如何合併具有相同ID的三個數組的元素？

Question

我有3個數據庫查詢（查詢結果概述見下文！）。

我該如何將這三個數組的元素合併在一起，並且沒有重複的數組鍵和值，如下例所示：

期望的結果：

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

    [...]  
) 

我曾嘗試使用數組函數array_merge，array_merge_recursive，array_replace，array_replace_recursive和array_unique。但我無法清楚地將所有陣列合併在一起。查詢1的

---

結果（> 200000總數）：

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

    [...] 

) 

查詢3（1820個）的結果：：

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

    [...] 
) 

查詢2（1450個）的結果

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

    [...] 

) 

注意：請不要提出建議ba sed on 「SQL JOIN子句」。

我使用PHP5.6，Symfony2，Doctrine2和MySQL 5.5.52。

Answer 1

我能找到的大型陣列更好的解決方案。

首先通過自定義庫中獲取數據：

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

二發現的productID在所有陣列由array_search和array_column手段：

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

第三合併陣列一起：

$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]; 
       } 

      } 

Answer 2

像這樣遍歷每個數組，保存結果數組$o中的元素。

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

Answer 3

可以使用array_merge功能合併的結果，以澄清事情來看看下面的例子：

<?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);