如何在URL中添加用戶名而不是cakephp 3中的user_id

Question

我希望我的網址像這樣的格式http：// localhost / blog / users / username而不是這個http：// localhost / blog / users / view / 6

我在用戶視圖index.ctp中有此代碼

<?php foreach ($users as $user): ?>
<?= $this->Html->link(__('View Profile'), ['action' => 'view', $user['user']['slug']]) ?>
<?php endforeach; ?>

routes.php文件

<?php
 $routes->connect('/user/*', array('controller' => 'users', 'action' => 'view'));
?>


//public function view($id = null)
public function view($username)
{

    $users = $this->Users->get($username, [
        'contain' => ['Subjects'] // i have relation
    ]);
    $this->set('users', $users);
    $this->set('_serialize', ['user']);
}

我嘗試了此鏈接，但沒有解決我的問題

public function edit($id = null)
{
  //$logged_user_id=$this->request->Session()->read('Auth.user.id');


  $logged_user_id=$this->Auth->user('id');
  if($logged_user_id==$id){
      $user = $this->Users->get($id, [
        'contain' => []
    ]);
      if ($this->request->is(['patch', 'post', 'put'])) {
        $user = $this->Users->patchEntity($user, $this->request->getData());


        if ($this->Users->save($user)) {
            $this->Flash->success(__('User profile successfuly  updated.'));
            return $this->redirect(['action' => 'index']);
        } else {
            $this->Flash->error(__('The user could not be saved. Please, try again.'));
        }

    }
    $this->set(compact('user'));
    $this->set('_serialize', ['user']);
} else {
    $this->Flash->error(__('You are not allowed to do this.'));
    return $this->redirect(['action' => 'index']);
}

}

Answer 1

在index.ctp中

<?php foreach ($users as $user): ?>
<?= $this->Html->link(__('View Profile'), ['action' => 'view', $user->username]) ?>
<?php endforeach; ?>

請根據您的結構更改$user->username 。

您無需在routs.php中做任何事情

用戶名將作為函數視圖的參數接收

function view($username){
    //Your code
}

Answer 2

get函數使用模型的主鍵字段。 可能可以將主鍵更改為username ，但是我懷疑這會給您帶來其他問題。 相反，請嘗試以下操作：

$users = $this->Users->find('first')
    ->where(['username' => $username])
    ->contain(['Subjects']);

另外，您的變量是否為復數（ $users ）是有原因的嗎？ 您應該只從中獲得一個用戶，對嗎？