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PHP警告–;yii\base\ErrorException缺少yii\db\Query::join()的参数2, PHP警告–yii\db\Query::join()的yii\base\ErrorException缺少参数2,_Php_Mysql_Yii2 - Fatal编程技术网

PHP警告–;yii\base\ErrorException缺少yii\db\Query::join()的参数2, PHP警告–yii\db\Query::join()的yii\base\ErrorException缺少参数2,

PHP警告–;yii\base\ErrorException缺少yii\db\Query::join()的参数2, PHP警告–yii\db\Query::join()的yii\base\ErrorException缺少参数2,,php,mysql,yii2,Php,Mysql,Yii2,代码中的错误是什么?请帮帮我 $query = new Query; $dataProvider = new ActiveDataProvider([ 'query'=> $query ->select(['vchr_name','vchr_actual_hours','vchr_worked_hours','fk_int_payroll_month','fk_int_payroll_ye

代码中的错误是什么?请帮帮我

     $query = new Query; 
     $dataProvider = new ActiveDataProvider([
             'query'=> $query
                        ->select(['vchr_name','vchr_actual_hours','vchr_worked_hours','fk_int_payroll_month','fk_int_payroll_year'])
                        ->from('tbl_payroll')
                         ->join(['INNER JOIN','tbl_employee', 'TblEmployee.pk_int_emp_id=TblPayroll.fk_int_emp_id'])
                        ->where(['pk_int_payroll_id'=> $id])
                        ->One(),       

         ]);

根据的文档,join()应该有4个参数。你的情况应该是这样的
->join('internal join','tbl_employee','TblEmployee.pk_int_emp_id=TblPayroll.fk_int_emp_id',[])
或者您可以省略最后的第四个非强制参数。

我如何加入3个表在您拥有的表之后再使用一个->join