Php 在模型内创建虚拟字段时出错

我很难在我的一个模型中设置虚拟场。我一直在我所有的模型中设置这样的虚拟场，没有任何问题。然而，由于某种原因，我不明白，它在这个模型中根本不起作用

错误：

SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'databases WHERE id = `DataSourceName`.`database_id`) AS `DataSourceName__databa' at line 1

型号：

<?php
App::uses('AppModel', 'Model');
/**
 * DataSourceName Model
 *
 * @property Database $Database
 */
class DataSourceName extends AppModel {
    public $virtualFields = array(
            'database' => 'SELECT name FROM databases WHERE id = DataSourceName.database_id',
    );

    //The Associations below have been created with all possible keys, those that are not needed can be removed

/**
 * belongsTo associations
 *
 * @var array
 */
    public $belongsTo = array(
        'Database' => array(
            'className' => 'Database',
            'foreignKey' => 'database_id',
            'conditions' => '',
            'fields' => '',
            'order' => ''
        )
    );
}

更新1

SQL失败：

(SELECT name FROM databases WHERE id = `DataSourceName`.`database_id`) AS `DataSourceName__database`

---

以下是我最终用于SQL的内容：

SELECT name FROM my_db.databases WHERE id = `DataSourceName`.`database_id`

注意表数据库前面的数据库名称前缀，即my_db。可能是因为数据库是一个保留字

最终型号：

<?php
App::uses('AppModel', 'Model');
/**
 * DataSourceName Model
 *
 * @property Database $Database
 */
class DataSourceName extends AppModel {

    public function __construct($id = false, $table = null, $ds = null) {
        parent::__construct($id, $table, $ds);

        $fields = get_class_vars('DATABASE_CONFIG');

        $this->virtualFields['database'] = sprintf(
                'SELECT name FROM %s.databases WHERE id = DataSourceName.database_id', $fields['default']['database']
        );


    }

    //The Associations below have been created with all possible keys, those that are not needed can be removed

/**
 * belongsTo associations
 *
 * @var array
 */
    public $belongsTo = array(
        'Database' => array(
            'className' => 'Database',
            'foreignKey' => 'database_id',
            'conditions' => '',
            'fields' => '',
            'order' => ''
        )
    );
}

---

我建议您阅读更多关于虚拟字段的内容。分享失败的实际SQL代码。我猜问题在于您语句中的id列不明确。请尝试“WHERE databases.id=DataSourceName.database_id`@AgRizzo查看失败SQL的更新。我尝试了你的建议，但没有解决我的问题。

<?php
App::uses('AppModel', 'Model');
/**
 * DataSourceName Model
 *
 * @property Database $Database
 */
class DataSourceName extends AppModel {

    public function __construct($id = false, $table = null, $ds = null) {
        parent::__construct($id, $table, $ds);

        $fields = get_class_vars('DATABASE_CONFIG');

        $this->virtualFields['database'] = sprintf(
                'SELECT name FROM %s.databases WHERE id = DataSourceName.database_id', $fields['default']['database']
        );


    }

    //The Associations below have been created with all possible keys, those that are not needed can be removed

/**
 * belongsTo associations
 *
 * @var array
 */
    public $belongsTo = array(
        'Database' => array(
            'className' => 'Database',
            'foreignKey' => 'database_id',
            'conditions' => '',
            'fields' => '',
            'order' => ''
        )
    );
}