Hi,
I had read an script from "The Yii book".
protected/controllers/SiteController.php
public function actionContact() {
[indent]$model = new ContactForm;
$form = new CForm(‘application.views.site.contactForm’,
$model);
if ($form->submitted() && $form->validate()) {
// Send the email!
$this->render(‘emailSent’);
} else {
$this->render(‘contact’, array(‘model’ => $model,
‘form’ => $form));
}[/indent]
}
As you can see, the $model and the $form were initially set as an new object.
I am wondering how does the actionContact know the form were submitted via POST method?
And also the page did not render emailSent after I submitted the form.Thanks
jkofsky
May 12, 2014, 4:04am
2
forpizinmind:
Hi,
I had read an script from "The Yii book".
protected/controllers/SiteController.php
public function actionContact() {
[indent]$model = new ContactForm;
$form = new CForm(‘application.views.site.contactForm’,
$model);
if ($form->submitted() && $form->validate()) {
// Send the email!
$this->render(‘emailSent’);
} else {
$this->render(‘contact’, array(‘model’ => $model,
‘form’ => $form));
}[/indent]
}
As you can see, the $model and the $form were initially set as an new object.
I am wondering how does the actionContact know the form were submitted via POST method?
And also the page did not render emailSent after I submitted the form.Thanks
Q: POST is the default action for CForm(). It can be changed via one of the options of CForm (sorry not sure which at this time)
Is the if($form->submitted() && $form->validate()) correct? Could it be $model->validate()?
joblo
May 12, 2014, 6:55pm
3
Take a look at the submitted() method.
This will check $_POST for the submitbutton name: default=‘submit’ and will assign the attributes from $_POST if $_POST[‘submit’] isset.
Maybe your button name is not ‘submit’.
Do a var_dump($_POST) or debug to see what values are submitted.
Thanks jkofsky and Joblo,I had solved the issue.
It failed since id did not pass the validation
I had another stupid problem now.(Sorry,its my first time using MVC php)
How does the $model know the user post data simply by "$model = new ContactForm"?
I found that it just automatically bind the user provided data to the $model.How?
[indent]public function actionContact() {
$model = new ContactForm;[/indent]
As in procedure programming,there must be an script like:
if(isset($_POST[‘contactform’])){$model->name=$_POST[‘contactform’][‘name’] …}
But I cant find the similar parts in the actionContact.Whats the mechanism underlying?
Thanks.
joblo
May 13, 2014, 6:09am
5
My advice:
Use a debugger to step through the Yii core or study the Yii source (you can take a quick look in the online documentation).
As I told above, the submitted() method will assign the attributes from $_POST .
submitted($buttonName='submit',$loadData=true)
clicked()
to check if the buttonname is submitted
loadData()
with
model->setAttributes($_POST[$class]);
The models setAttributes() only the safe attributes rules()
Study the tutorials/wiki articles:
For example Working with forms
and the best tutorial - the Yii source: there you will find the answers for all your questions
jkofsky
May 14, 2014, 5:33am
6
AND the attributes that pass validation
