Question

所以这就是我正在做的事情，我在表单中创建了一个依赖的下拉菜单，我在我的控制器上验证它：

public function actionCreate()
    {
        $model=new Management;

        if(isset($_POST['Management']))
        {
            if($_POST['Management']['company_id']==''){
                //error please select a company
                Yii::app()->user->setFlash('company_error',"Please select a company.");
                $this->redirect(array('create'));
            }else{
                if($_POST['Management']['site_id']==''){
                    //error please select a site
                    Yii::app()->user->setFlash('site_error',"Please select a site.");
                    Yii::app()->user->setFlash('cid',$_POST['Management']['company_id']);
                    $this->redirect(array('create'));
                }else{
                    //save
                    $model->attributes=$_POST['Management'];
                    if($model->save())
                        $this->redirect(array('view','id'=>$model->id));
                }
            }

        }

        $this->render('create',array(
            'model'=>$model,'update'=>''
        ));
    }

我正在将公司ID传递给我的表单视图，如果验证失败，我想保留上一个下拉列表。这就是我的下拉列表：

<div class="row">
        <?php echo $form->labelEx($model,'company_id'); ?>
        <?php echo $form->dropDownList($model, 'company_id', $company, /*array('options'=>array($cid=>array('selected'=>true))),*/
                array(
                        'ajax' => array(
                                'type'=>'POST',
                                'url'=>CController::createUrl('Management/dropDownSite'),
                                'update'=>'#SectorManagement_site_id',
                                'data'=>array('CompanyId'=>'js:this.value'),
                ))); 
        ?>
        <?php echo $form->error($model,'company_id'); ?>
    </div>

现在当我取消注释代码时：

array('options'=>array($cid=>array('selected'=>true))),

从上面的下拉列表中显示预先选择的值，但随后依赖下拉列表不起作用。任何想法我应该如何解决这个问题。提前致谢。：）

Answer 1

我认为你正在以错误的方式看待这个问题。如果您的验证失败，您可以重定向，而您已经在页面上进行重定向。

我不是100％肯定，但我认为你在行动中寻找的是

public function actionCreate()
{
    $model = new Management;

    if (isset($_POST['Management'])) {
        $model->attributes = $_POST['Management'];

        if ($_POST['Management']['company_id'] == '') {
            //error please select a company
            Yii::app()->user->setFlash('company_error', "Please select a company.");
        } elseif ($_POST['Management']['site_id'] == '') {
            //error please select a site
            Yii::app()->user->setFlash('site_error', "Please select a site.");
            Yii::app()->user->setFlash('cid', $_POST['Management']['company_id']);
        } elseif ($model->save()) {
            $this->redirect(array('view', 'id' => $model->id));
        }
    }

    $this->render(
        'create',
        array(
            'model' => $model,
            'update' => ''
        )
    );
}

您的观点可能会保持原样。

为了进一步改善这一点，您应该在模型中处理验证，如：

public function rules()
{
    return array(
        array('company_id, site_id', 'required'),
    );
}

然后你需要做的就是：

public function actionCreate()
{
    $model = new Management;

    if (isset($_POST['Management'])) {
        $model->attributes = $_POST['Management'];
        if ($model->validate() && $model->save()) {
            $this->redirect(array('view', 'id' => $model->id));
        }
    }

    $this->render(
        'create',
        array(
            'model' => $model,
            'update' => ''
        )
    );
}

Answer 2

我终于想通了，这就是它对我有用的方式：

public function actionCreate()
    {
        $model=new Management;

        if(isset($_POST['Management']))
        {
            if($_POST['Management']['company_id']==''){
                //error please select a company
                Yii::app()->user->setFlash('company_error',"Please select a company.");
                $this->redirect(array('create'));
            }else{
                if($_POST['Management']['site_id']==''){
                    //error please select a site
                    Yii::app()->user->setFlash('site_error',"Please select a site.");
                    $model->company_id = $_POST['Management']['company_id'];
                    $this->render('create',array(
                    'model'=>$model,
                    'update'=>'',
                ));
                exit;
                }else{
                    //save
                    $model->attributes=$_POST['Management'];
                    if($model->save())
                        $this->redirect(array('view','id'=>$model->id));
                }
            }

        }

        $this->render('create',array(
            'model'=>$model,'update'=>''
        ));
    }

保持在yii中验证后选择从属下拉列表

2 个答案: