如何用一种语言编写这个查询？
如何用一种语言编写这个查询？
提问于 2015-11-17 00:23:47
SELECT business.id,business.business_name,address.city FROM business INNER JOIN address ON business.id=address.business_id WHERE business.business_name like '%monal%' and address.city='islamabad' 在这里，monal和伊斯兰堡将是一种形式上的价值。当我用变量代替伊斯兰堡时，它会给我带来错误。我在伊伊的问题。
$user = Yii::app()->db->createCommand()
->select('business.id,business_name,business.image,business.business_description,address.city')
->from('business')
->join('address', 'business.id=address.business_id')
//->where(array('like', 'business.business_name', '%'.$name.'%'))
->where(array('and', 'address.city=$city', array('like', 'business.business_name', '%'.$name.'%')))
->queryALL();
CDbCommand failed to execute the SQL statement: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'Islamabad' in 'where clause'. The SQL statement executed was: SELECT `business`.`id`, `business_name`, `business`.`image`, `business`.`business_description`, `address`.`city`
FROM `business`
JOIN `address` ON business.id=address.business_id
WHERE (address.city=Islamabad) AND (`business`.`business_name` LIKE '%nan%') 回答 1
Stack Overflow用户
回答已采纳
发布于 2015-11-17 01:15:25
将此作为查询进行尝试。
$results = Yii::app()->db->createCommand()
->select('b.id, b.business_name, a.city')
->from('business b')
->join('address a', 'b.id = a.business_id')
->where('b.business_name LIKE :businessName AND a.city = :city', array(
':businessName' => '%' . $businessNameVariable . '%',
':city' => $cityVariable,
))
->queryAll();页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/33747096
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