I have a content type, f.e mycontentype, which has titles, like:
Title - library1
another title - library1
yet another title - library1
so i went in the db and with a MySQL command, i updated all titles to
Title
another title
yet another title
My issue here is that in a view, i have as a field, the
Content: mycontenttype
which i have set to the formatter as title link.
This brinks as the title the old title Title - library1, but when this link is clicked, we go the the node where the title is shown correctly.
if in the formatter i choose full content, i get again all data updated, including the correct title.
Of course i have cleared all caches, i have run update.php,
any ideas?
So you manually updated the title field in node table, and you probably forgot to update node_revision as well.
You may also have to run entity_get_controller('node')->resetCache($nids); $nids being the affected nodes.
Don't update the database manually, in this situation you could do something like this :
$nodes = node_load_multiple(array(), array('type' => 'mycontentype'));
foreach ($nodes as $nid => &$node) {
$node->title = preg_replace('/^(.+?)\s+-\slibrary1$/', '$1', $node->title);
$node->save();
}
Related
I'm creating a webform to build up a subscription form, on Drupal 7.65
Goal
What I need to do is: to select a role from a list, and automatically to display the associated name of that role, in a text field.
As I said, the name should be displayed into a not modifiable text field just below it.
Suppose valid, the following list (key => value)
Field: Department
business_manager|Business Manager
hr_consultant|Human Resources
training_developer|Training Developer
and from the time going on, the associated names, are respectively
Options can appear into text field hr_business_partner
Steve Abc
Gertrude Def
Sven Hgj Klm
Thus when the trainee selects "Human Resources", the name of "Gertrude Def" should appear into the text field below the select one.
I've attached a mokup to better understand what I do need.
IMPORTANT
I can't put the names into the list as value, because the association can change but old records should keep the previously registered associations
You can use hook_form_alter() and add a new select field with the paired key value list you need to the webform. And then use javascript to update which field value gets shown in the HR Business Partner field on change, which by the way would also need to be added via your hook_form_alter. You could use a taxonomy to maintain a list of Departments/Business partners which would populate your department and business partners.
Write some javascript to dynamically update your original fields not added through the form_alter, on change. I would suggest making two textfields in your webform components which will hold the value from your form alter added fields. So that these values selected by the user gets saved in your form.
function MODULENAME_form_alter(&$form, &$form_state, $form_id) {
if($form_id == "webform_client_form_####"){
$form['#attached']['js'] = array(drupal_get_path('module','MODULENAME') . '/js/webform.js');
$form['hr_dept'] = array(
"#type" => "select",
"#options" => array("business_manager"=>"Business Manager", "hr_consultant"=>"Human Resources"),
);
$partners = taxonomy_get_tree(#); //the VID of the taxonomy
$list = array("0"=>"None"); //first option
foreach($partners as $tid => $partner){
$list[$partner->tid] = $partner->name;
}
$form['hr_partner'] = array(
'#type' => 'select',
'#options' => $list,
);
}
}
In your javascript file, /js/webform.js you can include all your logic to check for which value is selected on the Department field and then display the correct value in the Partners fields. At the same time, updating the original fields you've added as textfields in the webform components UI.
I'm in the process of converting a 10 year old PHP application. After my boss hired a php consultant, he has set up a CakePHP application environment and we are learning as we go. (fun, I know). Also, I come from a javascript/sharepoint background and have not had a lot of php experience.
As a test, I created a basic address table with these fields: firstname, lastname, state, phonenumber. I've been using justice league members as names and other test data to populate my table. Baked it just fine, default bootstrap pages are working.
I decided I wanted to add a dropdown field called current status, and for now just to keep it simple I wanted the choices: alive, dead.I created the column in my address table.
I created a second table called statuses and pointed the status column in my first table to the status table, using the status id as the foriegn key.
Baked my new table and rebaked my old one.
The status drop down does not give my choices of dead or alive, If I click in the field I get an up or down arrow, and based on which one you click, it either increments or decrements by 1. So the first time I click it inserts a 0. If I go up or down, it adds or takes away one.
I'm not sure what I'm doing wrong, I'm guessing there is some additional code I need to add to the MVC?
ok, if this works, then a lot is working :-). Now to the following: set in the Status Model a query like this:
public function getStatus()
{
$opt = $this->Status->find('list', array(...));
return $opt;
}
Then get the list over to the Adress Controller like this:
$this->loadModel('Status');
$opt => $this->Status->getStatus();
$this->set('opt', $opt);
Now you are able to access the $opt in the view file.
Just delete this line in the view:
$opts = array('0' => __('dead'), '1' => __('alive'));
And it should work.
Keep it simple. Ad to your table this row (only to understand how it works): 'status' as typ "tinyint(1)". Then set this in your view file:
$opts = array('0' => __('dead'), '1' => __('alive'));
When you create the inputfield, do it like that:
echo $this->Form->input('Address.status', array('options' => $opts, 'label'
=> __('Status')));
This should work.
Just started using ATK4 and appreciating it very much so far, but not sure how to do this...
What I am trying to accomplish:
I am outputting a query's results to a grid, one of the fields is 'status', the data will either be '-1' or '1'.
Instead of outputting -1 or 1 to the column, how do I output an HTML snippet (or whatever I need to to get what I want) instead that shows a different icon for each value?
In short:
In column 'status':
if the value is -1, display iconDown.gif;
if the value is 1, display iconUp.gif
Code so far:
class page_showlist extends Page {
function init(){
parent::init();
$q=$this->api->db->dsql();
$q->table('remote_system')
->join('customers.id','customer_id')
->field('customer_id')
->field('ip')
->field('nickname')
->field('name','customers')
->field('status')
;
$grid = $this->add('Grid');
$grid->addColumn('text','status')->makeSortable();
$grid->addColumn('text','name')->makeSortable();
$grid->addColumn('text','ip');
$grid->addColumn('text','nickname');
$grid->addButton('Reload Grid')->js('click',$grid->js()->reload());
$grid->addQuickSearch(array('name'));
$grid->setSource( $q );
}
}
Any pointers/tips?
To add column with icons in Grid you can use custom template.
In one of my projects I do like this:
$url = $this->api->pm->base_url . $this->api->locateURL('template', 'images/');
$grid->addColumn('template', 'type', false)
->setTemplate('<img src="' . $url . 'icon_object_<?$type?>.png">');
It'll use model field named type (in your case use status) and show icons in that column. Icon source URL is generated dynamically and it'll search for image files in your template/images directory named icon_object_XXX.png where XXX value will be taken from field type value.
In my case type is like this: array('building','apartment','land','garage') etc.
And one more thing - you should start using Models whenever possible! That way you'll ease your life later when your project becomes bigger. Also can have extra security (conditions, etc.) with them.
Puling my hair out with this, I know its something very simple but I've been struggling to resolve it. For the record I'm using cakePHP 2.2.
I'm trying to populate a select input with related model data.
Basically I have users and user_statuses. The user model belongsTo the UserStatus model. Within the user/add() function I want to have a drop down select box populated with the data from the UserStatus model.
Heres the code:
UsersController/Add()
$groups = $this->User->Group->find('list');
$UserStatus = $this->User->UserStatus->find('list');
$this->set(compact('groups', 'UserStatus'));
View/Users/add.ctp
echo $this->Form->input('user_status_id', array('class' => 'span9'));
The field in the users table making the relationship is user_status_id which links to the id in the user_statuses table.
Based on the above the select box will NOT populate with the statuses from the user_statuses table. Interestingly enough, it will if I change the view code to just :
echo $this->Form->input('user_status_id', array('class' => 'span9'));
However, when I do this is will NOT write to the database.
Any help will be gratefully received.
Thanks in advance.
The options param should solve the problems for display and as long as there is a user_status_id in your User table, the below code should save the value in your User model on save.
echo $this->Form->input('User.user_status_id', array('options'=>$UserStatus,'class' => 'span9'));
Should be:
$userStatus = $this->User->UserStatus->find('list');
(note the lowercase 'u')
I'm developing with cakePhP and I have the following problem:
When a user logs in with his name and password to the account system that I've created, he can save items (images) as favorites. This is saved in a text field into the database. What is saved is the image ID.
The saving process works perfectly, the user clicks on the images and they're added to that field (it actually saves all the IDs as a text array that I process later).
The problem comes when removing images. When the user does it (I'll post the code below), the images is removed correctly from the database (I go to PHP MyAdmin and I see it). This means that the array that holds the favorite images IDs is updated instantly. However, when I reload that array from the website, it hasn't been updated. It's like it's stored in the caché or something. Then, if the user logs out and logs in again, then he can see the correct one. The thing is that I have other things in my website that work in a similar way and they all get updated instantly, so I can't see why this doesn't.
This is the code that I use to remove the ID from the database:
function remove_favorite($pictureID) {
$this->User->id = $this->Auth->User('id'); //We get the ID of the current user
$favoritesArray = $this->User->deleteFavoritePicture($this->User->id, $pictureID); //This function retrieves the array (string) of pictures from the user's table, and deletes all the images with the ID passed as parameter, returning the updated array (string)
$fields = array('images_favorites' => $favoritesArray, 'modified' => true); //We indicate the field that we're going to update in the users table
//We save the new string that doesn't contain the deleted image anymore
if($this->User->save($fields, false, array('images_favorites'))) {
$this->Session->setFlash(__('The image has been removed from your favorites', true));
} else {
$this->Session->setFlash(__('Error removing image from favorites, please try again', true));
}
$this->redirect(array('action' => 'manage_favorites',$this->User->id));
}
This is how the deleteFavoritePicture function looks like:
function deleteFavoritePicture($userID, $pictureID) {
$userInfo = $this->find("id = $userID");
$favoritePicturesString = $userInfo['User']['images_favorites'];
$favoritePicturesArray = explode(",", $favoritePicturesString); //Array
$i = 0;
while ($i < count($favoritePicturesArray)) {
//We remove from the array the images which ID is the one we receive to delete
if ($favoritePicturesArray[$i] == $pictureID) unset($favoritePicturesArray[$i]);
$i++;
}
$favoritePicturesString = implode(",", $favoritePicturesArray); //String
return ($favoritePicturesString);
}
That's it. Does anyone now what can be going on? Thanks so much in advance for any clue!
EDIT
Ok, I think I found something that may give a clue of what's going on here:
This is the code for the manage_favorites action:
function manage_favorites($id) {
//$user = $this->User->find("id = $id");
$user = $this->Auth->user();
$this->set('user', $user);
}
That is the action that is called for the page when a user wants to modify his favorites. The same action is called once the user removes a favorite. Here's the thing:
If I use the $id parameter in the manage_favorites function and the $user = $this->User->find("id = $id"); line (the one quoted now), then the problem does not exist! This is how I used to have it. HOWEVER, I had to change it because it was a big security flaw, since the user id ($id) was a visible parameter who anyone could change, and then access other users accounts. What I did was changing the way I obtain the user array of favorite images, using the following line: $user = $this->Auth->user();. This is how I have it now (well, and also without the $id parameter in the function header), so the user information (including the favorites array) comes from the Auth component, instead directly from the database.
So, the problem is clear: when the user deletes a favorite, it's doing it on the array in the database. WHen I show the result of that operation, the array I'm retrieving is not the one in the DB, it's the one in the session. That's why it's not showing the changes.
How can I avoid this without using a non-secure method like the one I had before?
When you save, the array passed to the save method of the model should look like this:
[User] => array(
[field] => value,
[field2] => value2,
...
)
In your example, you clearly haven't added the [User] key.
Also, is your modified field actually the default Cake modified field? That is, the DATETIME field which changes to the current time when the row is updated?
Lastly, maybe you have debugging set to 2 in config.php. try changing this to 0 (as in production) and see if caching persists.
Hope some of the points I have mentioned above will solve your problem. Please let me know!
There could be two things wrong with this.
What does your deleteFavoritePicture method look like? There could be something being done wrong there.
You're passing false as the second parameter to the User::save method, which means that you don't want to validate. Unless there is a SQL error, then this will return true even if it doesn't validate properly, I believe. Try changing this false to true and see if your results differ.