基于开关情况动态加载模型时CakePHP保存不工作
这是我的示例代码:基于开关情况动态加载模型时CakePHP保存不工作,php,cakephp,Php,Cakephp,这是我的示例代码: switch($topic_type) { case 1: //post $model = 'Userposts'; $field_name = 'user_post_id'; $select_name = 'post_title'; break; } $model = ClassRegistry::init($model); $model->alias = 'TP'; $model->id
switch($topic_type) {
case 1: //post
$model = 'Userposts';
$field_name = 'user_post_id';
$select_name = 'post_title';
break;
}
$model = ClassRegistry::init($model);
$model->alias = 'TP';
$model->id = $topic_id;
$result = $model->find('first',
array(
'conditions'=>array($field_name=>$topic_id),
'fields'=>array($select_name.' as topic',$field_name.' as id'),
));
if(!empty($result)){
$this->updata[$model_name][$select_name] = $modified_topic;
$this->updata[$model_name][$field_name] = $topic_id;
if($model->save($this->updata)){
$respArr['token'] = $token;
$respArr['status'] = 1;
$respArr['msg'] = 'Success';
echo json_encode(array('token' => $token, 'status' => 1, 'msg' => 'Success'));exit;
}
}
$model->find正在工作,正在获取结果,其中$model->save after find不工作
字段\u name是主键,选择\u name是修改标题后要更新的保存。使用变量
$model
而不是$model\u name
进行更新data@Sadikhasan我最初使用它,但ti抛出错误,因为$model是model对象,而不是字符串model\u name是字符串,它包含与“Userposts”您的模型应该是$model='Userpost'
而不是$model='Userposts'代码>。请更正您的型号名称。