PHP-执行查询时无法加载模型
作者:互联网
我有这样的查询:
$phsql = "
SELECT s.id AS siteId, s.name
FROM site s
INNER JOIN profiles p ON s.id = p.siteId
INNER JOIN users_profiles up ON up.profilesId = p.id
AND p.name = 'admin'
AND up.usersId = 2
";
我在模型方法中将其转换如下:
$sites = Site::query()
->innerJoin('Profiles', 'Sites.id = Profiles.siteId')
->innerJoin('UsersProfiles', 'UsersProfiles.profilesId = Profiles.id')
->andWhere('Profiles.name = name')
->andWhere('UsersProfiles.usersId = :usersId:', ['userId' => $admin_id])->execute();
在运行它会给出错误:
Model Profiles could not be loaded
请注意,我正在网站模型中运行它.
更新资料
我尝试了这个:
$sites = $this->modelsManager->createBuilder()
->from('myApp\Models\Site')
->innerJoin('myApp\Models\Profiles','myApp\Models\Site.id = myApp\Models\Profiles.siteId')
->andWhere("myApp\Models\Profiles.name = 'admin' ")
->where("myApp\Models\UsersProfiles.profilesId = 2")
->getQuery()
->execute();
现在,它给出了错误:
Unknown model or alias ‘myApp\Models\UsersProfiles’ (11), when preparing: SELECT [myApp\Models\Site].* FROM [myApp\Models\Site] INNER JOIN [myApp\Models\Profiles] ON myApp\Models\Site.id = myApp\Models\Profiles.siteId WHERE myApp\Models\UsersProfiles.profilesId = 2
解决方法:
查看您的代码,我看到两个问题:
1)您第二行的-> execute()应该抛出解析错误?
->innerJoin('Profiles', 'Sites.id = Profiles.siteId')->execute();
2)您必须向模型添加名称空间,请参见下面的代码.
查询的工作示例:
Objects::query()
->columns([
'Models\Objects.id AS objectID',
'Models\ObjectLocations.id AS locationID',
'Models\ObjectCategories.category_id AS categoryID',
])
->innerJoin('Models\ObjectLocations', 'Models\Objects.id = Models\ObjectLocations.object_id')
->innerJoin('Models\ObjectCategories', 'Models\Objects.id = Models\ObjectCategories.object_id')
->where('Models\Objects.is_active = 1')
->andWhere('Models\Objects.id = :id:', ['id' => 2])
->execute();
您可以在关系中添加第三个参数(别名)以减少名称空间并提高代码的可读性:
->innerJoin('Models\ObjectLocations', 'loc.object_id = obj.id', 'loc');
更多信息在这里:https://docs.phalconphp.com/en/latest/api/Phalcon_Mvc_Model_Criteria.html
还要注意:使用where()和andWhere()将where子句添加到查询中.在第一个查询示例中,子句位于第二个join语句内,而在Phalcon查询中,where子句将添加到整个查询中.如果您确实只希望那些条件用于第二个联接,则将它们添加到第二个联接参数中,如下所示:
->innerJoin(
'Models\ObjectCategories',
'Models\Objects.id = Models\ObjectCategories.object_id AND ... AND ... AND ...'
)
标签:phalcon,phalcon-orm,php 来源: https://codeday.me/bug/20191118/2028329.html