我正在渲染ajax表单
public function actionUser()
{
$model = new UserInfoForm();
$model->user_id = $this->user->id;
$validation = $this->performAjaxValidation($model);
if(null !== $validation){
return $validation;
}
$user = Yii::$app->user->identity;
return $this->renderAjax('user.php',[
'error' => $error,
'user' => $user,
'model' => $model
]);
}
在user.php中,我有以下行来获取所有用户公司和工作
<a href="<?= Url::to(['user/details', 'id' => $user_id]) ?>">Show User Companies and Jobs</a>
现在处于用户详细信息操作
public function actiondetails()
{
$model = new UserJobsForm();
$validation = $this->performAjaxValidation($model);
if(null !== $validation){
return $validation;
}
$companies = Companies::getUserCompanies();
$jobs = BlogPost::getUserJobs();
return $this->renderAjax('user_info.php',[
'error' => $error,
'model' => $model,
'companies' => $companies,
'jobs' => $jobs
]);
}
在我的user_info.php视图页面中,我可以看到所有细节。我也看到了user.php视图页面,这是因为,我在user.php上呈现user_info页面。我的要求是不要在新页面中打开。所以我试图在user.php上呈现。我想在user_info.php渲染后立即关闭user.php。我怎么能这样做?
答案 0 :(得分:0)
如果您只想关闭另一个模态,可以在user_info视图中调用下面的脚本
<script type="text/javascript">
$(function(){
$('#id_of_user_modal').modal('hide');
});
</script>