I want to parse my .browserslistrc in the middleware of TYPO3 to decide which build should be loaded. Yes, I know that I can do it in JavaScript but I want to make the decision earlier.
This is what I return in the process() function:
$response = $handler->handle($request->withAttribute('isModernBrowser', $isModern));
return $response;
isModern is just a boolean which contains the information if a modern or an "old" browser calls the page.
How can I catch isModern in my template?
ServerRequest is available in $GLOBALS['TYPO3_REQUEST']. Add this object or the specific attribute with $GLOBALS['TYPO3_REQUEST']->getAttribute('isModernBrowser') to view.
Related
My middleware need is to:
add an extra query param to requests made by a REST API client derived from GuzzleHttp\Command\Guzzle\GuzzleClient
I cannot do this directly when invoking APIs through the client because GuzzleClient uses an API specification and it only passes on "legal" query parameters. Therefore I must install a middleware to intercept HTTP requests after the API client prepares them.
The track I am currently on:
$apiClient->getHandlerStack()-push($myMiddleware)
The problem:
I cannot figure out the RIGHT way to assemble the functional Russian doll that $myMiddleware must be. This is an insane gazilliardth-order function scenario, and the exact right way the function should be written seems to be different from the extensively documented way of doing things when working with GuzzleHttp\Client directly. No matter what I try, I end up having wrong things passed to some layer of the matryoshka, causing an argument type error, or I end up returning something wrong from a layer, causing a type error in Guzzle code.
I made a carefully weighted decision to give up trying to understand. Please just give me a boilerplate solution for GuzzleHttp\Command\Guzzle\GuzzleClient, as opposed to GuzzleHttp\Client.
The HandlerStack that is used to handle middleware in GuzzleHttp\Command\Guzzle\GuzzleClient can either transform/validate a command before it is serialized or handle the result after it comes back. If you want to modify the command after it has been turned into a request, but before it is actually sent, then you'd use the same method of Middleware as if you weren't using GuzzleClient - create and attach middleware to the GuzzleHttp\Client instance that is passed as the first argument to GuzzleClient.
use GuzzleHttp\Client;
use GuzzleHttp\HandlerStack;
use GuzzleHttp\Command\Guzzle\GuzzleClient;
use GuzzleHttp\Command\Guzzle\Description;
class MyCustomMiddleware
{
public function __invoke(callable $handler) {
return function (RequestInterface $request, array $options) use ($handler) {
// ... do something with request
return $handler($request, $options);
}
}
}
$handlerStack = HandlerStack::create();
$handlerStack->push(new MyCustomMiddleware);
$config['handler'] = $handlerStack;
$apiClient = new GuzzleClient(new Client($config), new Description(...));
The boilerplate solution for GuzzleClient is the same as for GuzzleHttp\Client because regardless of using Guzzle Services or not, your request-modifying middleware needs to go on GuzzleHttp\Client.
You can also use
$handler->push(Middleware::mapRequest(function(){...});
Of sorts to manipulate the request. I'm not 100% certain this is the thing you're looking for. But I assume you can add your extra parameter to the Request in there.
private function createAuthStack()
{
$stack = HandlerStack::create();
$stack->push(Middleware::mapRequest(function (RequestInterface $request) {
return $request->withHeader('Authorization', "Bearer " . $this->accessToken);
}));
return $stack;
}
More Examples here: https://hotexamples.com/examples/guzzlehttp/Middleware/mapRequest/php-middleware-maprequest-method-examples.html
Im trying to style my login page. My login url is website/Security/login. Im trying to locate the 'login' piece of the url. What have i done wrong below?
public function DisplayPageType() {
$param = $this->request->param('Action');
if ($param === 'login')
{
return 'Login';
}
Thanks
I think that won't work since the controller during render is the Page_Controller and not the Security controller. So the $Action param is not equal to login. It could be index, I'm not sure.
If you just want to check if you're in the login page, you can add this to your Page_Controller:
public function getIsLoginPage()
{
return $_REQUEST['url'] == '/Security/login';
}
Then in your template:
<body class="<%if $IsLoginPage %>login-page<% end_if %>">
A bit dirty but it's the quickest way I know.
Another way is to leverage SilverStripe's legacy support. You can add a css file called tabs.css at mysite/css/tabs.css. If this file exists, SilverStripe will include this in the page.
You can also create templates that SilverStripe will automatically use if they exist:
themes/<theme_name>/Security.ss - If you want your login page to use an entirely different layout.
themes/<theme_name>/Layout/Security_login.ss - If you want to change just the content part (the $Layout section)
I hope this helps.
#gpbnz is right, the $Action param is not equal to login, it actually returns null as accessing $this->request from the Page_Controller when accessing the Security/login returns a NullHTTPRequest.
To get the action, you will want to get the current controller using Controller::curr(). It is then as simple as calling getAction on this controller.
To confirm that the action isn't from a random controller that happens to have an action called login, you can check the instanceof the controller like so: Controller::curr() instanceof Security
This check will still allow it to work for any controller that extends Security though which may/may not happen depending on the project.
I would stick away from actually reading the URL for the information manually though as that can create issues with maintainability in the future.
To bring this to a nice little function:
public function isLoginPage()
{
$controller = Controller::curr();
return $controller instanceof Security && $controller->getAction() == 'login';
}
Otherwise #gpbnz had a good suggestion of using the template system to your advantage for overriding not only the styles but the HTML around it.
I want to invoke an XPage from a rest client e.g http://myserver.com/myDB/myApp.nsf/mypage.xsp?attr=first
and have the XPage mypage.xsp to print the attr value.
var value1 = context.getUrlParameter("attr");
print(value1);
Is it possible?
Yes. Google "XAgents", there are plenty of resources about how to call an XPage and return non-XPage code.
Also, make sure you set viewState property of the XPage to "nostate", because there's no point storing a component tree server-side for something that can't be partially refreshed.
I am a beginner and I am creating some forms to be posted into MySQL using Zend, and I am in the process of debugging but I don't really know how to debug anything using Zend. I want to submit the form and see if my custom forms are concatenating the data properly before it goes into MySQL, so I want to catch the post data to see a few things. How can I do this?
The Default route for zend framework application looks like the following
http://www.name.tld/$controller/$action/$param1/$value1/.../$paramX/$valueX
So all $_GET-Parameters simply get contenated onto the url in the above manner /param/value
Let's say you are within IndexController and indexAction() in here you call a form. Now there's possible two things happening:
You do not define a Form-Action, then you will send the form back to IndexController:indexAction()
You define a Form action via $form->setAction('/index/process') in that case you would end up at IndexController:processAction()
The way to access the Params is already defined above. Whereas $this->_getParam() equals $this->getRequest()->getParam() and $this->_getAllParams() equals $this->getRequest->getParams()
The right way yo check data of Zend Stuff is using Zend_Debug as #vascowhite has pointed out. If you want to see the final Query-String (in case you're manually building queries), then you can simply put in the insert variable into Zend_Debug::dump()
you can use $this->_getAllParams();.
For example: var_dump($this->_getAllParams()); die; will output all the parameters ZF received and halt the execution of the script. To be used in your receiving Action.
Also, $this->_getParam("param name"); will get a specific parameter from the request.
The easiest way to check variables in Zend Framework is to use Zend_Debug::dump($variable); so you can do this:-
Zend_Debug::dump($_POST);
Zend framework is built on the top of the PHP . so you can use var_dump($_POST) to check the post variables.
ZF has provided its own functions to get all the post variables.. Zend_Debug::dump($this->getRequest()->getPost())
or specifically for one variable.. you can use Zend_Debug::dump($this->getRequest()->getPost($key))
You can check post data by using zend
$request->isPost()
and for retrieving post data
$request->getPost()
For example
if ($request->isPost()) {
$postData = $request->getPost();
Zend_Debug::dump($postData );
}
I am developing a website using zend framework.
i have a search form with get method. when the user clicks submit button the query string appears in the url after ? mark. but i want it to be zend like url.
is it possible?
As well as the JS approach you can do a redirect back to the preferred URL you want. I.e. let the form submit via GET, then redirect to the ZF routing style.
This is, however, overkill unless you have a really good reason to want to create neat URLs for your search queries. Generally speaking a search form should send a GET query that can be bookmarked. And there's nothing wrong with ?param=val style parameters in a URL :-)
ZF URLs are a little odd in that they force URL parameters to be part of the main URL. I.e. domain.com/controller/action/param/val/param2/val rather than domain.com/controller/action?param=val¶m2=val
This isn't always what you want, but seems to be the way frameworks are going with URL parameters
There is no obvious solution. The form generated by zf will be a standard html one. When submitted from the browser using GET it will result in a request like
/action/specified/in/form?var1=val1&var2=var2
Only solution to get a "zendlike url" (one with / instead of ? or &), would be to hack the form submission using javascript. For example you can listen for onSubmit, abort the submission and instead redirect browser to a translated url. I personally don't believe this solution is worth the added complexity, but it should perform what you're looking for.
After raging against this for a day-and-a-half, and doing my best to figure out the right way to do this fairly simple this, I gave up and did the following. I still can't believe there's not a better way.
The use case that necessitates this is a simple record listing, with a form up top for adding some filters (via GET), maybe some column sorting, and Zend_Paginate thrown in for good measure. I ran into issues using the Url view helper in my pagination partial, but I suspect with even just sorting and a filter-form, Zend_View_Helper_Url would still fall down.
But I digress. My solution was to add a method to my base controller class that merges any raw query-string parameters with the existing zend-style slashy-params, and redirects (but only if necessary). The method can be called in any action that doesn't have to handle POSTs.
Hopefully someone will find this useful. Or even better, find a better way:
/**
* Translate standard URL parameters (?foo=bar&baz=bork) to zend-style
* param (foo/bar/baz/bork). Query-string style
* values override existing route-params.
*/
public function mergeQueryString(){
if ($this->getRequest()->isPost()){
throw new Exception("mergeQueryString only works on GET requests.");
}
$q = $this->getRequest()->getQuery();
$p = $this->getRequest()->getParams();
if (empty($q)) {
//there's nothing to do.
return;
}
$action = $p['action'];
$controller = $p['controller'];
$module = $p['module'];
unset($p['action'],$p['controller'],$p['module']);
$params = array_merge($p,$q);
$this->_helper->getHelper('Redirector')
->setCode(301)
->gotoSimple(
$action,
$controller,
$module,
$params);
}