如何在 PHP 中迭代数组时从单个类创建对象的不同实例？

Question

我有这个数组用于处理多对多形式并将其保存到数据库，但我的代码不起作用，因为只创建了一个对象实例，然后在迭代数组时重写：

$i = '1';
foreach ($myarray as $key => $value) {
  foreach ($value as $key2 => $value2) {
    $addressObject = new \Entities\Clientaddress();
    foreach ($value2 as $key3 => $value3) {
      $addressObject ->$key3 = $value3;
      $account->getAddresses()->add($addressObject);
      $this->em->persist($addressObject );
      $i = $i + '1';
}}}

如果我的方法是错误的，那么正确的方法是什么创建一个对象而不显式定义它？

Answer 1

您可以在循环中创建一个新的构造函数，这对于下一个编码器来说似乎“更干净”/更清晰，除非构造函数非常繁重，不需要重复。那会得到我的投票。如果您想要对象的副本，请使用clone：

  ....
  foreach ($value2 as $key3 => $value3) {
      $curAdress = clone $addressObject;
      $curAdress->$key3 = $value3;
      $account->getAddresses()->add($curAdress);
      $this->em->persist($curAdress);

Answer 2

创建这些对象的数组：

    $addressObject = array();
    $i = '1';
    foreach ($myarray as $key => $value) {
         foreach ($value as $key2 => $value2) {

             // create array of objects here
             $addressObject[$key2] = new \Entities\Clientaddress();

             foreach ($value2 as $key3 => $value3) {
                 $addressObject[$key2] ->$key3 = $value3;
                 $account->getAddresses()->add($addressObject[$key2]);
                 $this->em->persist($addressObject[$key2] );
                 $i = $i + '1';
             }
         }
    }

Answer 3

在最后的嵌套循环中创建对象。

$i = '1';

 foreach ($myarray as $key => $value)
{ 
 foreach ($value as $key2 => $value2)
 { 
  foreach ($value2 as $key3 => $value3)
 {
   //create your object here
   $addressObject = new \Entities\Clientaddress();  
   $addressObject ->$key3 = $value3; 
   $account->getAddresses()->add($addressObject); 
   $this->em->persist($addressObject ); 
   $i = $i + '1'; 
  }
 }
}