Php “我有数据”;test1@gmail.com,test2@gmail.com&引用;如何使结果看起来像这样

Php “我有数据”;test1@gmail.com,test2@gmail.com&引用;如何使结果看起来像这样,php,codeigniter,Php,Codeigniter,我的代码: $read = $this->m_crud->read_data("subscribe","*"); $newData=array(); foreach($read as $key){ $newData[] = $key["email"]; } $to = implode(",",$newData); echo json_encode($to) 结果: "test1@gmail.com, t

我的代码:

  $read = $this->m_crud->read_data("subscribe","*");

   $newData=array();

      foreach($read as $key){
          $newData[] = $key["email"];
       }

    $to = implode(",",$newData);

    echo json_encode($to)
结果:

"test1@gmail.com, test2@gmail.com"
问题:如何使结果看起来像这样:

"test1@gmail.com", "test2@gmail.com"
只需删除内爆()函数。内爆()是PHP中的内置函数,用于连接数组的元素。建议内爆()的分隔符(,)参数始终使用两个参数以实现向后兼容性

<?php 
   $read = $this->m_crud->read_data("subscribe","*");

   $newData=array();

      foreach($read as $key){
          $newData[] = $key["email"];
       }
    echo json_encode($newData);
?> 


删除此
$to=内爆(“,”,$newData)
并更改此
echo json_encode($newData)
~结果
[”test1@gmail.com", "test2@gmail.com“]
在JS中,它是一个数组而不是字符串,因此您无需对其进行解析<代码>数据[0]或
数据[1]
等。如果您不是真的要求JSON,但实际上希望结果完全如图所示,那么答案将是使用分隔符
“,”
”对数组进行内爆,并在整个事件前后输出一个额外的