CakePHP内部联接不工作

我试图将这个mysql查询转换成cakephp，但它不起作用

这是我的mysql查询

SELECT r.id, p.name, r.paid, p.created
       FROM patients p
       INNER JOIN reports r ON p.id = r.patient_id
       AND doctor_name LIKE '%dr.saidul%'

在cakephp中，我尝试了下面的代码

$query_options = array();
            $query_options['fields'] = array( 'Report.id', 'Report.paid','Patient.name','Patient.created' );
            $query_options['conditions'] = array( 'Patient.doctor_name'=>'%dr.saidul%');
            $query_options['joins'] = array('table' => 'Report',
                                    'type' => 'INNER',
                                    'conditions' => array(
                                        'Patient.id = Report.id',
                                    )
                                );

            $patientlist=$this->Patient->find('all', $query_options);

检查sql\u转储后，它会给我以下结果

 SQL Query: SELECT `Report`.`id`, `Report`.`paid`, `Patient`.`name`, `Patient`.`created` FROM `diagnosis`.`patients` AS `Patient` LEFT JOIN `diagnosis`.`upazilas` AS `Upazila` ON (`Patient`.`upazila_id` = `Upazila`.`id`) LEFT JOIN `diagnosis`.`zilas` AS `Zila` ON (`Patient`.`zila_id` = `Zila`.`id`) LEFT JOIN `diagnosis`.`doctors` AS `Doctor` ON (`Patient`.`doctor_id` = `Doctor`.`id`) INNER JOIN `diagnosis`.`Report` ON (`Patient`.`id` => `Report`.`id`) WHERE `Patient`.`doctor_name` = '%dr.saidul%' 

---

hereerror:SQLSTATE[42000]：语法错误或访问冲突：1064您的SQL语法有错误；检查与您的MySQL服务器版本对应的手册，以了解在“=>

报告

id

）附近使用的正确语法，其中

患者

医生姓名='%dr.saidul%''在第1行

您得到了什么错误？数组（数组（'table'=>'Report'，'type'=>'internal'，'conditions'=>array（'Patient.id=Report.id'，））；您是否加载了报告模型？我已经提交了答案是的。我认为这里的问题是它还可以获取患者关系数据，在输出中，它还可以在这里获取zila。

I have submitted the cakephp inner join code please check it

$query_options = array();
            $query_options['fields'] = array( 'reports.id', 'reports.paid','Patient.name','Patient.created','Patient.doctor_name' );
            $query_options['conditions'] = array( 'Patient.doctor_name'=>'%dr.saidul%');
            $query_options['joins'] = array(
                                array('table' => 'reports',
                                'type' => 'INNER',
                                'conditions' => array(
                                    'Patient.id = reports.id',
                                    )
                                )
                            );
            $patientlist=$this->Patient->find('all', $query_options);