Question

来自网址：Count number of a group array in array in php

我问问题：

 $arr1 = [61,41,41,61,89,90]
 $arr2 = [61,41]
 $result = 2    

 //found 61 and 41 in $arr1  2 time; 
 //This mean : found 61 and 41 in $arr1[0] and $arr1[1] 
 //and found 61 and 41 in $arr1[3] and $arr1[2] again
 //So  $result = 2

我按照答案。那是代码：

 $arr1 = array(61,41,41,61,89,90);
 $arr2 = array(61,41);

 $count = array_count_values($arr1); //count values from arr1

 $result = array();
 foreach($arr2 as $row) {
    $result[$row] = array_key_exists($row, $count) ? $count[$row] : 0;
 }

 echo min($result);

但它有一个错误。如果我假设

 $arr1 = array(5,6,5,6,5,7);
 $arr2 = array(5,5);
 $result = 1; 

 //This mean : found 5 and 5 in $arr[0] and $arr[2]
 // but is'not found 5 and 5 again

真正的结果是1.但是这个结果是3。请帮我解决这个问题。

Answer 1

您可以使用array-intersect()和array_unique()

执行此操作

http://php.net/manual/en/function.array-intersect.php

http://php.net/manual/en/function.array-unique.php

$tmp = array_intersect($arr1, $arr2);
$tmp = array_unique($tmp);
$result = count($res);

Answer 2

只需添加条件，因为在这种情况下你必须匹配两次

  echo ($arr2[0] == $arr2[1]) ? floor($result[0] /2) : min($result);

修复bug：在php中计算数组中组数组的数量

2 个答案: