如何合并具有相同 ID 的三个数组的元素？

Question

我有 3 个数据库查询（查询结果概览见下文！）。

如何合并这三个数组的元素（DB 中有超过 200.000 个产品），而不会出现重复的数组键和值，如下例所示？

想要的结果：

Array
(
    [0] => Array
        (
            [id] => 42710
            [imageName] => somthing
            [image] => https://www.somthing.com/image.jpg
            [loyality] => 122                
            [p_num] => 8
            [cpc] => 2
        )

    [...]    
)

我尝试过使用数组函数array_merge、array_merge_recursive、array_replace、array_replace_recursive 和 array_unique。但我无法将所有数组清晰地合并在一起。

---

查询 1 的结果（>200000 总）：

Array
(
    [0] => Array
        (
            [id] => 42710
            [imageName] => somthing
            [image] => https://www.somthing.com/image.jpg
        )

     [...]
)

查询 2 的结果（共 1450 个）：

Array
(
    [0] => Array
        (
            [id] => 42710
            [loyality] => 122
        )

    [...]

)

查询 3 的结果（共 1820 个）：

Array
(
    [0] => Array
        (
            [id] => 42710
            [p_num] => 8
            [cpc] => 2
        )

    [...]

)

注意：请不要根据“SQL JOIN 子句”提出建议。

我正在使用 PHP5.6、Symfony2、Doctrine2 和 MySQL 5.5.52。

Answer 1

你可以使用array_merge函数来合并结果，看下面的例子来澄清一下：

<?php

$resultSet1=[['id' => '32','name' => 'john'], ['id' => '15','name' => 'amin']];
$resultSet2=[['id' => '32','lastname' => 'doe'],['id' => '15','lastname' => 'alizade']];
$resultSet3=[['id' => '32','age' => '28'], ['id' => '15','age' => '25']];

$resultSet = array_merge($resultSet1, $resultSet2, $resultSet3);

$result = [];
foreach ($resultSet as $record) {   
    $key = $record['id'];
    if (array_key_exists($key, $result)) {
        $result[$key] = array_merge($result[$key], $record);
    }else{
        $result[$key] = $record;
    }
}

var_dump($result);

Answer 2

这样遍历每个数组，将元素保存在结果数组$o中。

$o = [];
array_walk($array1, function($v) use (&$o) {
    foreach($v as $key => $value) {
        $o[$v['id']][$key] = $value;
    }
});

Answer 3

我可以为大型数组找到更好的解决方案。

首先通过自定义存储库获取数据：

$this->productAttributes = $this->manager->getRepository('ProductBundle:Rating')->findAllAttributeOfProducts();
$this->productOnlineDate = $this->manager->getRepository('ProductBundle:Rating')->findOnlineDate();
$this->numberOfClicks    = $this->manager->getRepository('ProductBundle:Rating')->findNumberOfClicks();

第二次通过array_search和array_column在所有数组中查找productID：

$findNumberOfClicks = array_search($this->arrayKey['id'], array_column($this->numberOfClicks, 'id'));
$findOnlineDate     = array_search($this->arrayKey['id'], array_column($this->productOnlineDate, 'id'));

第三次合并数组：

$this->productAttributes = $this->manager->getRepository('ProductBundle:Rating')->findAllAttributeOfProducts();
$this->productOnlineDate = $this->manager->getRepository('ProductBundle:Rating')->findOnlineDate();
$this->numberOfClicks    = $this->manager->getRepository('ProductBundle:Rating')->findNumberOfClicks();

$this->arrayMergedResult = [];

            foreach ($this->productAttributes as $this->product => $this->arrayKey) {

                // find product ID in all arrays
                $findNumberOfClicks = array_search($this->arrayKey['id'], array_column($this->numberOfClicks, 'id'));
                $findOnlineDate     = array_search($this->arrayKey['id'], array_column($this->productOnlineDate, 'id'));

                // merge arrays
                switch (true) {
                    case ($findOnlineDate === false && $findNumberOfClicks === false):
                        $this->arrayMergedResult = $this->arrayKey;
                        break;
                    case ($findOnlineDate === false):
                        $this->arrayMergedResult = $this->arrayKey + $this->numberOfClicks[$findNumberOfClicks];
                        break;
                    case ($findNumberOfClicks === false):
                        $this->arrayMergedResult = $this->arrayKey + $this->productOnlineDate[$findOnlineDate];
                        break;
                    default:
                        $this->arrayMergedResult = $this->arrayKey + $this->numberOfClicks[$findNumberOfClicks] + $this->productOnlineDate[$findOnlineDate];
                }

            }