after lot of google searches and going through prestashop's official documentation over and over again, I still couldn't find an example of backoffice controller. I even looked into the modules folder of prestashop's installation, but couldn't find any.
I need to implement 3 different back-office pages, each served by its own controller and view.
Can anyone provide me any hint, or even one working example...Just hello world is more than enough.
Thanks in advance...
notice:i write this article for prestashop1.5 and i don't check it for prestashop 1.6.perhaps it is works for 1.6 too.
You should know every thing in this way have special structure.
step 1: Create a folder in your module folder call that 'controllers' (notice:this name is static)
step 2: Create a folder in controller folder call that 'admin' (notice:this name is static too)
step 3: Create a php class file call that AdminMyclassnameController (notice:in this name Admin at first and controller at last is the key word and thay are static but Myclassname is dynamic .attention to first words all of should be uppercase A for Admin, M for Mclass,... )
step 4: Then you should write your class in AdminMyclassnameController and this class should extent with AdminController or AdminControllerCore.
for know how does it class work you can search about helper forums in internet.
step 5: When you create the class you want a tab to show that controller notice:when act to step 1,2,3,4 this controller take an automatic URL for access to this URL you should create a tab.in yourmodule.php in install() you should add this codes
$tab = new Tab();
$tab->class_name = 'AdminTest';
$tab->module = 'test';
$tab->id_parent = 9;
$tab->position = 11;
then you can see the tab in admin office that redirect to your controller.
i am tired to continue ... but if you want more send a message and i continue this article
this is use full site really clear:
http://doc.prestashop.com/display/PS15/Using+helpers+to+overload+a+back-office+template
http://presthemes.com/prestashop-news/modules-classes-and-controller-override-by-julien-breux-4.html
http://doc.prestashop.com/display/PS15/Diving+into+PrestaShop+Core+development
http://www.prestashop.com/forums/topic/270177-solved-making-own-objectmodel-class-doesnt-work/
http://doc.prestashop.com/display/PS15/New+Developers+Features+In+PrestaShop+1.5
http://blog.belvg.com/how-to-implement-a-controller.html
best regards
Related
The main goal that I want to achieve is to add a voucher code if the user has clicked on a specific external link that point to my shop.
So the (javascript?) script will analyze the url and if it contains any selected keyword will add the voucher.
Anyone knows how to do that?
Thanks
I found a pretty simple solution.
I made a module.
You can find it here: GITHUB
Was pretty easy:
Generated a basic module HERE
Selected Header Hook for my new module
Modified the header hook function in the modulename.php file in root with this one:
public function hookHeader()
{
$this->context->controller->addJS($this->_path.'/views/js/front.js');
$this->context->controller->addCSS($this->_path.'/views/css/front.css');
if (Tools::getValue('voucher')){
$cartVoucher = Tools::getValue('voucher');
$idDiscount = Discount::getIdByName($cartVoucher);
Context::getContext()->cart->addDiscount($idDiscount);
}
}
Hope will help someone.
Thanks to all.
Good afternoon,
Please let me know if this question is not clear enough, I'll try my best to make as straight-forward as possible.
How can I add access control to objects that are generated by an end-user using my data object?
Example: I have a class that extends a DataObject. Someone logs in the back-end; fills out the form that's generated by the CMS for the data object. A record is then created in the database by the CMS.
I would like to add an access control to that newly created record in the database.
For a code scenario you can take a look at one of my posts: Silverstripe 3 - Unable to implement controller access security from CMS
The only other way I can think of asking this question is: How to Dynamically (or programmatically) create permissions for records that are created by a DataObject extension via the CMS?
Thanks for your assistance.
Update - Sample Code
///>snippet, note it also has a Manager class that extends ModelAdmin which manages this!
class component extends DataObject implements PermissionProvider{
public static $db = array(
'Title' => 'Varchar',
'Description' => 'Text',
'Status' => "Enum('Hidden, Published', 'Hidden')",
'Weight' => 'Int'
);
///All the regular permission checks (overrides), for the interface goes here, etc...
///That is: canView, canDelete, canEdit, canCreate, providePermissions
}
Now, from the back-end an end-user can add components using the Manager Interface that's generated by extending ModelAdmin. How can I add individual permissions to those added components by the end-user?
Thanks.
Update 2
Example: Add Process Data Object that extends ModelAdmin will give you this in the back end
Then, when you click on the generated 'Add Process' button, you'll get this:
Finally, someone fills out the form and clicks on the 'Create' button, which saves the data in the database. That looks like this:
Now, on that record thats created in MySQL I'd like to add granular permissions to that record. Meaning, for every record created I want to be able to Deny/Allow access to it via a Group/Individual, etc.
Is that even possible with the SilverStripe framework? Thanks.
Implement the functions canView, canEdit, canDelete, and/or canCreate on your DataObject.
Each function will return true or false depending on the conditions you set - any conditions, not just what is defined in the CMS.
See the example code on the tutorial site.
Good day! Tell me, how can I add a quick link on the home page of the administrator to configure my module?
Follow the following steps:
1) In admin section to go Administration at top menu and then click on Quick Access.
2) IN next page click on Add new and you will see a form
3) Now open admin panel in another tab and to the module page or section, of which you want to place a link in quick access.
4) Copy that complete link in note pad and remove the token section of the link. It is required to remove the token section according to Prestashop.
5) Now come back to the add new form for quick access, give your link a name and then copy that modified link to the Url field.
6) Save it and you will have that link in quick access.
The above is method is used to add it at admin. Now if you want to add it pro-grammatically you can follow the following steps.
1) In your module in the install function, user a code like below
Db::getInstance()->insert('quick_access', array('new_window' => 0, 'link' => 'link_to_your_module_page'));
//an entry is made in quick_access table, get the quick_access id to insert lang data
$id = Db::getInstance()->Insert_ID(); //this will give you last inserted ID from quick_access table which is your current quick_access id.
//now make insertions in quick_access_lang table for multi language data.
//get all your site languages, and place a foreach loop and in that loop insert
//data into the quick_access_lang table using below code
Db::getInstance()->insert('quick_access_lang', array('id_quick_access' => $id, 'id_lang' => 'lang_id', 'name' => 'name of your link'));
//Now for uninstalling module, you want to delete the link, so you need to store the quick access link id in configuration table so you can use it later.
Configuration::updateValue('MY_QUICK_ACCESS_LINK_ID', $id);
2) Now in your uninstall function in your module class, place the below code
$id = Configuration::get('MY_QUICK_ACCESS_LINK_ID'); //get id of your quick access link
Db::getInstance()->delete('quick_access', 'where id_quick_access = '.$id);
Db::getInstance()->delete('quick_access_lang', 'where id_quick_access = '.$id);
//now delete the id from config table
Configuration::deleteByName('MY_QUICK_ACCESS_LINK_ID');
Note : The above code is not tested, it may / may not need some adjustments.
Thank you
There is a hook in Prestashop DisplayAdminHomeQuickLinks it will help you to add quick link on prestashop admin panel. I have use this in my theme. http://goo.gl/0S3mn And it will help you to solve the quick link.
In Prestashop 1.6.1 (maybe earlier too) in Admin view, at the top of the page, Quick Access has the option to "Add current page to QuickAccess". So just navigate to the Configuration page you need and use it.
I need to develop an application that extracts all the contents in Content Tab of the Ektron Workarea and I have to keep tree structure of folders (taxonomies,collections,forms,etc.) also.When I click the content I need to get the Content ID in the code behind also.I need to do all these in a single function.
I tried this requirement with the concept of content block widget in workarea.When we drag that widget and edit it a pop up will come and it displays the folders of work area in tree structure.But when I created an aspx page, put the same code and I browse that page I didn't get the tree structure of all contents.Only the main tabs(Folders,Taxonomies and search ) are visible.Then I drag the user control in the aspx page .But it also doest work.
So how will I solve the above problem.
Can I pull all the contents in tree structure from work area from the root using API codes?.Then can anyone please give the API code to solve?
Please anyone reply!
Assuming you are using 8.6 look here to start with:
http://reference.ektron.com/developer/framework/content/contentmanager/getlist.aspx
Update:
I think I misread your question the first time around. Allow me to expand on my answer a bit. My original answer with the web services assumes that you are rendering the content tree from some sort of "presentation tier" -- a different web site, a console app, or a WPF/WinForms app, etc.
You can get the recursive folder structure with something like this:
private FolderData GetFolderWithChildren(long folderId)
{
var folderApi = new Ektron.Cms.API.Folder();
var folderData = folderApi.GetFolder(folderId);
// This next method is marked as obsolete in v9.0;
// a newer overload is available in v9.0, but I
// don't know if it's available in v8.0
folderData.ChildFolders = folderApi.GetChildFolders(folderId, true);
}
I'm a little confused as to what exactly you're trying to accomplish. If you want to show the entire tree structure graphically, have you tried taking the code and markup from the edit view of the content widget and using it on your non-edit view?
I must say, your requirement that "I need to do all these in a single function" worries me a bit. Workarea content trees can get really large very quickly. If you're trying to load all of the folders and all the taxonomies and all the collections, etc. Then the user will likely be waiting a long time for the page to load, and you risk running into timeout issues.
-- Original Answer --
Ektron v8.0 doesn't have the 3-tier option, which is too bad because that would really make your job a lot easier. In v8.0, there are ASMX web services that you can reference, including:
/workarea/webservices/content.asmx
/workarea/webservices/webserviceapi/user/user.asmx
There are lots more than this; browse through the folders within /workarea/ to see what's available.
It's been a while since I've worked with these services, so I'm a little rusty...
Suppose you add references to those two services I listed above and name them ContentService and UserService. The first thing you'll want to do is set the authentication headers. Then you can call the service methods in much the same way as the old legacy apis.
var contentApi = new ContentService.Content();
contentApi.AuthenticationHeaderValue = new ContentService.AuthenticationHeader();
contentApi.AuthenticationHeaderValue.Username = username;
contentApi.AuthenticationHeaderValue.Password = password;
contentApi.AuthenticationHeaderValue.Domain = domain;
var userApi = new UserService.User();
userApi.AuthenticationHeaderValue = new UserService.AuthenticationHeader();
userApi.AuthenticationHeaderValue.Username = username;
userApi.AuthenticationHeaderValue.Password = password;
userApi.AuthenticationHeaderValue.Domain = domain;
var ud = userApi.GetUserbyUsername("jimmy456");
long folderID = 85;
bool recursive = true;
ContentData[] folderContent = contentApi.GetChildContent(folderID, recursive, "content_id");
Drupal has a function called "drupal_is_front_page". Does YII have something similar to deal with navigation in this way?
Unfortunately not. And while the information needed to piece this together is available, doing so is really more pain than it should be.
To begin with, the front page is defined by the CWebApplication::defaultController property, which can be configured as discussed in the definitive guide. But there's a big issue here: defaultController can in reality be any of the following:
a bare controller name, e.g. site
a module/controller pair, e.g. module/site
a controller/action pair, e.g. site/index
a module/controller/action tuple, e.g. module/site/index
If you have specified the defaultController as #4 (which is the same as #3 if your application does not include any modules) then everything is easy:
function is_home_page() {
$app = Yii::app();
return $app->controller->route == $app->defaultController;
}
The problem is that if defaultController is specified as #1 or #2 then you have to examine a lot of the runtime information to convert it to form #3 or #4 (as appropriate) so that you can then run the equality check.
Yii of course already includes code that can do this: the CWebApplication::createController method, which can accept any of the valid formats for defaultController and resolve that to a controller/action pair (where controller is dependent on the module, if applicable). But looking at the source doesn't make you smile in anticipation.
To sum it up: you can either assume that defaultController will always be fully specified and get the job done with one line of code, or borrow code from createController to determine exactly what defaultController points to (and then use the one line of code to check for equality).
I do not recommend looking into solutions based on URLs because the whole point of routes is that different URLs can point to the same content -- if you go that way, can never be sure that you have the correct result.
In my experience, there is no such function in Yii. However, you can retrieve the followings:
base url: Yii::app()->request->baseUrl
current URL : Yii::app()->request->requestUri.
current page controller with Yii::app()->getController()->getAction()->controller->id .
With these APIs, it should be possible to find out whether the current page is front page.
another simple idea:
in your action (that one you use to present your 'main front page'), you could set up a variable using a script in its view:
Yii::app()->getClientScript()->registerScript("main_screen",
"var main_front_page = true;",CClientScript::POS_BEGIN);
put that code in the "main view", (the rest view pages dont have this piece of code).
so when you need to check if a page is the "main page" you could check for it using javascript, quering for:
if(main_front_page){..do something..}.
if you need to recognize the main page in php (in server side), use the method proposed by Jon.
another solution, based on a common method for your controller:
Your controllers all of them must extend from CController, but, when you build a new fresh yii application Gii creates a base Controller on /protected/components/Controller.php so all your controllers derives from it.
So, put a main attribute on it, named:
<?php
class Controller extends CController {
public $is_main_front_page;
public function setMainFrontPage(){ $this->is_main_front_page = true; }
public function getIsMainFrontPage(){ returns $this->is_main_front_page==true; }
}
?>
well, when you render your main front page action, set up this core varible to true:
<?php
class YoursController extends Controller {
public function actionPrimaryPage(){
$this->setMainFrontPage();
$this->render('primarypage');
}
public function actionSecondaryPage(){
$this->render('secondarypage');
}
}
next, in any view, you could check for it:
<?php // views/yours/primaryview.php
echo "<h1>Main Page</h1>";
echo "is primary ? ".$this->getIsMainFrontPage(); // must say: "is primary ? true"
?>
<?php // views/yours/secondaryview.php
echo "<h1>Secondary Page</h1>";
echo "is primary ? ".$this->getIsMainFrontPage(); // must say: "is primary ? false"
?>