Question

更新＃2 - 解决方案： 结果我使用了这个查找：

$this->User->Group->find(....)

不是我需要的。要拉出我需要使用的用户组：

$this->User->find('all',array('conditions' => array('User.id' => $user_id)));

＆LT; / UPDATE＃2＆gt;＆lt;问题＆GT;

我正在尝试在Users表和Groups表之间建立HABTM关系。问题是，当我发出这个电话时我：

$this->User->Group->find('list');

发出的查询是：

SELECT [Group].[id] AS [Group__id], [Group].[name] AS [Group__name] FROM [groups] AS [Group] WHERE 1 = 1

我只能假设我已经定义了我的关系错误，因为我希望行为使用按照约定在数据库上定义的groups_users表。我的人际关系：

class User extends AppModel {
        var $name = 'User';
        //...snip...
    var $hasAndBelongsToMany = array(
        'Group' => array(
            'className'             => 'Group',
            'foreignKey'            => 'user_id',
            'associationForeignKey' => 'group_id',
            'joinTable'             => 'groups_users',
            'unique'                => true,
        )
    );
        //...snip...
}

class Group extends AppModel {
    var $name = 'Group';
    var $hasAndBelongsToMany = array ( 'User' => array(
        'className'             => 'User',
        'foreignKey'            => 'group_id',
        'associationForeignKey' => 'user_id',
        'joinTable'             => 'groups_users',
        'unique'                => true,
    ));
}

我对HABTM的理解是错误的吗？我如何实现这种多对多关系，我可以使用CakePHP查询groups_users表，以便返回当前经过身份验证的用户所关联的组列表？

更新

应用 ndm 建议的更改后，我仍会收到一个返回所有组的大数组返回（太大而无法发布），如果用户具有该成员资格，则会返回'User'元素组。我查看了CakePHP再次使用的查询：

SELECT 
    [User].[id] AS [User__id], 
    [User].[username] AS [User__username], 
    [User].[password] AS [User__password], 
    [User].[email] AS [User__email], CONVERT(VARCHAR(20), 
    [User].[created], 20) AS [User__created], CONVERT(VARCHAR(20), 
    [User].[modified], 20) AS [User__modified], 
    [User].[full_name] AS [User__full_name], 
    [User].[site] AS [User__site], 
    [GroupsUser].[user_id] AS [GroupsUser__user_id], 
    [GroupsUser].[group_id] AS  [GroupsUser__group_id], 
    [GroupsUser].[id] AS [GroupsUser__id] 
FROM 
    [users] AS [User] JOIN 
        [groups_users] AS [GroupsUser] ON (
        [GroupsUser].[group_id] IN (1, 2, 3, 4, 5) AND 
        [GroupsUser].[user_id] = [User].[id]
    )

是否有一种简单的方法来改进，以便我只收到组ID＆amp;我有会员资格的条目的名称？我在考虑使用：

array('conditions'=>array('GroupsUser.user_id'=>$user_id))

...但我在groups表上收到sql错误：

SELECT TOP 1 [Group].[name] AS [Group__name], CONVERT(VARCHAR(20), [Group].[created], 20) AS [Group__created], CONVERT(VARCHAR(20), [Group].[modified], 20) AS [Group__modified], [Group].[id] AS [Group__id] FROM [groups] AS [Group] WHERE [GroupsUser].[user_id] = 36 ORDER BY (SELECT NULL)

Answer 1

我认为你误解了list找到类型的内容。

查询完全正常，list查找类型仅用于检索单个模型的记录列表，其中模型primary key用作索引，{{3}作为价值。

display field

HABTM选择似乎忽略了joinTable

1 个答案: