如何返回具有相同 id 的所有值的数组？

Question

这是我的代码：

function getPhoto($primary_key , $row){

   $this->db->select('id_report, path_photo');
   //$this->db->join('report', 'report.id = id_report');
   //$this->db->where('id_report', 'id_report');
   $this->db->from('report r, photo p');
   $where = "p.id_report = r.id AND id_report=".$row->id."";
   $this->db->where($where);
   $query = $this->db->get()->result_array();


  foreach ($query as $row2) {

     return " ".$row2['id_report']." - ".$row2['path_photo']."<br>";
     //echo " ".$row2['id_report']." - ".$row2['path_photo']."<br>";


}

function admin(){

      $crud->callback_column('POTHOS', array($this,'getPhoto')); 

.......

我需要从我的数据库中获取照片的路径，但一个报告可以有多张照片，所以我使用返回和回显：

（对于这种情况，我有更明确的 id 和 url）

回声：

1 - a.jpeg
1 - b.jpeg
2 - ....
3 - ....
4 - ....
5 - ....
6 - ....
7 - ....
8 - pic1.jpeg
8 - pic2.jpeg
9 - .....
10 - ....

返回：

1 - b.jpeg (Where is the other photo? a.jpeg)
2 - ......
3 - .....
4 - .....
5 - .....
6 - .....
7 - .....
8 - pic2.jpeg (pic1.jpeg?)
9 - .....
10 - .....

Answer 1

循环内不能有“return”语句...否则循环只会运行一次。

相反，创建一个连接字符串（使用 .=）或循环内的值数组，然后在循环后返回它：

$myString = '';

foreach ($query as $row2) {

    $myString .= " ".$row2['id_report']." - ".$row2['path_photo']."<br>";

}

return $myString;