PHP foreach遍历数组追加键值后原数组无新增字段问题排查

现有代码如下：

$campaignUserModel = new CampaignUserModel();
$subcontractors = $campaignUserModel
    ->select('campaigns_users.*, users.fullname user_fullname')
    ->join('users', 'users.id = campaigns_users.user_id')
    ->where('campaign_id', $id)
    ->findAll();

$postModel = new PostModel();
foreach ($subcontractors as $subcontractor) {
    $subcontractor_posts = $postModel
        ->where('user_id', $subcontractor['id'])
        ->where('campaign_id', $id)
        ->countAllResults();

    $subcontractor['posts_count'] = $subcontractor_posts;

    echo "<pre>";
    var_dump($subcontractor);
    echo "</pre>";
}

echo "<pre>";
var_dump($subcontractors);
echo "</pre>";

现象：循环内打印单个$subcontractor时能看到新增的posts_count字段，循环结束后打印整个$subcontractors数组，所有元素都没有该字段。

PHP的foreach默认按值传递遍历，循环中使用的$subcontractor是原数组对应元素的拷贝副本，不是原元素本身。你在循环内对副本的所有修改，都不会同步到原数组上，所以循环结束后原数组自然不会保留新增的字段。

有两种常用修复方式，按需选择即可：

• 方案1：遍历使用引用传递，直接操作原数组元素
  注意遍历结束后必须unset掉引用变量，避免后续逻辑误修改数组最后一个元素。

  foreach ($subcontractors as &$subcontractor) {
      $subcontractor_posts = $postModel
          ->where('user_id', $subcontractor['id'])
          ->where('campaign_id', $id)
          ->countAllResults();
  
      $subcontractor['posts_count'] = $subcontractor_posts;
  }
  unset($subcontractor); // 释放引用
  

• 方案2：通过键名直接修改原数组，不需要用引用，更不容易出隐性bug

  foreach ($subcontractors as $key => $subcontractor) {
      $subcontractor_posts = $postModel
          ->where('user_id', $subcontractor['id'])
          ->where('campaign_id', $id)
          ->countAllResults();
      // 直接定位原数组对应键的元素修改
      $subcontractors[$key]['posts_count'] = $subcontractor_posts;
  }
  

内容的提问来源于stack exchange，提问作者Lidia K.