Question

我在if/else中有一个Yii Framework条件。当用户是admin时，数据库查询不应该有过滤器，如果用户不是admin，则数据库查询应该有过滤器。

 if($ut=='Admin'){ // if the user is Admin the query should not have a filter
    $queryCustomer = Yii::app()->db->createCommand()
			->select('cust.mobile_number as customer,sum(trans.price) as sales,trans.datetime')
		 	->from('customers cust')
		    	->join('transactions trans', 'cust.id=trans.customer_id')
		    	->where('transaction_type_id=:id', array(':id'=>1))	
		    	->group('customer_id')
		    	->order('trans.datetime DESC')
		    	->queryAll();
		    	 $per_customer=$this->renderTableCustomer($queryCustomer);
        } else { // if the user is not Admin the query should have a filter
        	$queryCustomer = Yii::app()->db->createCommand()
			->select('cust.mobile_number as customer,sum(trans.price) as sales,trans.datetime')
		 	->from('customers cust')
		    	->join('transactions trans', 'cust.id=trans.customer_id')
		    	->where('transaction_type_id=:id', array(':id'=>1))
		    	->andWhere('user_id=:userID',array(':userID'=>$filter))		
		    	->group('customer_id')
		    	->order('trans.datetime DESC')
		    	->queryAll();
    $per_customer=$this->renderTableCustomer($queryCustomer);

除了我在else查询中添加了where语句之外，查询是相同的。由于查询是相同的，是否有可能将它放在函数上，当我调用它时，我会将where语句添加为参数？

Answer 1

你真的不需要一个功能。你可以这样做：

＆＃13;

$queryCustomer = Yii::app()->db->createCommand()
			->select('cust.mobile_number as customer,sum(trans.price) as sales,trans.datetime')
		 	->from('customers cust')
		    ->join('transactions trans', 'cust.id=trans.customer_id')
		    ->where('transaction_type_id=:id', array(':id'=>1))	

if($ut !== 'Admin') // if the user is Admin the query should not have a filter
            ->andWhere('user_id=:userID',array(':userID'=>$filter))	

            ->group('customer_id')
            ->order('trans.datetime DESC')
            ->queryAll();

$per_customer=$this->renderTableCustomer($queryCustomer);

＆＃13;

这样，如果用户不是Admin，那么->addWhere()方法将不会执行。所以函数是针，所以作为单个方法调用的双查询...

希望这有帮助！：d

sql查询放到php函数

1 个答案: