检查如果一个数组包含另一个数组的所有元素

问题描述：

我设计的电气工程应用。不过，我被困在此：
我有以下阵列

I'm designing an electrical engineering application. However, i'm stuck on this: I have the following array

<?php 
// Static Array
$GroupOfEight = array (
                      array(0,1,3,2,4,5,7,6),
                      array(4,5,6,7,16,12,13,14),
                      array(12,13,15,14,8,9,11,10),
                      array(2,6,14,10,3,7,15,11),
                      array(1,3,5,7,13,15,9,11),
                      array(0,4,12,8,1,5,13,9),
                      array(0,1,3,2,8,9,11,10)
                      );
?>

和我有另一个数组，但是这一个是一维的。

And I have another array, but this one is one dimensional.

<?php
$myStack = array(0,1,3,2,4,5,7,6); //Dynamic, gets value by POST method.
?>

我想要做的是检查是否$ myStack等于$ GroupOfEight数组的任何子阵列。（号码顺序并不重要。脚本应该只是检查是否每一个元素包含。如果它们的顺序是相同或不这并不重要。）

What I want to do is to check if $myStack is equal to any sub array of $GroupOfEight array. ( Number ordering is not important. The script should just check if every elements contained. It's not important if their order is same or not. )

下面是我做了什么来解决这个问题至今：

Here is what I've done to solve the issue so far:

<?php
//Check if stackArray contains 8group
for($i=0; $i<count($GroupOfEight);$i++)
for($j=0; $j<count($GroupOfEight[$i]); $j++){
    //$containsSearch = count(array_intersect($search_this,$all)) == count($search_this);
    $containsSearch = count(array_intersect($stackArray,$GroupOfEight[$j])) == count($stackArray);
    echo $containsSearch;
}
?>

请帮我纠正我的code或引进我这个问题的解决，
谢谢你。

Please help me correct my code or introduce me the solution of this issue, Thanks.

编辑：这应该只给1索引号。例如stackArray是0,1,3,2,4,1,2,3并且它应该发现GroupOfEight [N]匹配相同的号码，而不管数字的顺序的。我应该得到N个如果有匹配的情况。

It should give only 1 index number. for example stackArray is 0,1,3,2,4,1,2,3 and it should find GroupOfEight[N] that matches the same numbers, regardless of the order of the numbers. I should get the N if there is a matching case.

答

由于样品阵列，这样做的结果应该是：

Given your sample arrays, the output of this will be:

> 0

在情况下，你必须有唯一的一个数字输出，这应该做到这一点：

In case you HAD to have only one number output, this should do that:

<?php
//Check if stackArray contains 8group
$check=false;
for($i=0; $i<count($GroupOfEight);$i++){
    //$containsSearch = count(array_intersect($search_this,$all)) == count($search_this);
    $containsSearch = (count(array_intersect($stackArray,$GroupOfEight[$i])) == count($stackArray) && count(array_intersect($stackArray,$GroupOfEight[$i])) == count($GroupOfEight[$i]));
    if($containsSearch && !$check){
        echo $i; //This specifies which index in GroupOfEight contains a matching array
        $check=true;
    }
}
?>

编辑：制成的功能。返回第一个匹配的索引或-1为没有匹配：

Made a function. Returns first matched index or -1 for no matches:

function searcheight($stackArray,$GroupOfEight){
    for($i=0; $i<count($GroupOfEight);$i++){
        $containsSearch = (count(array_intersect($stackArray,$GroupOfEight[$i])) == count($stackArray) && count(array_intersect($stackArray,$GroupOfEight[$i])) == count($GroupOfEight[$i]));
        if($containsSearch){
            return $i; //This specifies which index in GroupOfEight contains a matching array
        }
    }
    return -1;
}
echo searcheight($stackArray,$GroupOfEight);

检查如果一个数组包含另一个数组的所有元素

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