Question

我有一群看起来像这样的人：

$people =
    Array
    (
        [0] => Array
            (
                [email] => NameSurname@example.com
                [name] => Name Surname
                [count] => 0
            )

        [1] => Array
            (
                [email] => Name2Surname@example.com
                [name] => Name2 Surname
                [count] => 0
            )
    )

我有一个数组，它是MySQL查询的结果，如下所示：

$query=
    Array
    (
        [0] => Array
            (
                [email] => NameSurname@example.com
                [name] => Name Surname
            )

        [1] => Array
            (
                [email] => Name2Surname@example.com
                [name] => Name2 Surname
            )

        [2] => Array
            (
                [email] => NameSurname@example.com
                [name] => Name Surname
            )

    )

对于$ people中的每个电子邮件地址，我希望$people['count']等于$ query中该电子邮件地址的出现次数。

我已经尝试了许多方法来做到这一点，而且我没有得到理想的结果。

为避免疑义，基于上述示例的最终结果应如下所示：

$people =
    Array
    (
        [0] => Array
            (
                [email] => NameSurname@example.com
                [name] => Name Surname
                [count] => 2
            )

        [1] => Array
            (
                [email] => Name2Surname@example.com
                [name] => Name2 Surname
                [count] => 1
            )
    )

Answer 1

foreach ($people as $key => $man) { // iterating through each man
  $_occurences = 0;

  foreach ($query as $_item) // iterating through each result in the result set
    if ($_item['email'] == $man['email']) // comparing current man to each result item
      $_occurences ++;

  $people[$key]['count'] = $_occurences; // saving number of occurrences in the `count` key 

 }

UPD：使用array_map，array_reduce和不错ternary运算符的另一个解决方案。它比foreach慢，但紧凑和专业两倍。

它更慢（在性能方面）因为。函数调用开销，但在“小”迭代量上，这种下降可以忽略不计。它将更新$people数组而不重新分配它。我们通过引用传输了每个&$man。

array_map(function (&$man) use ($query) {
    $man['count'] = array_reduce($query, function ($count, $row) use ($man) {
        return ($row['email'] === $man['email']) ? ++$count : $count ;
    }, 0);
}, $people);

Answer 2

不确定这是否正是您所需要的。但您可以更改正在运行的查询：

SELECT COUNT(name) as cnt, name,email FROM table_name GROUP BY name;

然后$query数组将如下所示：

$query=
    Array
    (
        [0] => Array
            (
                [email] => NameSurname@example.com
                [name] => Name Surname
                [cnt] => 2
            )

        [1] => Array
            (
                [email] => Name2Surname@example.com
                [name] => Name2 Surname
                [cnt] => 1
            )                  
    )

然后您可以在数组上运行，只需获取cnt数组中每个项目的$query。

计算出现在另一个数组中的一个数组中值的出现次数

2 个答案: