Warning: No view partial provided and
no default set in
/Applications/MAMP/htdocs/getv/library/Zend/Paginator.php
on line 465
This is the warning message that I get when loading the paginator; can someone give me a solution or tips where my problem could be?
public function getPaginator() {
if ($this->view === null) {
$this->view = $this->getActionController()->view;
}
$db = Zend_Db_Table::getDefaultAdapter();
/* #var $searcher ZendX_Searcher_Abstract */
foreach ($this->searchers as $searcher) {
$searcher->setRequest($this->getRequest())
->setView($this->view)
->setSelect($this->select)
->perform();
}
$paginator = new Zend_Paginator(new Zend_Paginator_Adapter_DbSelect($this->select));
$paginator->setCurrentPageNumber($this->getRequest()->getParam('page', 1));
$paginator->setPageRange(7);
if ($this->perPage > 0) {
$paginator->setItemCountPerPage($this->perPage);
} else {
$paginator->setItemCountPerPage(PHP_INT_MAX);
}
return $paginator;
}
Hey I was getting the same problem. I am using zend pagination view helper. The solution I got is little strange. In the view(.phtml) file I was checking like
if(isset($this->records) && $this->records!='')
but later than i change this to
if(isset($this->records) && sizeof($this->records) > 0)
and my problem solved. Hope, it may be helpful to you and others.
Correct solution is IMO what RobertPitt proposes:
Zend_View_Helper_PaginationControl::setDefaultViewPartial ('paginator.phtml' );
This error is not triggered from your controller but rather, from your view.
Show the script where you call <?php echo $this->paginationControl(...) ?>
For reference, you need to supply at least two things to the PaginationControl view helper:
A Zend_Paginator object. Supplied as the first argument to the helper or by setting it as the paginator property of your view.
A partial view path. Supplied as the third argument to the view helper (after scrollingStyle) or via the static method RobertPitt mentioned in his comment.
In my case I don't give to view an empty paginator object but an empty string, this to avoid an initial not filtered search.
I solved in this way in my view.phtml:
if ( is_a($this->paginator,'Zend\Paginator\Paginator') ) {
// paginator print
}
I hope this help.
Related
In my web file, I have a route that accepts a $id as a value to be passed to a function within my PagesController. However, I want the function to still execute and show the intended form even when the $id is not passed to the function.
web.php file
Route::get('/request/{id}', 'PagesController#makeRequest');
PagesController.php file
public function makeRequest($id)
{
if(!empty($id)){
$target = Partner::find($id);
}
return view('pages.makeRequest')->with('target', $target);
}
makeRequest.blade.php
<input type="text" class="form-control" value="{{$target->inst_name}}" required disabled>
I want the page to display details from the database with the $id when provided or have empty spaces when the $id isn't provided.
As the Laravel Documentation states: Use Optional Parameters like this:
Route::get('/request/{id?}', 'PagesController#makeRequest'); //Optional parameter
Controller
public function makeRequest($id = null)
{
if(!empty($id)){
$target = User::find($id);
return view('pages.makeRequest')->with('target', $target);
} else {
return view('pageslist'); ///set default list..
}
}
This is the way I did it:
Route::get('index', 'SeasonController#index');
// controller
public function index(Request $request )
{
$id= $request->query('id');
}
The way you call it:
localhost/api/index?id=7
All your solutions were helpful. The main thing was that when I called just the view without passing $target to the view, the page displayed an error. So this is what I did.
Route::get('/request/{id?}', 'PagesController#makeRequest');
Then in the controller,
public function makeRequest(Request $request, $id=null)
{
if ($id != null) {
$target = Partner::find($id);
return view('pages.makeRequest')->with('target', $target);
}
return view('pages.makeNullRequest');
}
If you didn't understand what happened, I created a new view which had this instead of what I had posted in the question.
<input type="text" class="form-control" value="" required readonly>
Sorry I didn't update you guys in time. I think Jignesh Joisar came closest to helping me solve this issue. really appreciate all you guys. You're just awesome
You can use optional parameter :
Route::get('/request/{id?}', 'PagesController#makeRequest');
Now, as the parameter is optional, while defining the controller function you need to assign its default value to null in argument declaration.
<?php
public function makeRequest($id = null)
{
if($id){
$target = Partner::findOrFail($id);
return view('pages.makeRequest')->with(compact('target'));
}
// Return different view when id is not present
// Maybe all targets if you want
$targets = Partner::select('column1', 'column2')->get();
return view('pages.all')->with('targets');
}
I am using findOrFail instead of find. Its Laravel's very handy function which automatically throws a ModelNotFound exception and for frontend user throws a simple 404 page.
So if anyone is accessing www.url.com/request/2, its a valid id then it will show a valid page with data. If the accessed url is www.url.com/request/blahblah then it will throw 404. It avoids efforts of handling this manually.
For optional parameter pass id with ? in route and give $id = null in your function's parameter like this:
Route::get('/request/{id?}', 'PagesController#makeRequest'); //Optional parameter
makeRequest($id = null) {
// Code here...
...
}
in your routes file (web.php , as mentioned in your question)
Route::get('/request/{id?}', 'PagesController#makeRequest');
and in your controller PagesController.php
public function makeRequest($id = null)
{
}
To read more about this, just read https://laravel.com/docs/5.7/routing#parameters-optional-parameters
For me the answer was in the order that I listed the Routes in the routes file.
The routes file will call the first one that matches the pattern.
Route::get('/ohmy/{id?}', 'OhMyController#show');
Route::get('/ohmy/all', 'OhMyController#all'); //never gets called
Instead, put optional parameters at end of list:
Route::get('/ohmy/all', 'OhMyController#all');
Route::get('/ohmy/{id?}', 'OhMyController#show');
the answer has been said. just a side note: optional parameters won't work if you are using resource routes.
for example:
Route::resource('items',itemController::class)->except([
'create',
]);
Route::get('/items/create/{category_id?}',function($category_id = 'abc'){
dd($category_id);
});
if i go to " items/create/1 ", the result will be "1".
if i go to " items/create ", it will return 404. ( but we expect it to say "abc".)
this happens because other routes that start with "items" are expected to be generated from "resource" functionality.
so if you use resource routes, you should consider that.
I doing tutorial follow http://book.cakephp.org/3.0/en/development/errors.html#exception-renderer but it is not working and display blank page.
In config/bootstrap.php
use App\Error\AppError;
$errorHandler = new AppError();
$errorHandler->register();
In src/Error/AppError.php
<?php
namespace App\Error;
use Cake\Error\BaseErrorHandler;
class AppError extends BaseErrorHandler
{
public function _displayError($error, $debug)
{
return 'There has been an error!';
}
public function _displayException($exception)
{
return 'There has been an exception!';
}
public function handleFatalError($code, $description, $file, $line)
{
return 'A fatal error has happened';
}
}
I create my_error.ctp in src/Template/Layout/my_error.ctp. And in my src/Template/Error/error404.ctp I change layout to my_error.ctp.
$this->layout = 'my_error';
Finally, In my controller
use Cake\Network\Exception\NotFoundException;
$staff = $this->Staff->find()->where(['Staff.StaffId = '=> $id, 'Staff.PartnerId = ' =>$this->partnerId])->first();
if (empty($staff)) {
throw new NotFoundException(__('Staff not found'));
}
Whenever encountering blank pages, enabled debug mode, visit the URL again, and check your error logs.
However, problem in this case is most likely that the docs are incorrect/misleading, as the example app error won't do anything at all. The _ prefixed methods are ment to be protected, having them return something has no effect, and handleFatalError is ment to return a boolean.
Just look at the source of Cake\Error\BaseErrorHandler and the core error handler Cake\Error\ErrorHandler, the methods that you are overwriting are ment to generate output!
You may want to report that as an issue over at GitHub.
If all you want to do, is create a custom 4xx error page, then all you need to do is to edit the src/Template/Error/error400.ctp template accordingly.
I found my mistake. :(
Because in bootstrap.php I copy below code at the end of file. Therefore Cake cannot understand it. Please close this issue. Thank you for support.
use App\Error\AppError;
$errorHandler = new AppError();
$errorHandler->register();
I'm using Cakephp with json parse extension and the RequestHandler component in order to create Web services using json.
I created a controller named Ws
In this controller I have a named userSubscribe
In order to avoid a lot of If else statements in the next methods, I thought about using a private function inside this controller that will check somes conditions and stop the script normaly BUT ALSO render the json normaly. I just want to do a DRY way !
My question is :
How could I render the json view in a sub function (called by the userSubscribe) ?
To make it clear, here is the style code that would like
public function userSubscribe() {
$this->check();
// Following code only executed if check didn't render the json view
// $data = ...
$code = 1;
$i = 2;
}
private function check() {
$input = &$this->request->data;
if ($_SERVER["CONTENT_TYPE"] != "application/json") { // For example
$result = "KO";
$this->set(compact("result"));
$this->set('_serialize', 'result');
$this->render(); // HERE, it will stop the 'normal behaviour' and render the json with _serialize
}
if (!isset($input["input"])) {
$result = "KO";
$this->set(compact("result"));
$this->set('_serialize', 'result');
$this->render(); // HERE, it will stop the 'normal behaviour' and render the json with _serialize
}
}
It's seems to be quite simple to do, but why can't I find the answer ?!
Thanks in advance for clue/advise/anything !
I'm trying to figure out how to use one of my view elements inside of a controller...
I know, I know: "Don't do that!" (99% of the time this is the correct answer)
But I think I actually have a good reason. The action is handling an AJAX request which returns markup. The returned markup is a list which I display everywhere else using an element. So in an effort to keep my code DRY, I think it's appropriate to do this here.
Is this possible?
Easy:
$view = new View($this, false);
$content = $view->element('my-element', $params);
Also:
DON'T DO THAT ANYMORE!!!
Sometimes, you need to render a CakePhp element from a view and inject its content into the page using AJAX the same time. In this case rendering element as a regular view from controller is better than creating a dedicated view that just contains <?php echo $this->element('some_element') ?>, and may be done this way:
<?php
public function ajax_action() {
// set data used in the element
$this->set('data', array('a'=>123, 'b'=>456, 'd'=>678));
// disable layout template
$this->layout = 'ajax';
// render!
$this->render('/Elements/some_element');
}
I know this is an old question and other people have already given basically the same answer, but I want to point out that this approach (provided by Serge S.) ...
<?php
public function ajax_action() {
// set data used in the element
$this->set('data', array('a'=>123, 'b'=>456, 'd'=>678));
// disable layout template
$this->layout = 'ajax';
// render!
$this->render('/Elements/some_element');
}
...is not a hacky workaround, but is in fact the recommended approach from the CakePHP docs for this common and legitimate use case:
If $view starts with ‘/’, it is assumed to be a view or element file
relative to the /app/View folder. This allows direct rendering of
elements, very useful in AJAX calls.
(Again: Credit to Serge S. for the code above)
$this->view = '/Elements/myelement';
You should use a client-side template. You should never return mark-up from a web service or API, just data. Have your JavaScript take the data, and then format it how you wish.
For example:
function getItems() {
$.get('/some/url', function(response) {
if (response.data.length > 0) {
for (var i = 0; i < response.data.length; i++) {
var item = response.data[i];
$('.results').append('<li>' + item.title + '</li>');
}
}
});
};
This is just an example written off the cuff. Obviously you’ll need to write your own implementation.
The way I did any ajax handling in Cake was to have my own AjaxController. Any interaction of ajax-kind goes there, which in-turn uses their own views (and view partials / elements). That way you can keep your code DRY and isolate and propagate all ajax use-cases there.
Example excerpt:
<?php
class AjaxController extends AppController {
/**
* (non-PHPdoc)
* Everything going to this controller should be accessed by Ajax. End of story.
* #see Controller::beforeFilter()
*/
public function beforeFilter() {
parent::beforeFilter();
$this->autoRender = false;
$this->layout = false;
if (!$this->request->is('ajax')) {
$this->redirect('/');
}
}
public function preview() {
if ($this->request->is('ajax')) {
$this->set('data', $this->data);
$this->render('/Elements/ajaxpreview');
}
}
?>
Here's the source: https://github.com/Sobient/dosspirit/blob/master/app/Controller/AjaxController.php
I am trying to attack this problem from a completely different angle, because it doesn't look like I can achieve my goal that way.
I want to loop over the item stack in the HeadScript View Helper, and make modifications to it. The documentation for this and some of the other view helpers makes this statement:
HeadScript overrides each of append(),
offsetSet(), prepend(), and set() to
enforce usage of the special methods
as listed above. Internally, it stores
each item as a stdClass token, which
it later serializes using the
itemToString() method. This allows
you to perform checks on the items in
the stack, and optionally modify these
items by simply modifying the object
returned.
So, where is this "object returned"? I am missing a piece of the puzzle here.
Thanks for your help!
In the toString() method of Zend_View_Helper_HeadScript I noticed a foreach() loop on $this, so I tried that and it worked. Here's a HeadScript extension I wrote that illustrates the solution:
class My_View_Helper_HeadScript extends Zend_View_Helper_HeadScript
{
public function toString($indent = null)
{
$files = array();
foreach ($this as $key => $item) {
if (!empty($item->attributes)
&& array_key_exists('src', $item->attributes)
&& ('scripts' == substr($item->attributes['src'], 1, 7))) {
$files[] = $item->attributes['src'];
unset($this[$key]);
}
}
if (0 < count($files)) {
$this->prependFile('/combo.php?type=scripts&files=' . implode(',', $files));
}
return parent::toString($indent);
}
}
In Bootstrap.php the following lines to point to my helpers:
$this->bootstrap('view');
$view = $this->getResource('view');
$view->addHelperPath('My/View/Helper', 'My_View_Helper');
In my layout, I have this line:
<?php echo $this->headScript(); ?>
If my solution is unclear in any way, let me know and I'll update it to clarify.