I need to find a user by email.
So i have try to route on lumen like
/user/email/abc#test.com
http://127.0.0.1:8888/user/email/abc#test.com
routes/web.php
$router->get('/user/email/{email}', ['middleware' => ['cors','auth'], 'uses' => 'UserController#getUserByEmail']);
When i have include a DOT(.) the result shows like
"The requested resource /user/email/abc#test was not found on this server."
Otherwise the result is fine.
Please advice how do i route like these scenarios or it is possible or not.
Sorry for my Bad English
I think it's a good idea to take email from request body and not from url.
In your function get the email from request. see this example:
public function getUserByEmail(Request $request){
$this->validate($request, [
'email' => 'required',
]);
$email = $request->email;
//then the rest of your code logic
}
route will be now like this:
$router->get('/user/email', ['middleware' => ['cors','auth'], 'uses' => 'UserController#getUserByEmail']);
Related
I've been banging my head on this trying to find a solution, searching all around for something that would work, but I got no chance.
I have a "dashboard" where users have a list of event they took part in where they can rate/comment the event. I'ts basically a custom comment form for a node that is not displaying on the node page itself. The user click on an icon in their dashboard next to the event they want to comment, they get to the form, fill it and it returns them back to the dashboard. The return is adding parameters with a custom submit function and using the redirect function to make sure the user return to the proper tab in their dashboard.
function custom_form_alter(&$form, &$form_state, $form_id) {
if ($form_id == 'comment_node_event_form') {
$form['#submit'][] = 'customcomment_form_submit';
}
}
function customcomment_form_submit($form, &$form_state) {
if($form['#form_id']=='comment_node_event_form'){
$pos = strpos($_SERVER['HTTP_REFERER'], 'qt-dashboard');
if ($pos !== FALSE) {
$form_state['redirect'] = array(
'dashboard',
array(
'query' => array(
'qt-dashboard' => '2',
'qt-dashboard_event' => '2',
),));
}else{
$form_state['redirect'] = array(
'dashboard',
array(
'query' => array(
'qt-dashboard' => '2',
'qt-dashboard_event' => '1',
),));
}
}
}
This portion is working as it should and expected. The problem is when form validation fails, it send the comment form error message and form to refill to the node page instead of staying where it is.
I found that if I set the #action with the link where my comment form is, it does send the fail to the proper page
$form['#action']='/rating_comment/'.$form['#node']->vid.'?destination=dashboard&qt-dashboard=2&qt-dashboard_event=2';
But, doing so break the redirect when successfully submitting the form and it doesn't take the parameter in the redirect..it basically send the user directly to dashboard and scrapes the parameter. Now there might be a better solution for form validation fail to stay on the same page and that is pretty much what I am looking for.
Thanks
Looks like this form isn’t in your module - and you’re altering the other module.
Now, when the validate function gets invoked at the end you can check for failure and if there is failure cancel processing/redirect etc.
$form_state['no_redirect'] = FALSE:
Also, you can use the error function to check for errors and if so cancel the rest. This goes inside validate method.
if (form_get_errors()) { return FALSE ; }
// .. Otherwise, process validation
Check out the following
https://drupal.stackexchange.com/questions/170815/is-it-possible-to-stop-a-webform-form-during-submission
https://drupal.stackexchange.com/questions/5861/how-to-redirect-to-a-page-after-submitting-a-form
I need to keep SEO links active so I'm trying to 301 redirect google trafic to new CakePHP route.
I go to:
http://localhost/jakne/someCategory/item-slug
And I want it to 301 redirect to:
http://localhost/product/item-slug
So I tried with route::redirect but I can't make it work. Doc on this is also non existent :(
$routes->redirect(
'/jakne/:subcategory/:item',
['controller' => 'Catalog', 'action' => 'product'],
['status' => 301, 'pass' => ['item']]
);
My Catalog::product looks like:
public function product($productId) {
}
I always get error that no parameter was passed to the action.
What am I missing? :(
The option for retaining parameters in redirect routes isn't pass (that's for regular routes and defines which parameters to pass as function arguments), it's persist, ie your route would need to be something like:
$routes->redirect(
'/jakne/:subcategory/:item',
['controller' => 'Catalog', 'action' => 'product'],
['status' => 301, 'persist' => ['item']]
);
This should work fine, assuming you have a proper target route connected that has a parameter named item, something like.
$routes->connect(
'/product/:item',
['controller' => 'Catalog', 'action' => 'product'],
['pass' => ['item']]
);
Generally you may want to consider doing such redirects on server level instead (for example via mod_rewrite on Apache), performance wise that's much better.
ps. Browsers do cache 301 redirects, so when making changes to such redirects, make sure that you clear the cache afterwards.
See also
Cookbook > Routing > Redirect Routing
So it turns out this is quite simple. I use this to dynamically generate a list of redirects based on what admins enter in the control panel. We use this to keep google traffic when the URL changes and is not rescanned by the google bot yet.
$builder->redirect('/from-url', '/to-url', ['status' => 301]);
Try this ways it is working for me:
Example request like: localhost:08080/get-username?id=%3Cid%3E
Routes :
$routes->connect('/get-username', ['controller' => 'Users', 'action' => 'getUserName']);
Controller :
class UsersController extends AppController {
public function initialize() {
parent::initialize();
$this->loadComponent('RequestHandler');
}
public function beforeFilter(Event $event) {
parent::beforeFilter($event);
$this->set('_serialize', false);
$this->Auth->allow([
'getUserName'
]);
}
public function getUserName() {
$id = $this->request->getQuery('id');
}
}
I'm using CakePHP 2.x for a RESTful API, I want to be able to handle requests in the following form
api/activity/17?page=1&limit=10
Typically CakePHP I think likes each param to be separated by the forward slash char and then each of these is mapped into the variables defined in the 2nd array argument of router::connect above. For exampple:
api/activity/17/1/10
In my case though this won't work so I am trying to pass a custom query string which I will then decode in my controller. My router connect is as follows:
So I am using router::connect as follow:
Router::connect('/api/activity/:queryString', [
'controller' => 'users',
'action' => 'activity',
'[method]' => 'GET',
'ext' => 'json',
],
[
'queryString' => '[0-9]+[\?]...[not complete]'
]);
I can't get the regular expression to accept the '?' which I am exscaping in the regex above. How can I achieve this or otherwise is there a better or easier way of sending the URL in the format I require.
You can get the URL parameters (among other methods) via $this->request->query;
So, in your example, add the following in method view() of app/Model/Activity.php:
<?php
// file app/Model/Activity.php
public function view($id)
{
// URL is /activity/17?page=1&limit=10
echo $this->request->query['page']; // echo's: 1
echo $this->request->query['limit']; // echo's: 10
}
See 'Accessing Querystring parameters' in the CakePHP book
My login form may be called with a re-direct query and I am wondering if there is a simple way to include this in the subsequent post action.
The use case is for SSO login.
My normal login route is:
/customer/login
and when called from a third party client becomes:
/customer/login?redirectTo=http://www.example.com
My login action:
public function loginAction()
{
$prg = $this->prg();
if ($prg instanceof Response) {
return $prg;
} elseif ($prg === false) {
return new ViewModel(['form' => $this->loginForm]);
}
This loads my view and I currently define my action as so:
$form = $this->form;
$form->setAttribute('action', $this->url());
Now when the action is called, I am losing the redirectTo parameter.
So my question is this, is it possible to update the action to include the re-direct url so that when a user clicks to login, it is posted back to my form?
thanks!
EDIT -
Obviously I can create a redirectTo route in the configs and test on the initial call to the page for the existence of such a route and include this in the form. My question however is whether or not this can be done automagically simply from the viewscript.
To generate query string arguments from the view helper, you need to assign them as the third argument using the query key. Please refer to the ZF2 docs http://framework.zend.com/manual/current/en/modules/zend.view.helpers.url.html
$form->setAttribute('action', $this->url('application', array('action' => 'login'), array('query' => array('redirectTo' => 'http://www.example.com,))));
$form->setAttribute('action', $this->url('login', [
'query' => [
'redirectTo' => $this->params()->fromQuery('redirectTo')
]
]);
Where 'login' is the name of the login route.
See Url View Helper
Well my solution is not as elegant as I hoped it would be. I wanted to avoid using the controller for the query params. As #Stanimir pointed out, the view helpers are in fact, to help with view so my original idea was unfounded.
This is an end to end of what I have put together:
Controller:
$redirect_url = $this->params()->fromQuery('redirectTo',null);
Returns this to view on initial load:
return new ViewModel( ['form' => $this->loginForm , 'redirect_url' => $redirect_url] );
View
$form->setAttribute(
'action',
$this->url(
'customer/login', [] ,
[ 'query'=>
[ 'redirectTo' => $this->redirect_url ]
]
)
);
I'm building a CakePHP website that sends an e-mail like this:
$email = new CakeEmail('default');
$email->template('test');
$email->emailFormat('html');
$email->to(array('john_doe#example.com' => 'John Doe'));
$email->subject('Test E-mail');
$email->helpers(array('Html', 'Text'));
$email->viewVars(
array(
...
)
);
if ($email->send()) {
$this->Session->setFlash('The e-mail was sent!', 'default', array('class' => 'alert alert-success'));
}
else {
$this->Session->setFlash('An unexpected error occurred while sending the e-mail.', 'default', array('class' => 'alert alert-error'));
}
I'd like to be able to capture the HTML rendered by the e-mail in a variable in addition to actually sending the e-mail. This way, I can record in the database the exact content of the e-mail's body. Is this doable?
Per line 50 of the MailTransport class, it appears the actual send() function returns the message and the header. So instead of:
if($email->send()) {
Try:
$mySend = $email->send();
if($mySend) {
//...
Then, $mySend should be an array:
array('headers' => $headers, 'message' => $message);
Thats what I do in my EmailLib:
https://github.com/dereuromark/tools/blob/2.0/Lib/EmailLib.php
it logs email attempts and captures the email output into a log file (email_trace.log) in /tmp/logs/ - if you are in debug mode it will only log (no emails sent - this has been proven quite useful for local delopment).
you can write a similar wrapper for your case.
but if you want to write it back into the DB Dave's approach seems to fit better.