Question

我有一个项目，我已经继承并且设计很差，我只是想为它带来一些理智。这是我遇到的问题。我有一个表存储项目，联系人和另一个存储附加到该项目的作者的表（尽管还有更多的表）。我设法编写了以下脚本。

$conditions = $this->paginate = array('joins' => array(
                                        'table' => 'wp3_order_writers',
                                        'alias' => 'OrderWriter',
                                        'type' => 'LEFT',
                                        'conditions' => array(
                                            'OrderWriter.order_id=Order.id'
                                            )
                                    ),
            'conditions' => array(
                         array('Order.status_id' => array(5,6,7,8),
                          'OR' => array('Order.assigned_to_writer' => $this->_user['Contact']['id'],
                              'OrderWriter.writer_id' => $this->_user['Contact']['id'],
                              )

                             ),
            ),
            'fields' => array(
            'Order.id', 'Order.urgent', 'Order.writer_deadline', 'Order.subject_title', 'Client.name', 'Client.email', 'Service.*', 'Order.writer_fee', 'Order.count_of_words', 'Status.*', 'Order.pay_status', 'Subject.*'
            )
            );
$this->paginate['limit'] = 100000;
$orders = $this->paginate($this->Order);

当我检查生成的sql脚本时，它包含错误。以下是生成的内容

SELECT `Order`.`id`, `Order`.`urgent`, `Order`.`writer_deadline`, `Order`.`subject_title`, `Client`.`name`, `Client`.`email`, `Service`.*,`Order`.`writer_fee`, `Order`.`count_of_words`, `Status`.*, `Order`.`pay_status`, `Subject`.* FROM `wp3_orders` AS `Order` wp3_order_writers OrderWriter LEFT Array LEFT JOIN `wp3_contacts` AS `Client` ON (`Order`.`client_id` = `Client`.`id`) LEFT JOIN `wp3_keys_values` AS `Service` ON (`Order`.`service_id` = `Service`.`id` AND `Service`.`key` = 'services') LEFT JOIN `wp3_keys_values` AS `PayMethod` ON (`Order`.`pay_method_id` = `PayMethod`.`id` AND `PayMethod`.`key` = 'pay_methods') LEFT JOIN `wp3_subjects` AS `Subject` ON (`Order`.`subject_id` = `Subject`.`id`) LEFT JOIN `wp3_keys_values` AS `Status` ON (`Order`.`status_id` = `Status`.`id` AND `Status`.`key` = 'statuses') LEFT JOIN `wp3_clients_sessions` AS `ClientsSession` ON (`ClientsSession`.`current_order_id` = `Order`.`id`)  WHERE ((`Order`.`status_id` IN (5, 6, 7, 8))  AND  (((`Order`.`assigned_to_writer` = 1087) OR (`OrderWriter`.`writer_id` = '1087')))) AND   `Order`.`completed` = 1    LIMIT 100000

出现此错误“警告（512）：SQL错误：1064：您的SQL语法中有错误;请查看与您的MySQL服务器版本对应的手册，以便在'wp3_order_writers'附近使用正确的语法OrderWriter LEFT Array LEFT JOIN第1行wp3_contacts AS Client ON'[CORE \ cake \ libs \ model \ datasources \ dbo_source.php，第684行]“

非常感谢您的帮助

Answer 1

我会用Containable制作它（假设你有像Order，OrderWriter，Writer和里面定义的关系这样的模型）

然后您的查询将如下所示：

$this->paginate['Order'] = array('contain' => array('Writer'));
$this->paginate['conditions'] = array(...);
$this->paginate['fields'] = array(...);

$orders = $this->paginate($this->Order);

我不确定订单部分是否应该如上所述：

$this->paginate['Order'] = array('contain' => array('OrderWriter'=>array('Writer')));

另外我建议你不要使用100000限制，因为它可能是一种跳过分页的方法，但只是使用find（）而不是paginate。

虽然避免分页会降低请求速度，因为它会对每行订单进行查询，因此具有默认限制（20）或一些合理的数字可以加速网站/应用。

在cakephp中使用paginate加入

1 个答案: