Question

我正在尝试在 Phalcon框架中加入两个表（MYSQL）。

＆＃34; 表格＆＃34;名称：

user
contact_details

＆＃34;的模型＆＃34;名称：

User
ContactDetails

初始化：

In "User" model ----->$this->hasMany("id", "ContactDetails", "user_id"); 
In "ContactDetails" model----->$this->belongsTo("user_id", "User", "id");

我的代码：

public function userDetails(){             
      $phql = "SELECT User.*, ContactDetails.* FROM User LEFT JOIN ContactDetails";
      $user = $this->modelsManager->executeQuery($phql);
      -------
      -------Remaining code---------
      --------------
      }
Thanks in advance.

Answer 1

感谢您的快速回复。

我实际上需要运行MySQL查询而不是使用内置函数。

这是我最终得到的解决方案：

$query = "SELECT u.*, c.* FROM user u "
              . "LEFT JOIN ContactDetails c "
              . "ON c.user_id = u.id ORDER BY u.id ";
      $cars = $this->modelsManager->executeQuery($query);
      $user = array();      
      foreach ($cars as $car) {
        //print_r($car);exit;
          $user[$car->u->id]['firstname'] = $car->u->firstname;
          $user[$car->u->id]['lastname'] = $car->u->lastname;
          $user[$car->u->id]['username'] = $car->u->username;
          $user[$car->u->id]['emailid'] = $car->u->emailid;
          $user[$car->u->id]['address'] = $car->c->address;
          $user[$car->u->id]['street'] = $car->c->street;
          $user[$car->u->id]['city'] = $car->c->city;
      }

Answer 2

这没有经过测试，但它可能会给你一个想法，你实际上并不需要使用$ phql，因为模型应该能够容纳你需要的一切。

$users = User::find();

foreach($users as $user) {
    // whatever you want
    echo $user->name;

    foreach($user->contactdetails as $contactDetails) {
        echo $contactDetails->telephone;
    }
}

您可以从文档中获取所有这些信息 - http://docs.phalconphp.com/en/latest/reference/models.html

如何在PHALCON FRAMEWORK中连接多个表

2 个答案: