我正在开发 Yii 驱动的应用程序。我想转换这个 SQL:
SELECT m.sendDate, m.status, c.name, c.email, mt.name, mt.subject, CONCAT( op.firstName, ' ', op.lastName ) operator
FROM `mail` m, client c, mailTemplate mt, operator op
WHERE m.customerID = c.id
AND m.operatorID = op.id
AND m.templateID = mt.id
AND c.name LIKE '%L%'
AND c.email LIKE '%@gmail.com%'
AND m.sendDate < '2012-07-31'
AND m.sendDate > '2012-06-30'
进入 CDBcriteria 但不知道如何。
最佳答案
首先,您必须创建 4 个模型:Mail、Client、MailTemplate、Operator。
在邮件模型中定义关系:
public function relations()
{
return array(
'client' => array(self::BELONGS_TO, 'Client', 'customerID'),
'operator' => array(self::BELONGS_TO, 'Operator', 'operatorID'),
'mailTemplate' => array(self::BELONGS_TO, 'MailTemplate', 'templateID'),
);
}
还有...
$criteria = new CDbCriteria;
$criteria->select = new CDbExpression('t.sendDate, t.status, client.name, client.email, mailTemplate.name, mailTemplate.subject, CONCAT( operator.firstName, " ", operator.lastName ) operator');
$criteria->addSearchCondition('client.name', 'L', true);
$criteria->addSearchCondition('client.email', '@gmail.com', true);
$criteria->compare('mailTemplate.sendDate', '<2012-07-31');
$criteria->compare('mailTemplate.sendDate', '>2012-06-30');
好了,现在我们可以找到项目了:
$mails = Mail::model()->with(array('client', 'operator', 'mailTemplate'))->findAll($criteria);
第二个问题更新:
在邮件模型中定义方法搜索:
public function search()
{
/* Criteria from my first answer */
return new CActiveDataProvider('Mail', array(
'criteria' => $criteria,
));
}
在gridview中,列必须使用关系 $data是Mail模型中具有关系的一项。
'columns' => array(
array(
'name' => 'prop_defined_in_mail_model',
'value' => '$data->client->id',
),
),
关于php - Yii CDbCriteria sql,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11734801/