PHP：比较数组的值

Question

$a1 = array('A', 'A', 'A', 'A', 'B', 'C');
$a2 = array('A', 'A', 'A', 'A', 'B', 'C');

我想将 $a1 中的第一个 'A' 与 $a2 中的最后一个 'A' 进行匹配。以及 $a1 中的第二个 'A' 以及 $a1 中的 倒数第二个 'A' >$a2。

$a1 和 $a2，如下所示：

0 => 'A' with 3 => 'A' 
1 => 'A' with 2 => 'A'
2 => 'A' with 1 => 'A'
3 => 'A' with 0 => 'A' 

我需要一个 if 语句来确定 中是否存在另一个 'A' 实例$a1 将其与其在 $a2 中的相应值进行匹配

Answer 1

function myMatch($vala, $valb) {
/*this can make your custom matching equality, length in case of strings, types what ever
returns true if values $vala and $valb match and false if they do not
*/
}

$matches = array();

foreach($a1 as $key=>$value) {

   if(isset($a2[$key])) {

   $matches[$key] = myMatch($a1[$key],$a2[$key]) ? 
"matches for ".$key." with value ".$value : "no match for ".$key." with ".$value;

   } else {

      $matches[$key] = "Missing pair in a2 for key ".$key;

   }

}


print_R($matches);

Answer 2

如果您只想按照您描述的相反顺序匹配“A”，您可以尝试以下操作：

$a1 = array('A', 'A', 'A', 'A', 'B', 'C');
$a2 = array('A', 'A', 'A', 'A', 'B', 'C');

$j=count($a2)-1;
for($i=0;$i<count($a1);$i++){
    while($j>=0){
        if($a2[$j]=='A'){
            echo "Match ($i,$j)";
            $j--;
            break;
        }
        else
            $j--;
    }
}

Answer 3

我的建议：

<?php

$a1 = array('A', 'A', 'A', 'A', 'B', 'C');
$a2 = array('A', 'A', 'A', 'A', 'B', 'C');

$a3 = getInversedArrayIndexMatch($a1, $a2);
print_r($a3);

if(getInversedArrayIndexMatch($a1, $a2, 3)) {
    echo 'Value with index 3 exists in a2';
}

/** Function **/
function getInversedArrayIndexMatch($a1, $a2, $indexSearch = null)
{
    $arrWithInversedMatches = array();
    $inversedArray = array_reverse($a2);
    ksort($inversedArray);

    $len1 = count($a1)-1;
    $len2 = count($a2)-1;
    for($i=0; $i<=$len1; $i++) {
        $valNeedle = $a1[$i];

        $foundKeys = array_keys($inversedArray, $valNeedle);
        if(is_array($foundKeys) && count($foundKeys) > 0) {
            $valueIndex = count($foundKeys)-1;
            $inversedIndex = $foundKeys[$valueIndex];
            $removeIndex = array_search($valNeedle, $inversedArray);
            array_splice($inversedArray, $removeIndex, 1, "");
            if(!isset($indexSearch)) {
                $arrWithInversedMatches[] = "Index $i with value $valNeedle in Array1 matches the $valueIndex. $valNeedle in inversed Array2 at index ". ($len2 - $removeIndex) ."";
            } elseif ($indexSearch == $i) {
                return true;
            }         
        } else {
            $arrWithInversedMatches[] = "Index $i with value $valNeedle in Array1 has no match in inversed Array2";
        }
    }
    if(isset($indexSearch)) {
        return false;        
    }
    return $arrWithInversedMatches;
}

？>

输出：

Array
(
    [0] => Index 0 with value A in Array1 matches the 3. A in inversed Array2 at index 3
    [1] => Index 1 with value A in Array1 matches the 2. A in inversed Array2 at index 2
    [2] => Index 2 with value A in Array1 matches the 1. A in inversed Array2 at index 1
    [3] => Index 3 with value A in Array1 matches the 0. A in inversed Array2 at index 0
    [4] => Index 4 with value B in Array1 matches the 0. B in inversed Array2 at index 4
    [5] => Index 5 with value C in Array1 matches the 0. C in inversed Array2 at index 5
)

Value with index 3 exists in a2

Answer 4

这正是你所解释的。它不太完整，但它比较了您所解释的两个数组并且更紧凑。

$a1 = array('A', 'A', 'A', 'A', 'B', 'C');
$a2 = array('A', 'A', 'A', 'A', 'B', 'C');

$a2_aux = array_reverse($a2);
//reset internal pointer
reset($a1);
reset($a2);

while (list($pos_a1, $value_a1) = each($a1)) 
{
    list($pos_a2, $value_a2) = each($a2_aux);

    if($value_a1 == $value_a2)
        echo 'At position '.$pos_a1. ' the value '.$value_a1.' in a1 array, is in a2 array. Because'.$value_a1. '='.$value_a2."\n" ;
    else
        echo 'At position '.$pos_a1. ' the value '.$value_a1.' in a1 array, is NOT in a2 array. Because '.$value_a1. '!='.$value_a2."\n";
}