I have a module named attendance but value in it is comes from other tables with some conditions. so how can i get id of other tables. my code is like below:
models are:
if($type == 2)
{
$model = new Supplier();
}
elseif($type==3)
{
$model = new TruckingCompany();
}
elseif($type==4)
{
$model = new ServiceProvider();
}
elseif($type==5)
{
$model = new Landowner();
}
elseif($type==6)
{
$model = new Refiner();
}
elseif($type==8)
{
$model = new Worker();
}
elseif($type==9)
{
$model = new Trainer();
}
now i want to update this record
array(
'header'=>'Action',
'class' => 'CButtonColumn',
'template'=>'{update}',
'updateButtonUrl'=>'Yii::app()->createUrl("attendance/update", array("id"=>$data->id))',
);
now i want to get id of these diffent table for update record and these are different in all tables. i can get it by below line:
'updateButtonUrl'=> 'Yii::app()->createUrl("attendance/update", array("id"=>$data->supplier_master_id))';
but it is not a proper way to define by these for all tables so what to do in this case... any idea... i am new in yii.
Why do not you create the URL along with the models like
if($type == 2)
{
$model = new Supplier();
$updateUrl = 'Yii::app()->createUrl("attendance/update", array("id"=>$data->id))';
}
elseif($type==3)
{
$model = new TruckingCompany();
$updateUrl = 'Yii::app()->createUrl("attendance/update", array("id"=>$data->supplier_master_id))';
}
Pass it to the view while rendering
$this->render('view',array('updateUrl'=>$updateUrl));
And then finally use it in view file like
array(
'header'=>'Action',
'class' => 'CButtonColumn',
'template'=>'{update}',
'updateButtonUrl'=>$updateUrl,
);
In addition to let me see's answer. I would put it in array form:
$types = array(
1 => array("modelName"=>"Supplier","url"=>'Yii::app()->createUrl("attendance/update", array("id"=>$data->id))'),
2 => array("modelName"=>"TruckingCompany","url"=>'Yii::app()->createUrl("attendance/update", array("id"=>$data->supplier_master_id))'),
);
Of course you can improve on this by dividing the "url" key for example like this:
"url"=>array(
"route"=>"attendance/update",
"params"=>'"id"=>$data->id',
)
This way you can pass it like this:
if(array_key_exists($type,$types)){ //Checks if you've defined the type.
$this->render('view',array(
'updateUrl'=>$types[$type]['url'],
'model'=>new $types[$type]['modelName'](), //Creates your object.
)
);
return; //just to be sure.
}
throw new CHTTPException(404, "Type not Found!");
Your view would look something like this with the improved "url" key:
array(
'header'=>'Action',
'class' => 'CButtonColumn',
'template'=>'{update}',
'updateButtonUrl'=>'Yii::app()->createUrl(' . $updateUrl['route'] . ', array(' . $updateUrl['params'] . '))',
);
This clears up your code a bit and gets rid of the huge if statement/switch.
The createUrl string is still a bit ugly, so you can improve on that further if you would like.
Related
I am working on zend framwork 2
I have created one module with two fields
1) Test1
2) Test2
Database structure for this :
db name : zend_test_db
db fields : config_key , config_value
I want to store like config_key = test1key and config_value : textbox
enter value
Multiple records store at a time.
below is my controller function :
public function indexAction()
{
$form = new ConfigurationForm();
$form->get('submit')->setValue('Save Settings');
$form->get('test1key')->setValue('test1key');
$form->get('test2key')->setValue('test2key');
$request = $this->getRequest();
if ($request->isPost()) {
$configuration = new Configuration();
$form->setInputFilter($configuration->getInputFilter());
$form->setData($request->getPost());
if ($form->isValid()) {
$configuration->exchangeArray($form->getData());
$this->getConfigurationTable()->saveConfiguration($configuration);
// Redirect to list of configuration
return $this->redirect()->toRoute('configuration');
}
}
return array('form' => $form);
}
Above code works fine on Add fields . I am able to insert those fields and stored as key and value
But i am not able to update this.
Hope its clear
Where do i make mistake ?
I am not able to comment as I have reputation less then 50. I think you are trying to say that you are able to insert the data in database but you are not able to update it.
This is happening because you are creating new model every time.
$configuration = new Configuration();
You should initialize it using id params.
$id = (int) $this->params()->fromRoute('id', 0);
if (!$id) {
return $this->redirect()->toRoute('configuration', array(
'action' => 'add'
));
}
try {
$configuration = $this->getConfigurationTable()->getConfiguration($id);
}
catch (\Exception $ex) {
return $this->redirect()->toRoute('configuration', array(
'action' => 'index'
));
}
Using this you will be able to update the data as well. For more reference you can check zend framework Album module. Here is the link
https://framework.zend.com/manual/2.2/en/user-guide/forms-and-actions.html
If this was not the problem then please let me know so that I can help you in this concern.
I have a master table jobs with multiple location in separate table job_location. Now I am not able to update/delete, if extra rows found from job_location. Now why I am saying DELETE is because sync() did this, but it's related to many-to-many relation. I am new to laravel, just trying to get eloquent approach to achieve this, otherwise deleting all rows and inserting can be done easily OR updating each and delete remaining is also an option but I wonder Laravel has something for this.
In every request I get multiple job locations(with unchanged/changed city,phone_number,address) which is creating trouble.
Some codeshots:
Model: [Job.php]
class Jobs extends Model
{
protected $fillable = [
'job_id_pk', 'job_name','salary'
];
public function joblocation() {
return $this->hasMany('\App\JobLocation', 'job_id_fk', 'job_id_pk');
}
}
Model:[JobLocation.php]
class JobLocation extends Model
{
protected $fillable = [
'jobl_id_pk', 'job_id_fk','city', 'address', 'phone_number'
];
public function job() {
return $this->belongsTo('\App\Jobs', 'job_id_fk', 'job_id_pk');
}
}
Controller:[JobController.php]
function jobDetail() {
if($params['jid']) {
// update
$obj = \App\Jobs::find($params['jid']);
$obj->job_name = $params['name'];
$obj->salary = $params['salary'];
$obj->save();
} else {
// create new
$data = array(
'job_name' => $params['name'],
'salary' => $params['salary'],
);
$obj = \App\Jobs::create($data);
}
// don't bother how this $objDetail has associative array data, it is processed so
foreach ($params['jobLocations'] AS $key => $objDetail) {
$jobLoc = new \App\JobLocation;
$jobLoc->city = $objDetail['city'];
$jobLoc->phone_number = $objDetail['phone_number'];
$jobLoc->address = $objDetail['address'];
$jobLoc->job()->associate($obj);
$obj->jobLoc()->save($jobLoc);
}
}
In this approach I am able to save all job locations, but I am using same function to update also. Please tell how I can update jobLocations if present. I am ok to loose previous entries, but it would be good if previous gets updated and new get entered OR if we have extra entries they get deleted. I know sounds weird but still guide me a way.
Yea, you cannot use the same function, do this
$jobs = \App\Jobs::find($params['jid']);
foreach ($params['jobLocations'] as $key => $objDetail) {
$joblocation = $jobs->joblocation->where('jobl_id_pk', $objDetail['some_id'])->first();
//here update you job location
$joblocation->save();
}
Something like this:
Controller:[JobController]
public function jobDetail() {
if( !empty($params['jid']) ) {
// update
$job = \App\Jobs::find($params['jid']);
$job->job_name = $params['name'];
$job->salary = $params['salary'];
$job->save();
} else {
// create new
$data = array(
'job_name' => $params['name'],
'salary' => $params['salary'],
);
$job = \App\Jobs::create($data);
}
$locationDetails = !empty($params['jobLocations']) ? $params['jobLocations'] : [];
$jobLocations = array_map(function($location) use($job) {
$location = array_merge($location, [ 'job_id_fk' => $job->job_id_pk ]);
return \App\JobLocation::firstOrNew($location);
}, $locationDetails);
$job->jobLocations()->saveMany($jobLocations);
}
I am using the MultiForm module to submit a long form with SilverStripe. The logic for this form is in 'CampaignBriefForm.php' whereas the gridfield CMS field is being added in 'CampaignBriefPage.php'. I have a Data Object for a CampaignBriefLead which is what the form creates.
Campaign Brief Page
private static $has_many = array(
'CampaignBriefLeads' => 'CampaignBriefLead'
);
public function CampaignBriefForm() {
return new CampaignBriefForm($this, 'CampaignBriefForm');
}
Campaign Brief Lead (DO)
private static $has_one = array( "Page" => "CampaignBriefPage" );
As you can see the Campaign Brief page has the correct relationship with the Data Object and also you can see the the form itself (done in a sepearate file) is correctly returning (as it's being saved in the DB). For some reason however, the gridfield will not show me what is in the database for that Data Object. The grid field code is as follows.
$fields = parent::getCMSFields();
$contactConfig = GridFieldConfig_RelationEditor::create();
$contactConfig->getComponentByType('GridFieldDataColumns')->setDisplayFields(
array(
'CompanyName' => 'Company Name',
'StartDate' => 'Start Date',
'Duration' => 'Duration',
'WebsiteURL' => 'Website',
'Budget' => 'Budget'
));
$contactGrid = new GridField(
'CampaignBrief',
'Campaign Enquiries',
$this->CampaignBriefLeads(),
$contactConfig
);
$fields->addFieldToTab("Root.Enquiries", $contactGrid);
To me this all looks correct and should work but for some reason it is not working.
Note
The link existing option on the gridfield allows me to link one of the entries from the DO with the gridfield weirdly?? So it saves one entry but I have to do it manually, this tells me it can see the DB but won't pull for some reason.
For reviewing reasons, here is the code for the multiform where the campaign brief lead is actually saved to the DB after the form is submitted.
public function finish($data, $form) {
parent::finish($data, $form);
$steps = DataObject::get(
'MultiFormStep',
"SessionID = {$this->session->ID}"
);
$enquiry = new CampaignBriefLead();
foreach($steps as $step) {
$data = $step->loadData();
foreach($data as $key => $value) {
if($key == 'url' || $key == 'MultiFormSessionID' || $key == 'action_finish') {
continue;
}
if(isset($data[$key])) {
$enquiry->$key = $data[$key];
error_log($data[$key]);
}
}
}
$enquiry->write();
$this->controller->redirect('/campaign-brief/');
}
If you need anything more let me know. Thanks.
I would take a guess that the CampaignBriefLead PageID is not being set on your form submission.
Check the CampaignBriefLead table in your database and check the PageID column. If it is blank, null or 0 for each row then it is not being set.
One way to fix this problem for any new submission is to set the PageID for the $enquiry:
public function finish($data, $form) {
// ...
$enquiry = new CampaignBriefLead();
if ($campaignBriefPage = CampaignBriefPage::get()->first()) {
$enquiry->PageID = $campaignBriefPage->ID;
}
// ...
}
For the existing entries you will need to update the entries to have the correct PageID.
I am building a simple mechanism where a user can like a post by clicking on a link. I'm using GET rather than POST as I want to allow the method to fire via the URL.
That been said how do I save data using GET? As the request data doesn't exist in this scenario... My model looks like:
class Like extends AppModel
{
public $name = 'Like';
public $belongsTo = array('User','Post');
}
and the method for adding looks like:
public function add( $id )
{
$post = $this->Post->find('first', array(
'conditions' => array('Post.id'=>Tiny::reverseTiny($id))
));
if (!$post)
{
throw new NotFoundException('404');
}
if($post['Post']['user_id'] == $this->Auth->user('id'))
{
$this->Session->setFlash('You can\'t like your own post... That\'s just silly!');
}
if ($this->Like->create())
{
$liked = $this->Like->find('first', array(
'conditions' => array('Like.id'=>Tiny::reverseTiny($id), 'Like.user_id'=>$this->Auth->user('id') )
));
if(!$liked){
$this->Like->saveField('user_id', $this->Auth->user('id'));
$this->Like->saveField('post_id', $post['Post']['id']);
$this->redirect(array('controller'=>'posts','action'=>'view','id'=>Tiny::toTiny($post['Post']['id']),'slug'=>$post['Post']['slug']));
} else {
$this->Session->setFlash('You already like this post!');
}
else
{
$this->Session->setFlash('Server broke!');
}
}
Can anyone help?
<?php echo $this->Html->link('1', array('controller'=>'followers','action'=>'add','id'=>Tiny::toTiny($post['Post']['id'])),
array('title'=>'Follow','class'=>'follow')); ?>
This part all works fine. It's saving a new row in the DB on GET that I'm struggling with.
Hi you just need to make a link to your controller action and pass you variable in the url.
to be clear the link on the post to like is in your post view :
$this->Html->link('like this post', array('controller' => 'like', 'action' => 'add', $postId))
It should render a link like this :
www.yourWebSite/likes/add/1 to like the postId 1,
variables after your action (add) are interpreted as variable for your controller action
if your fonction add had been
public function add($postId, $wathever){
}
the url should look like www.yourWebSite/likes/add/1/blabla
where 1 is the first var for the add action and blabla the second one and so on.
this is the equivalent of a non rewriting url : ?postId=1&whatever=blabla
EDIT :
if(!$liked){
//simulate the post behaviour
$this->request->data['Like']['user_id'] = $this->Auth->user('id');
$this->request->data['Like']['post_id'] = $post['Post']['id'];
//save the data
if ($this->Like->save($this->request->data)) {
$this->Session->setFlash(__('Thanks for your support !'));
$this->redirect(array('controller'=>'posts','action'=>'view','id'=>Tiny::toTiny($post['Post']['id']),'slug'=>$post['Post']['slug']));
} else {
$this->Session->setFlash('Server broke!');
}
}
How about using save with id=0 instead of create?
<?php
$like = array(
"Like" => array
(
"id" => 0,
"user_id" => $this->Auth->user("id"),
"post_id" => $post['Post']['id']
)
);
$result = $this->Like->save($like);
if(!$result){$this->Session->setFlash('Server broke!');}
$this->redirect(array('controller'=>'posts','action'=>'view','id'=>Tiny::toTiny($post['Post']['id']),'slug'=>$post['Post']['slug']));
?>
How can I pass the model in array format.
I want to pass models in this format from controller to view:-
Users[user_contact]=Contact
Users[user_contact][contat_city]=City
Users[user_contact][contact_state]=state
This is what I am doing
public function actionCreate() {
$user = new Users;
$presContact = new Contacts;
$presCity = new Cities;
$presState = new States;
$contactArr = array();
// Uncomment the following line if AJAX validation is needed
// $this->performAjaxValidation($model);
if (isset($_POST['Users'])) {
$transaction = CActiveRecord::getDBConnection()->beginTransaction();
$contactArr = CommonFunctions::saveContact($_POST['Users']['user_pres_contact'],'user_pres_contact',$errorArr);
$presContact = $contactArr['contact'];
$presCity = $contactArr['city'];
$presState = $contactArr['state'];
$user->attributes = $_POST['Users'];
$user->user_pres_contact_id = $presContact->contact_id;
if($user->save()){
$transaction->commit();
$this->redirect(array('view', 'id' => $user->user_id));
} else {
$transaction->rollback();
}
}
$this->render('createUser', array(
'Users' => $user,
'Users[\'user_pres_contact\']'=>$presContact,
'Users[\'user_pres_contact\'][\'contact_city\']'=>$presCity,
'Users[\'user_pres_contact\'][\'contact_state\']'=>$presState,
));
}
I am able to access only $users but
I m not able to access $Users['user_pres_contact'] in the view
That's because you are assigning them as strings...
The correct way of doing things would be (btw, what you are asking for can't done literally, it is impossible to assign 2 values to one key):
$user = array(
'user_press_contact' => array(
'contact' => $presContact,
'city' => $presCity,
'state' => $presState,
),
);
$this->render('createUser', array(
'Users' => $user,
));
It will give you $Users['user_press_contact']['contact'] for the name in the view, etc.
You can use
$user->getAttributes() //it returns an array of data.
Hope that's usefull
It is possible to solve this using model relations? You can define a relation from the User model to the City model (e.g. naming it relation_to_city), then you can just assign the user model in the controller
$this->render('view', 'user'=>$user);
and access the city (from the view)
$user->relation_to_city