Question

此时我可以在选定的组合框中显示人物的姓名，但我想将person_id存储在数据库中而不是人名。我怎样才能做到这一点。

控制器

公共职能指数（） {

  $this->Relation->save($this->request->data);
  $this->loadModel('Profile');
  $person= $this->Profile->find('all');
    if($this->request->is('post')){
    }
    $temp = array();
$temp1 = array();

    foreach($person as $person_name)
    {
      $temp[ $person_name['Profile']['id']] =  $person_name['Profile']['id'];
      $temp1[ $person_name['Profile']['first_name']] = $person_name['Profile']['first_name'];

    }
    $this->set('person_name', $temp);
    $this->set('person_id', $temp1);
}

查看index.ctp

<?php echo $this->Form->input('person_id',array('multiple'=>true,'label'=>false,'type'=>'select','options'=>$person_name,'selected'=>$person_id)); ?>

Answer 1

在您的视图中查看index.ctp

$data = array_combine($person_id,$person_name);

如果您将$ data打印为pr（$ data）;

Array
(
[54af7764-dcf8-4355-bcc7-3980c2f436a7] => sandeep
[54af77bb-fe28-4634-a3f9-3980c2f436a7] => prakash

)

所以你选择的组合框将是

<?php echo $this->Form->input('person_id',array('label'=>false,'type'=>'select','options'=>$data)); ?>

如何将值传递给cakephp中的选定组合框？

1 个答案: