Php yii多重内连接
我正在尝试用yii编写一个查询。我有以下几种方法可以奏效Php yii多重内连接,php,mysql,yii,Php,Mysql,Yii,我正在尝试用yii编写一个查询。我有以下几种方法可以奏效 $criteria = new CDbCriteria; $criteria->condition = "($column = :id)"; $criteria->params = array(":id" => $id ); $rows = Jobs::model()->with('pROJ')->findAll($criteria); 这将返回数组中作业的模型。我需要用yii编写以下查询以返回模型 SE
$criteria = new CDbCriteria;
$criteria->condition = "($column = :id)";
$criteria->params = array(":id" => $id );
$rows = Jobs::model()->with('pROJ')->findAll($criteria);
这将返回数组中作业的模型。我需要用yii编写以下查询以返回模型
SELECT jobs.JOBNO, jobs.STATUS, projects.ORDERNO, jobs.PROJID, jobs.NAME, jobs.SEQ, jobs.PCENTDONE, jobs.EARNED, jobs.VALUE, jobs.DATEIN, jobs.DATEDONE, jobs.DATEDUE, jobs.SENTBACK, jobs.ORIGTAPES, jobs.COMMENTS, projects.CATEGORY, orders.BIDNO
FROM (jobs INNER JOIN projects ON jobs.PROJID = projects.PROJID) INNER JOIN orders ON projects.ORDERNO = orders.ORDERNO
where jobs.projid = 3002001
ORDER BY jobs.JOBNO, jobs.PROJID
我尝试了以下方法,但不起作用
$rows = Yii::app()->db->createCommand()
->select('jobs.*, projects.ORDERNO, projects.CATEGORY, orders.BIDNO')
->from('jobs, projects, orders')
->join('projects p','jobs.PROJID = p.PROJID')
->join('orders o', 'p.ORDERNO = o.ORDERNO')
->where('jobs.projid=:id', array(':id'=>$id))
->queryRow();
我得到以下错误
CDbCommand failed to execute the SQL statement: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'jobs.PROJID' in 'on clause'. The SQL statement executed was: SELECT `jobs`.*, `projects`.`ORDERNO`, `projects`.`CATEGORY`, `orders`.`BIDNO`
FROM `jobs`, `projects`, `orders`
JOIN `projects` `p` ON jobs.PROJID=p.PROJID
JOIN `orders` `o` ON p.ORDERNO=o.ORDERNO
WHERE jobs.projid=:id
我已经更新到
$rows = Yii::app()->db->createCommand()
->select('jobs.*, projects.orderno, projects.category, orders.bidno')
->from('jobs')
->join('projects p','jobs.projid = p.projid')
->join('orders o', 'p.orderno = o.orderno')
->where('jobs.projid=:id', array(':id'=>$id))
->queryRow();
但我仍然得到了错误。mysql中的所有列都是大写的
CDbCommand failed to execute the SQL statement: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'projects.orderno' in 'field list'. The SQL statement executed was: SELECT `jobs`.*, `projects`.`orderno`, `projects`.`category`, `orders`.`bidno`
FROM `jobs`
JOIN `projects` `p` ON jobs.projid = p.projid
JOIN `orders` `o` ON p.orderno = o.orderno
WHERE jobs.projid=:id
我认为您应该从('jobs,Project,orders')中删除项目,订单,可能您应该降低
作业的大小写。PROJID
因为您的错误消息说它找不到列。我认为您应该从('jobs,Project,orders')中删除项目,订单并且可能是,您应该降低作业的大小写。PROJID,因为您的错误消息说它找不到列。正如@DCoder所说:您的选择现在应该是select('jobs.*,p.orderno,p.category,o.bidno')
。为了保持一致性,您还应将作业别名为jobs
,如下所示
$rows = Yii::app()->db->createCommand()
->select('j.*, p.orderno, p.category, o.bidno')
->from('jobs j')
->join('projects p','j.projid = p.projid')
->join('orders o', 'p.orderno = o.orderno')
->where('j.projid=:id', array(':id'=>$id))
->order('j.jobno,j.projid')
->queryRow();
正如@DCoder所说:您的选择现在应该是
select('jobs.*,p.orderno,p.category,o.bidno')
。为了保持一致性,您还应将作业别名为jobs
,如下所示
$rows = Yii::app()->db->createCommand()
->select('j.*, p.orderno, p.category, o.bidno')
->from('jobs j')
->join('projects p','j.projid = p.projid')
->join('orders o', 'p.orderno = o.orderno')
->where('j.projid=:id', array(':id'=>$id))
->order('j.jobno,j.projid')
->queryRow();
在某些操作系统上,如Linux,MySQL表/列名区分大小写。
SELECT。。。projects.orderno。。。加入项目p
。您将projects
别名为p
,因此无法再使用projects
。在某些操作系统上,如Linux,MySQL表/列名区分大小写。选择。。。projects.orderno。。。加入项目p
。您将projects
别名为p
,因此无法再使用projects
。