Php 修改yii中模型的objectname
我有一个模型Skill.php。在*SkillController.php*中,我使用了由crud生成的actionCreate方法Php 修改yii中模型的objectname,php,mysql,yii,Php,Mysql,Yii,我有一个模型Skill.php。在*SkillController.php*中,我使用了由crud生成的actionCreate方法 actionCreate(){ $model=new Skill; } 当我将$model更改为$modelSkill时,它会向我显示错误。为什么 public function actionCreate() { $model=new Skill; // Uncomment the following
actionCreate(){
$model=new Skill;
}
当我将$model更改为$modelSkill时,它会向我显示错误。为什么
public function actionCreate()
{
$model=new Skill;
// Uncomment the following line if AJAX validation is needed
// $this->performAjaxValidation($model);
if(isset($_POST['Skill']))
{
$model->attributes=$_POST['Skill'];
if($model->save())
$this->redirect(array('view','id'=>$model->skill_id));
}
$this->render('create',array(
'model'=>$model,
));
}
更改为$modelSkill后
public function actionCreate()
{
$modelSkill=new Skill;
// Uncomment the following line if AJAX validation is needed
// $this->performAjaxValidation($modelSkill);
if(isset($_POST['Skill']))
{
$modelSkill->attributes=$_POST['Skill'];
if($modelSkill->save())
$this->redirect(array('view','id'=>$modelSkill->skill_id));
}
$this->render('create',array(
'model'=>$modelSkill,
));
}
我想我发现你失败了。是否也可以在视图文件中替换它?如果是,您还需要更改此选项:
$this->render('create',array(
'model'=>$modelSkill,
));
为此:
$this->render('create',array(
'modelSkill'=>$modelSkill,
));
正如您在中看到的,数组将被放入函数中。因为我认为您不止一次使用$model。请发布整个actionCreate。我用$modelSkill替换了所有$model。您可以发布错误吗?您还可以用替换的模型发布代码吗?未定义变量:modelSkill