Php 如何将数组与子元素合并

我有一个具有相同

customerid

的数组。我想将所有相同的

customerid

数组合并到一个数组中，只需对数组进行少量修改

Array
(
    [0] => Array
        (
            [customerid] => 13
            [customer_fullname] => Chris
            [profession_id] => 8
            [profession_name] => Producer
         )

    [1] => Array
        (
            [customerid] => 1
            [customer_fullname] => John
            [profession_id] => 8
            [profession_name] => Producer

        )

    [2] => Array
        (
            [customerid] => 13
            [customer_fullname] => Chris
            [profession_id] => 7
            [profession_name] => Camera

        )

)

现在我想创建一个新数组，如下所示：

Array(
    [customerid] => 13
    [customer_fullname] => Chris
    [new_array] => array(
        [0]=>[profession_id] => 8, [profession_name] => Producer, 
        [1]=>[profession_id] => 7, [profession_name] => Camera
    )
)

---

花了一些时间在这上面，但没能把它弄好

如果要合并很多记录，有更好的方法，但是如果您想找到一种方法，只需按照说明合并两个记录，我会这样做：

$array1 = array(
      'customerid' => 13
      'customer_fullname' => 'John',
      'profession_id' => 8,
      'profession_name' => 'Producer'
);
$array2 = array(
      'customerid' => 13
      'customer_fullname' => 'John',
      'profession_id' => 7,
      'profession_name' => 'Director'
);

function merge_customers($customerA, $customerB)
{
    $newCustomer = array();
    if ($customerA['customerid'] == $customerB['customerid'])
    {
        $newCustomer['customerid'] = $customerA['customerid'];
        $newCustomer['customer_fullname'] = $customerA['customer_fullname'];

        $newCustomer['new_array'] = array(
            array(
                'profession_id' => $customerA['profession_id'],
                'profession_name' => $customerA['profession_name']
            ),
            array(
                'profession_id' => $customerB['profession_id'],
                'profession_name' => $customerB['profession_name']
            )
        );

        return $newCustomer;
    }

    /* We can't merge these if they're different customers. */
    return NULL;
}

---

如果要合并很多记录，有更好的方法，但如果您想找到一种方法，只需按照说明合并两个记录，我会这样做：

$array1 = array(
      'customerid' => 13
      'customer_fullname' => 'John',
      'profession_id' => 8,
      'profession_name' => 'Producer'
);
$array2 = array(
      'customerid' => 13
      'customer_fullname' => 'John',
      'profession_id' => 7,
      'profession_name' => 'Director'
);

function merge_customers($customerA, $customerB)
{
    $newCustomer = array();
    if ($customerA['customerid'] == $customerB['customerid'])
    {
        $newCustomer['customerid'] = $customerA['customerid'];
        $newCustomer['customer_fullname'] = $customerA['customer_fullname'];

        $newCustomer['new_array'] = array(
            array(
                'profession_id' => $customerA['profession_id'],
                'profession_name' => $customerA['profession_name']
            ),
            array(
                'profession_id' => $customerB['profession_id'],
                'profession_name' => $customerB['profession_name']
            )
        );

        return $newCustomer;
    }

    /* We can't merge these if they're different customers. */
    return NULL;
}

---

快速破解，也许有更好的解决方案。 注意：第二个“for each”循环仅在数组可能没有相同字段时才需要

function merge($array1, $array2){
     $result = array();
     foreach($array1 as $key => $value){
         if(isset($array2[$key]) && $array2[$key]!=$array1[$key]){
              $result[$key][]=$value;
              $result[$key][]=$array2[$key];
         }else{
              $result[$key]=$value;
         }
     }
     foreach($array2 as $key => $value){
        if(!isset($result[$key])){
             $result[$key] = $value;
        }
     }
     return $result;
}

print_r(merge($array1, $array2));

---

快速破解，也许有更好的解决方案。 注意：第二个“for each”循环仅在数组可能没有相同字段时才需要

function merge($array1, $array2){
     $result = array();
     foreach($array1 as $key => $value){
         if(isset($array2[$key]) && $array2[$key]!=$array1[$key]){
              $result[$key][]=$value;
              $result[$key][]=$array2[$key];
         }else{
              $result[$key]=$value;
         }
     }
     foreach($array2 as $key => $value){
        if(!isset($result[$key])){
             $result[$key] = $value;
        }
     }
     return $result;
}

print_r(merge($array1, $array2));

---

该扩展解决方案也非常适合查找和“合并”具有相同

customerid的多组条目。使用的功能：
array\u filter
，
array\u count\u value
，
array\u key
，
array\u walk
，
array\u chunk
和
array\u value
：

// supposing $arr is your initial array

// finds which 'customerid' has multiple entries
$dupIds = array_filter(array_count_values(array_column($arr, "customerid")), function($v) {
    return $v > 1;
});

$dupIds = array_keys($dupIds);
$result = [];

array_walk($arr, function($v) use(&$result, $dupIds) {
    if (in_array($v['customerid'], $dupIds)) {
        $parts = array_chunk($v, 2, true);
        if (!isset($result[$v['customerid']])) {
            $result[$v['customerid']] = $parts[0] + ['new_array' => [$parts[1]]];
        } else {
            $result[$v['customerid']]['new_array'][] = $parts[1];
        }
    }
});

print_r(array_values($result));

输出：

Array
(
    [0] => Array
        (
            [customerid] => 13
            [customer_fullname] => Chris
            [new_array] => Array
                (
                    [0] => Array
                        (
                            [profession_id] => 8
                            [profession_name] => Producer
                        )

                    [1] => Array
                        (
                            [profession_id] => 7
                            [profession_name] => Camera
                        )
                )
        )
)

该扩展解决方案也非常适合查找和“合并”具有相同
customerid的多组条目。使用的功能：
array\u filter
，
array\u count\u value
，
array\u key
，
array\u walk
，
array\u chunk
和
array\u value
：

// supposing $arr is your initial array

// finds which 'customerid' has multiple entries
$dupIds = array_filter(array_count_values(array_column($arr, "customerid")), function($v) {
    return $v > 1;
});

$dupIds = array_keys($dupIds);
$result = [];

array_walk($arr, function($v) use(&$result, $dupIds) {
    if (in_array($v['customerid'], $dupIds)) {
        $parts = array_chunk($v, 2, true);
        if (!isset($result[$v['customerid']])) {
            $result[$v['customerid']] = $parts[0] + ['new_array' => [$parts[1]]];
        } else {
            $result[$v['customerid']]['new_array'][] = $parts[1];
        }
    }
});

print_r(array_values($result));

输出：

Array
(
    [0] => Array
        (
            [customerid] => 13
            [customer_fullname] => Chris
            [new_array] => Array
                (
                    [0] => Array
                        (
                            [profession_id] => 8
                            [profession_name] => Producer
                        )

                    [1] => Array
                        (
                            [profession_id] => 7
                            [profession_name] => Camera
                        )
                )
        )
)

实际上是的……可能有多个阵列，而不仅仅是2个。有什么更好的方法吗？我会在几分钟内添加，当然。根据您的更新，我会按照@RomanPerekhrest建议的方式进行：）实际上是的……可能有多个数组，而不仅仅是2个。有什么更好的方法吗？我会在几分钟内添加它，当然。根据您的更新，我会按照@RomanPerekhrest建议的方式进行：）您想为第一个customerid omly还是为所有客户构建阵列？我正在合并所有具有相同customerid的阵列，所以在我们的例子中，有两个customerid具有相同的id，所以我们将其合并。那么[customerid]=>1是什么呢？我们忽略了这一点，因为它没有任何匹配的customerid。您是要为第一个customerid omly还是所有customerid构建阵列？我将合并所有具有相同customerid的阵列，在我们的例子中，有两个customerid具有相同的id，所以我们将其合并。那么[customerid]=>1是什么呢？我们忽略了这一点，因为它没有任何匹配的customerid…但是可能有n个相同的customerid…但是可能有n个相同的customerid