这是我当前使用的结构
模型用户:
class User extends Authenticatable implements MustVerifyEmail
public function groups()
{
return $this->belongsToMany('App\Group')
->withTimestamps();
}
模型组:
class Group extends Model
public function users() {
return $this->belongsToMany('App\User')
->withTimestamps();
}
数据透视表:
Schema::create('group_user', function (Blueprint $table) {
$table->integer('group_id');
$table->integer('user_id');
$table->integer('role_id')->nullable();
$table->timestamps();
});
我希望获得当前用户属于同一组的所有用户,并且不返回重复的用户(一个用户可以位于多个组中)。 理想的功能是:
$user->contacts()
// Return an object with (unique) contacts of all current user groups
AND
$user->groupContacts($group)
// Witch is the same as $group->users
// Return an object with all the users of the current group
我正在处理的功能不正常(模型User.php):
public function contacts()
{
$groups = $this->groups;
$contacts = new \stdClass();
foreach ($groups as $key => $group) :
$contacts->$key = $group->users;
endforeach;
return $contacts;
}
我真的不是表格结构方面的专家,因此,如果有更好的方法可以做到这一点,那么我只是在这个个人项目的开始,所以什么也没写。
可选内容:
答案 0 :(得分:1)
您可以在用户模型上尝试类似的操作
public function contacts()
{
// Get all groups of current user
$groups = $this->groups()->pluck('group_id', 'group_id');
// Get a collection of all users with the same groups
return $this->getContactsOfGroups($groups);
}
public function groupContacts($group)
{
// Get a collection of the contacts of the same group
return $this->getContactsOfGroups($group);
}
public function getContactsOfGroups($groups)
{
$groups = is_array($groups) ? $groups : [$groups];
// This will return a Collection Of Users that have $groups
return \App\User::whereHas('groups', function($query) use($groups){
$query->whereIn('group_id', $groups);
})->get();
}