Question

我正在尝试使用cakePHP更新数据库中的记录，因为它适合创建一个新记录。我的脚本使用“FIRST”进行检查，如果它没有返回false，它应该创建一个新记录，否则它应该根据我指定的ID更新现有记录。我在下面故意添加了$ this-＆gt; Asset-＆gt; id = 62;尝试强制更新，但它不起作用。有帮助吗？（另外，如果你能理解它想做什么，请随意指导我或许更好的代码:)非常新的蛋糕：

 public function index() {

        if($this->request->is("post")) {
            $aid = $this->request->data['Asset']['asset_identifier'];
            $type = $this->request->data['Asset']['asset_type_id'];
            $asset = $this->Asset->find("first",array("fields" => array("Asset.id", "Asset.user_id", "Asset.status_id", "Asset.financial_id", "Asset.insurance_id"),"conditions" => array("Asset.asset_identifier" => $aid, "Asset.asset_type_id" => $type)));
            if($asset != false) {
                $uid = $this->Auth->user("id");
                // UPDATE, depends on info:
                $this->Asset->id = $asset['Asset']['id'];
                foreach(array("financial_id", "insurance_id", "user_id") as $value) {
                    if($asset['Asset'][$value] != 0) {
                        $emails[] = $asset['Asset'][$value];
                    }
                }
                if(isset($emails) && count($emails)) {
                    $this->request->data['Asset']['status_id'] = 1;
                    $email = new CakeEmail();
                    $email->from(array('noreply@assetchase.co.za' => 'Assetchase.co.za'));
                    $email->subject('Assetchase.co.za result notification.');
                    foreach($emails as $value) {
                        $user = $this->User->find("first",array("fields" => array("username"),"conditions" => array("id" => $value)));
                        $email->to($user['User']['username']);
                        $email->send('A new notification, booyah!');
                        // Send an email with the username.
                    }
                }
                if($this->Auth->user("user_type_id") == 2) {
                    $this->request->data['Asset']['user_id'] = $uid;
                } elseif($this->Auth->user("user_type_id") == 3) {
                    $this->request->data['Asset']['financial_id'] = $uid;
                } elseif($this->Auth->user("user_type_id") == 4) {
                    $this->request->data['Asset']['insurance_id'] = $uid;
                }
            }
            $this->Asset->id = 62;
            if($this->Asset->saveAll($this->request->data)) {
                $this->Session->setFlash("A new asset has been loaded",'success');
                $this->redirect(array("controller" => "asset", "action" => "thankyou", "guest"));
            } else {
                $this->Session->setFlash('An error has occured.','default',array('class'=>'error'));
            }
        }
        $assetTypes = $this->Asset->AssetType->find('list');
        $this->set("assetTypes", $assetTypes);

    }

Answer 1

当您致电$this->Asset->saveAll($this->request->data)时，您正在告诉Cake获取POST数据并尝试保存。它不会查看您将ID设置为62的$this->Asset。您可以通过检查var_dump($this->request->data)的输出来验证这一点。请尝试使用$this->request->data['Asset']['id'] = 62之类的内容。

cakePHP没有更新记录

1 个答案: