

++++++++foreach+($item['options']+as+$index+=>+$n)+{
++++++++++++$newStore+=+$item['store'];
++++++++++++$newStore[]+=+$n;

++++++++++++//+every+combine+can+be+ordered
++++++++++++//+so+accept+only+options+which+is+greater+than+store+numbers
++++++++++++$newOptions+=+array_slice($item['options'],+$index+%2B+1);

++++++++++++//+push+new+items
++++++++++++array_push($stack,+array(
++++++++++++++++'store'+=>+$newStore,
++++++++++++++++'options'+=>+$newOptions,
++++++++++++));
++++++++}
++++}

++++return+$combinations;
}|code-block|syntax|javascript|926583|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Another solution that bases on stack. It's quit fast but eats much memory.

Hope that helps someone.

In detail:

<pre><code>function _combine($numbers, $length)
{
 $combinations = array();
 $stack = array();

 // every combinations can be ordered
 sort($numbers);

 // startup
 array_push($stack, array(
 'store' =&gt; array(),
 'options' =&gt; $numbers,
 ));

 while (true) {
 // pop a item
 $item = array_pop($stack);

 // end of stack
 if (!$item) {
 break;
 }

 // valid store
 if ($length &lt;= count($item['store'])) {
 $combinations[] = $item['store'];
 continue;
 }

 // bypass when options are not enough
 if (count($item['store']) + count($item['options']) &lt; $length) {
 continue;
 }

 foreach ($item['options'] as $index =&gt; $n) {
 $newStore = $item['store'];
 $newStore[] = $n;

 // every combine can be ordered
 // so accept only options which is greater than store numbers
 $newOptions = array_slice($item['options'], $index + 1);

 // push new items
 array_push($stack, array(
 'store' =&gt; $newStore,
 'options' =&gt; $newOptions,
 ));
 }
 }

 return $combinations;
}
</code></pre>

blocks|key|1141130|text|一种优化组合算法速度和内存的新方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1141131|思维方式:生成数字数组的组合K个数字。新的解决方案将使用K+'for‘语句。一个“代表”一个数字。例如：$K+=5表示使用了5个'for‘语句|1141132|$total+=+count($array);
$i0+=+-1;
for+($i1+=+$i0+%2B+1;+$i1+<+$total;+$i1%2B%2B)+{
++++for+($i2+=+$i1+%2B+1;+$i2+<+$total;+$i2%2B%2B)+{
++++++++for+($i3+=+$i2+%2B+1;+$i3+<+$total;+$i3%2B%2B)+{
++++++++++++for+($i4+=+$i3+%2B+1;+$i4+<+$total;+$i4%2B%2B)+{
++++++++++++++++for+($i5+=+$i4+%2B+1;+$i5+<+$total;+$i5%2B%2B)+{
++++++++++++++++++++$record+=+array();
++++++++++++++++++++for+($i+=+1;+$i+<=+$k;+$i%2B%2B)+{
++++++++++++++++++++++++$t+=+"i$i";
++++++++++++++++++++++++$record[]+=+$array[$$t];
++++++++++++++++++++}
++++++++++++++++++++$callback($record);
++++++++++++++++}
++++++++++++}
++++++++}
++++}
}|code-block|syntax|javascript|1141133|以及生成将由eval()函数执行的真实代码的代码细节|1141134|function+combine($array,+$k,+$callback)
{
++++$total+=+count($array);
++++$init+=+'
++++++++$i0+=+-1;
++++';
++++$sample+=+'
++++++++for($i{current}+=+$i{previous}+%2B+1;+$i{current}+<+$total;+$i{current}%2B%2B+)+{
++++++++++++{body}
++++++++}
++++';

++++$do+=+'
++++++++$record+=+array();
++++++++for+($i+=+1;+$i+<=+$k;+$i%2B%2B)+{
++++++++++++$t+=+"i$i";
++++++++++++$record[]+=+$array[$$t];
++++++++}
++++++++$callback($record);
++++';
++++$for+=+'';
++++for+($i+=+$k;+$i+>=+1;+$i--)+{
++++++++switch+($i)+{
++++++++++++case+$k:
++++++++++++++++$for+=+str_replace(['{current}',+'{previous}',+'{body}'],+[$i,+$i+-+1,+$do],+$sample);
++++++++++++++++break;
++++++++++++case+1:
++++++++++++++++$for+=+$init+.+str_replace(['{current}',+'{previous}',+'{body}'],+[$i,+$i+-+1,+$for],+$sample);
++++++++++++++++break;
++++++++++++default:
++++++++++++++++$for+=+str_replace(['{current}',+'{previous}',+'{body}'],+[$i,+$i+-+1,+$for],+$sample);
++++++++++++++++break;
++++++++}
++++}

++++//+execute
++++eval($for);
}|1141135|如何组合K个数组|1141136|$k+=+5;
$array+=+array(1,+2,+3,+4,+5,+6,+7);
$callback+=+function+($record)+{
++++echo+implode($record)+.+"\n";
};
combine($array,+$k,+$callback);|1141137|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

New solution which optimizes speed and memory for combining algorithm

Mindset: generate combinations K numbers of Array of numbers. New solution will use K 'for' statements. One 'for' One number.
Such as: $K = 5 mean that 5 of 'for' statements is used

<pre><code>$total = count($array);
$i0 = -1;
for ($i1 = $i0 + 1; $i1 &lt; $total; $i1++) {
 for ($i2 = $i1 + 1; $i2 &lt; $total; $i2++) {
 for ($i3 = $i2 + 1; $i3 &lt; $total; $i3++) {
 for ($i4 = $i3 + 1; $i4 &lt; $total; $i4++) {
 for ($i5 = $i4 + 1; $i5 &lt; $total; $i5++) {
 $record = array();
 for ($i = 1; $i &lt;= $k; $i++) {
 $t = "i$i";
 $record[] = $array[$$t];
 }
 $callback($record);
 }
 }
 }
 }
}
</code></pre>

And detail of code that generated the real code that will be execute by eval() function

<pre><code>function combine($array, $k, $callback)
{
 $total = count($array);
 $init = '
 $i0 = -1;
 ';
 $sample = '
 for($i{current} = $i{previous} + 1; $i{current} &lt; $total; $i{current}++ ) {
 {body}
 }
 ';

 $do = '
 $record = array();
 for ($i = 1; $i &lt;= $k; $i++) {
 $t = "i$i";
 $record[] = $array[$$t];
 }
 $callback($record);
 ';
 $for = '';
 for ($i = $k; $i &gt;= 1; $i--) {
 switch ($i) {
 case $k:
 $for = str_replace(['{current}', '{previous}', '{body}'], [$i, $i - 1, $do], $sample);
 break;
 case 1:
 $for = $init . str_replace(['{current}', '{previous}', '{body}'], [$i, $i - 1, $for], $sample);
 break;
 default:
 $for = str_replace(['{current}', '{previous}', '{body}'], [$i, $i - 1, $for], $sample);
 break;
 }
 }

 // execute
 eval($for);
}
</code></pre>

How to combine K numbers of Array

<pre><code>$k = 5;
$array = array(1, 2, 3, 4, 5, 6, 7);
$callback = function ($record) {
 echo implode($record) . "\n";
};
combine($array, $k, $callback);
</code></pre>

blocks|key|926643|text|改进了此answer，使其也能与关联数组一起使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|926644|function+uniqueCombination($values,+$minLength+=+1,+$maxLength+=+2000)+{
++++$count+=+count($values);
++++$size+=+pow(2,+$count);
++++$keys+=+array_keys($values);
++++$return+=+[];

++++for($i+=+0;+$i+<+$size;+$i+%2B%2B)+{
++++++++$b+=+sprintf("%250"+.+$count+.+"b",+$i);
++++++++$out+=+[];

++++++++for($j+=+0;+$j+<+$count;+$j+%2B%2B)+{
++++++++++++if+($b[$j]+==+'1')+{
++++++++++++++++$out[$keys[$j]]+=+$values[$keys[$j]];
++++++++++++}
++++++++}

++++++++if+(count($out)+>=+$minLength+&&+count($out)+<=+$maxLength)+{
+++++++++++++$return[]+=+$out;
++++++++}
++++}

++++return+$return;
}|code-block|syntax|javascript|926645|例如：|926646|print_r(uniqueCombination([
++++'a'+=>+'xyz',
++++'b'+=>+'pqr',
]);|926647|结果：|926648|Array
(
++++[0]+=>+Array
++++++++(
++++++++++++[b]+=>+pqr
++++++++)

++++[1]+=>+Array
++++++++(
++++++++++++[a]+=>+xyz
++++++++)

++++[2]+=>+Array
++++++++(
++++++++++++[a]+=>+xyz
++++++++++++[b]+=>+pqr
++++++++)

)|926649|它仍然适用于非关联数组：|926650|print_r(uniqueCombination(['a',+'b']);|926651|926652|Array
(
++++[0]+=>+Array
++++++++(
++++++++++++[1]+=>+b
++++++++)

++++[1]+=>+Array
++++++++(
++++++++++++[0]+=>+a
++++++++)

++++[2]+=>+Array
++++++++(
++++++++++++[0]+=>+a
++++++++++++[1]+=>+b
++++++++)

)|926653|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/38871855/14066311^0|4|6|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|D|3|E|5|F|7|19|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1A|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|1B|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|1C|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|1D|8|@]|9|@]|C|$G|H]]|$1|Q|3|R|5|6|7|1E|8|@]|9|@]|C|$]]|$1|S|3|T|5|F|7|1F|8|@]|9|@]|C|$G|H]]|$1|U|3|N|5|6|7|1G|8|@]|9|@]|C|$]]|$1|V|3|W|5|F|7|1H|8|@]|9|@]|C|$G|H]]|$1|X|3|-4|5|6|7|1I|8|@]|9|@]|C|$]]]|Y|$Z|$5|10|11|12|C|$13|14]]]]

Improved this <a href="https://stackoverflow.com/a/38871855/14066311">answer</a> to work with associative array as well:
<pre class="lang-php prettyprint-override"><code>function uniqueCombination($values, $minLength = 1, $maxLength = 2000) {
 $count = count($values);
 $size = pow(2, $count);
 $keys = array_keys($values);
 $return = [];

 for($i = 0; $i &lt; $size; $i ++) {
 $b = sprintf(&quot;%0&quot; . $count . &quot;b&quot;, $i);
 $out = [];

 for($j = 0; $j &lt; $count; $j ++) {
 if ($b[$j] == '1') {
 $out[$keys[$j]] = $values[$keys[$j]];
 }
 }

 if (count($out) &gt;= $minLength &amp;&amp; count($out) &lt;= $maxLength) {
 $return[] = $out;
 }
 }

 return $return;
}
</code></pre>
Eg:
<pre class="lang-php prettyprint-override"><code>print_r(uniqueCombination([
 'a' =&gt; 'xyz',
 'b' =&gt; 'pqr',
]);
</code></pre>
Result:
<pre><code>Array
(
 [0] =&gt; Array
 (
 [b] =&gt; pqr
 )

 [1] =&gt; Array
 (
 [a] =&gt; xyz
 )

 [2] =&gt; Array
 (
 [a] =&gt; xyz
 [b] =&gt; pqr
 )

)
</code></pre>
It will still work for non-associative arrays:
<pre class="lang-php prettyprint-override"><code>print_r(uniqueCombination(['a', 'b']);
</code></pre>
Result:
<pre><code>Array
(
 [0] =&gt; Array
 (
 [1] =&gt; b
 )

 [1] =&gt; Array
 (
 [0] =&gt; a
 )

 [2] =&gt; Array
 (
 [0] =&gt; a
 [1] =&gt; b
 )

)
</code></pre>

blocks|key|926679|text|我需要一个包含子集的组合函数，所以我采用了@Nguyen+Van+Vinh的答案，并根据需要对其进行了修改。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|926680|如果将[1,2,3,4]传递给函数，它将按顺序返回每个唯一的组合和子集：|926681|[
++[1,2,3,4],+[1,2,3],+[1,2,4],+[1,3,4],+[2,3,4],+[1,2],+[1,3],+[1,4],+[2,3],+[2,4],+[3,4],+[1],+[2],+[3],+[4]
]|code-block|syntax|javascript|926682|下面是函数：|926683|function+get_combinations_with_length(+$numbers,+$length+){
++++$result+=+array();
++++$stack+=+array();
++++//+every+combinations+can+be+ordered
++++sort($numbers);
++++//+startup
++++array_push($stack,+array(
++++++++'store'+=>+array(),
++++++++'options'+=>+$numbers,
++++));
++++while+(true)+{
++++++++//+pop+a+item
++++++++$item+=+array_pop($stack);
++++++++//+end+of+stack
++++++++if+(!$item)+break;
++++++++//+valid+store
++++++++if+($length+<=+count($item['store']))+{
++++++++++++$result[]+=+$item['store'];
++++++++++++continue;
++++++++}
++++++++//+bypass+when+options+are+not+enough
++++++++if+(count($item['store'])+%2B+count($item['options'])+<+$length)+{
++++++++++++continue;
++++++++}
++++++++foreach+($item['options']+as+$i=>$n)+{
++++++++++++$newStore+=+$item['store'];
++++++++++++$newStore[]+=+$n;
++++++++++++//+every+combine+can+be+ordered,+so+accept+only+options+that+are+greater+than+store+numbers
++++++++++++$newOptions+=+array_slice($item['options'],+$i+%2B+1);
++++++++++++//+array_unshift+to+sort+numerically,+array_push+to+reverse
++++++++++++array_unshift($stack,+array(
++++++++++++++++'store'+=>+$newStore,
++++++++++++++++'options'+=>+$newOptions,
++++++++++++));
++++++++}
++++}
++++return+$result;
}

function+get_all_combinations(+$numbers+){
++++$length+=+count($numbers);
++++$result+=+[];
++++while+($length+>+0)+{
++++++++$result+=+array_merge($result,+get_combinations_with_length(+$numbers,+$length+));
++++++++$length--;
++++}
++++return+$result;
}


$numbers+=+[1,2,3,4];
$result+=+get_all_combinations($numbers);

echo+'START:+'.json_encode(+$numbers+).' ';
echo+'RESULT:+'.json_encode(+$result+).' ';
echo+'('.count($result).'+combination+subsets+found)';|926684|entityMap^0|L|G|0|3|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

I needed a combining function that included subsets, so I took <code>@Nguyen Van Vinh</code>'s answer and modified it for my needs.
If you pass <code>[1,2,3,4]</code> to the function, it returns every unique combination and subset, sorted:
<pre><code>[
 [1,2,3,4], [1,2,3], [1,2,4], [1,3,4], [2,3,4], [1,2], [1,3], [1,4], [2,3], [2,4], [3,4], [1], [2], [3], [4]
]
</code></pre>
Here's the function:
<pre class="lang-php prettyprint-override"><code>function get_combinations_with_length( $numbers, $length ){
 $result = array();
 $stack = array();
 // every combinations can be ordered
 sort($numbers);
 // startup
 array_push($stack, array(
 'store' =&gt; array(),
 'options' =&gt; $numbers,
 ));
 while (true) {
 // pop a item
 $item = array_pop($stack);
 // end of stack
 if (!$item) break;
 // valid store
 if ($length &lt;= count($item['store'])) {
 $result[] = $item['store'];
 continue;
 }
 // bypass when options are not enough
 if (count($item['store']) + count($item['options']) &lt; $length) {
 continue;
 }
 foreach ($item['options'] as $i=&gt;$n) {
 $newStore = $item['store'];
 $newStore[] = $n;
 // every combine can be ordered, so accept only options that are greater than store numbers
 $newOptions = array_slice($item['options'], $i + 1);
 // array_unshift to sort numerically, array_push to reverse
 array_unshift($stack, array(
 'store' =&gt; $newStore,
 'options' =&gt; $newOptions,
 ));
 }
 }
 return $result;
}

function get_all_combinations( $numbers ){
 $length = count($numbers);
 $result = [];
 while ($length &gt; 0) {
 $result = array_merge($result, get_combinations_with_length( $numbers, $length ));
 $length--;
 }
 return $result;
}


$numbers = [1,2,3,4];
$result = get_all_combinations($numbers);

echo 'START: '.json_encode( $numbers ).'&lt;br&gt;&lt;br&gt;';
echo 'RESULT: '.json_encode( $result ).'&lt;br&gt;&lt;br&gt;';
echo '('.count($result).' combination subsets found)';
</code></pre>

I have an array of 7 numbers (1,2,3,4,5,6,7) and I want to choose 5 of the numbers like 
(1,2,3,4,5), (1,2,3,4,6), (1,2,3,4,7). 
Note that (1,2,3,4,5) is equal to (4,5,3,1,2), so only one of those should be included in the output.
I would like to know if there is a function in PHP or any algorithm that can do this ?
I have no idea where to start from.
Can you help me ?
I want all the combinations of 7 given numbers ( they are taken from an array ) put into 5 slots, disregarding order.

PHP array combinations

我有一个由7个数字(1,2,3,4,5,6,7)组成的数组，我想选择其中的5个数字，如下所示(1,2,3,4,5)，(1,2,3,4,6)，(1,2,3,4,7)。请注意，(1,2,3,4,5)等于(4,5,3,1,2)，因此输出中应该只包含其中的一个。我想知道PHP中是否有一个函数或任何算法可以做到这一点？我不知道从...

问PHP数组组合
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