I have merged admin into create file by copying the gridview into create.php.
<?php $this->renderPartial('_form', array('model'=>$model)); ?>
<?php $this->widget('zii.widgets.grid.CGridView', array(
'id'=>'center-grid',
'dataProvider'=>$model->search(),
'filter'=>$model,
'columns'=>array(
'name',
array(
'class'=>'CButtonColumn',
),
),
)); ?>
It is saving data but the search functionality of grid view is not functioning. Whereas, the same code of CGridView is functioning in admin.php.
here is my controller:
public function actionCreate()
{
$model=new Center;
// Uncomment the following line if AJAX validation is needed
// $this->performAjaxValidation($model);
if(isset($_POST['Center']))
{
$model->attributes=$_POST['Center'];
if($model->save())
$this->redirect(array('create','id'=>$model->id));
}
$this->render('create',array(
'model'=>$model,
));
}
After the model is saved you are redirecting to the same url i.e create but without the data from $_POST. The solution is to remove the call since you are essentially loading the same view, and would like to keep some of the data from the previous page.
public function actionCreate()
{
$model=new Center;
// Uncomment the following line if AJAX validation is needed
// $this->performAjaxValidation($model);
if(isset($_POST['Center']))
{
$model->attributes=$_POST['Center'];
/*if($model->save())
$this->redirect(array('create','id'=>$model->id));*/
}
$this->render('create',array(
'model'=>$model,
));
}
HOWEVER
Reusing the same model for searching and inserting isn't a good approach since you will retain the data for the model that has been entered into the database (including its primary key(s)). A better one would be to use separate variables for the search and create and different name and id prefixes for fields of both forms. Then you can pass both variables into the views, and something entered into one field will not affect the other. You can also reuse the code in actionAdmin to refresh your grid via ajax without having to send the create form too.
Related
everyone I am working on a project that allow user do upload and download data. I want to show only a user's own uploaded data in CGridView.
Here's my relation table
I have tried to edit my page --> views/file/myUploads.php
<?php
$isMine=Yii::app()->user->id; // I have tried this
if($isMine){
$this->widget('zii.widgets.CGridView',array(
'id'=>'file-grid',
'dataProvider'=>$model->search(),
//'filter'=>$model,
'columns'=>array(
'name'
'description',
'category'
'uploader'
'upload_date',
array(
'header'=>'Actions',
'class'=>'bootstrap.widgets.TbButtonColumn',
'template'=>'{delete}', //'visible'=> (Yii::app()->user->getLevel()==1),
'deleteConfirmation'=>"js: 'Are you want to delete '+$(this).parent().parent().children(':first-child').text()+ '?'",
'buttons'=>array(
'delete' => array(
'visible'=> '$data->pengunggah==Yii::app()->user->id',
),
)
),
),
)); }
?>
I have tried that codes above but the page still display all data, not user's own data.
I have tried to edit myUpload function in FileController.php too
public function actionMyUpload()
{
$isMine=Yii::app()->user->id; // I have tried this
if($isMine){
$model=new File('search');
$model->unsetAttributes(); // clear any default values
if(isset($_GET['File']))
$model->attributes=$_GET['File'];
$this->render('myUpload',array(
'model'=>$model,
));
}
}
But still display all data.
Can anyone help me? Thanks a lot.
Revert all your changes back, Open Your Model Class File ie "File.php" and add this line in your search() function:
$criteria=new CDbCriteria;
// Simply add this line
$criteria->addCondition('id_user='.Yii::app()->user->id);
I am a bit new to the Yii Framework. I am making a product selling website, which has 3 basic models
1. Users model containing the primary key id
2. Products model containing the primary key id
3. Orders model which is basically a mapping between the products and orders. It contains the fields product_id and user_id as foreign keys.
I have made a page where all the products are populated and the logged in user can click on a button on product box to order a particular product.
the code of the link is like this
<?php echo CHtml::link('Order Now',array('order',
'product_id'=>$model->id,
'user_id'=>Yii::app()->user->id)); ?>
(Q1) This is sending a GET request but I want to sent the details as post request. How to do this?
My default controller is the site controller. I have made an actionOrder method in this controller.
The code is:
if(Yii::app()->user->isGuest){
$this->redirect('login');
}else{
$model=new Orders;
if(isset($_POST['products_id']))
{
$model->attributes->products_id=$_POST['product_id'];
$model->attributes->users_id=Yii::app()->user->id;
if($model->save())
$this->redirect(array('index'));
}
$this->render('index');
}
But this code is showing bunch of errors. Also, (Q2) how can I put both products_id and users_id in a single array Orders so that I just have to write $_POST['orders']
Also, (Q3) how can I display a flash message after the save is successful?
Kindly help me to solve my 3 problems and sorry if you feel that the questions are too stupid.
Q1: If you want to use POST request, you're going to have to use a form of sorts, in this case the CActiveForm.
Controller:
public function actionOrder()
{
if(Yii::app()->user->isGuest)
$this->redirect('login');
else
{
$model=new Orders;
if(isset($_POST['Orders']))
{
$model->product_id=$_POST['Orders']['products_id'];
$model->users_id = Yii::app()->user->id;
if($model->save())
{
// Q3: set the flashmessage
Yii::app()->user->setFlash('ordered','The product has been ordered!');
$this->redirect(array('index'));
}
}
$this->render('index', array('model'=>$model)); //send the orders model to the view
}
}
View:
<!-- Q3: show the flash message if it's set -->
<?php if (Yii::app()->user->hasFlash('ordered')): ?>
<?php echo Yii::app()->user->getFlash('ordered'); ?>
<?php endif ?>
...
<?php $form=$this->beginWidget('CActiveForm', array('id'=>'order-form')); ?>
<?php echo $form->hiddenField($model,'products_id',array('value'=>$product->id)); ?> // please note the change of variable name
<?php echo CHtml::submitButton('Order Now'); ?>
<?php $this->endWidget(); ?>
Please note that I have changed the name of the product model variable $model to $product, because we will be using $model for the Orders model for the form.
Q2: In this case I set the users_id value in the controller, so $_POST['Orders'] only contains the value for products_id. In yii you can also mass assign your attributes with:
$model->attributes = $_POST['Orders']
Which basicly means $_POST['Orders'] is already an associative array containing the attribute names and values that are in your form.
Q3: The code shows you how to set and show a flash message after an order is succesfull.
First you have to declare forms send method, if you're using bootsrap it'll be like mine:
<?php $form = $this->beginWidget('bootstrap.widgets.TbActiveForm', array(
'action' => Yii::app()->createUrl($this->route),
'method' => 'post',
'id' => 'activity_timeRpt',
));
?>
Second if you want to send custom inputs, you have to specify, otherwise it'll be like
i'll be back to finish this
For your questions 1 and 2 I'd recomend you to use a CActiveForm class. For example
<?php $form = $this->beginWidget('CActiveForm', array(
'action' => 'you_action_here'
'method'=>'post' // this is optinal parameter, as 'post' is default value
)); ?>
<?php echo $form->textField($model,'product_id'); ?>
<?php echo $form->hiddenField($model,'user_id', array('value'=>Yii::app()->user->id)); ?>
<?php $this->endWidget(); ?>
where $model is instance of Orders class, passed by variables thru controller, or set in view file. After that you can use it in way you wanted $model->attributes = $_POST['orders'] in your action method.
For flash message you can use Yii->app()->user->setFlash('orderStatus', 'Successful'), before redirect( or render ) in your actionOrder. To show it:
<?php if(Yii::app()->user->hasFlash('orderStatus')):?>
<div class="info">
<?php echo Yii::app()->user->getFlash('orderStatus'); ?>
</div>
<?php endif; ?>
I am just newbie to the yii framework. Currently I have two tables
one for sales and other for stores
Sales table is looking like this
==============
sales
==============
id
store_id
store table is looking like this
==============
Stores
==============
id
store_name
store_location
Now in sales view form(_form.php) I have rendered both sales and stores.
In sales controller the code for action create is like this
public function actionCreate()
{
$model=new Sales;
$stores = new Stores;
// Uncomment the following line if AJAX validation is needed
// $this->performAjaxValidation($model);
if(isset($_POST['Sales']$_POST['Stores']))
{
$model->attributes=$_POST['Sales'];
$stores->attributes = $_POST['Stores'];
$valid = $model->validate();
$valid = $stores->validate();
if($valid)
{
$stores->save(false);
$model->store_id = $stores->getPrimaryKey();
$model->save(false);
$this->redirect(array('view','id'=>$model->id));
}
}
$this->render('create',array(
'model'=>$model,
'stores' => $stores,
));
}
To get all the stores name in dropdown list I made my code like this
<div class="row">
<?php echo $form->labelEx($stores,'store_name'); ?>
<?php echo $form->dropDownList($stores,'store_name', CHtml::listData(Stores::model()->findAll(), 'store_name', 'store_name'), array('empty'=>'--Select--')) ?>
<?php echo $form->error($stores,'store_name'); ?>
</div>
But here I want the cjuiautocomplete field so that when someone will press any key then it will start to show the suggesions stores name. For that I just came through this link
and just like the docs I made the EAutoCompleteAction.php under protected/extension directory
Then I just made my controller code like this in sales controller
public function actions()
{
return array(
'aclist'=>array(
'class'=>'application.extensions.EAutoCompleteAction',
'model'=>'Stores', //My model's class name
'attribute'=>'store_name', //The attribute of the model i will search
),
);
}
and in view file of sales(_form.php) I made the code like this
<div class="row">
<?php echo $form->labelEx($stores,'store_name'); ?>
<?php
$this->widget('zii.widgets.jui.CJuiAutoComplete', array(
'attribute'=>'store_name',
'model'=>$stores,
'sourceUrl'=>array('stores/store_name'),
'name'=>'store_name',
'options'=>array(
'minLength'=>'3',
),
'htmlOptions'=>array(
'size'=>45,
'maxlength'=>45,
),
)); ?>
After all when I am doing the search by keywords it is showing 404 error in console panel of firebug.
The requested search url in firebug was like this (ads was my search query in store name field)
http://localhost/WebApp/index.php?r=stores/store_name&term=ads
any one here for help?
You forgot to change action name from sample aclist into store_name:
public function actions()
{
return array(
'store_name'=>array( // << Array key is action name
'class'=>'application.extensions.EAutoCompleteAction',
'model'=>'Stores', //My model's class name
'attribute'=>'store_name', //The attribute of the model i will search
),
);
}
I'm trying to merge 3 models to create a fourth one. I have model1, model2 and model3 and I want to merge them into modelMaster. I've also created controllers for all of them. When I call modelMaster/create action, I render the modelMaster/create view which renders the modelMaster/_form view. Inside this _form view, I also want to render model1/_form, model2/_form and a CHtml::dropDownList(), wich takes datas from model3. However, this doesn't work. How can I combine these three different views into one another?
If you try to skip the form generate from the _form views and use unique model names, I think you can use this manual: single form with more models
So the generate of the form definition handles always the parent view and the _form's only the inputs
The other way to use single model in views, create a form model by extend CFormModel, and handle the data binding between this model and the wrapped submodels
If you want to nest several forms into one master form you have to adjust the form templates accordingly. All of your modelMaster/create, model1/_form, model2/_form-views create and render a new CActiveForm (and thus several <form> tags).
Since you cannot nest form-elements in html (how should html know which action to pass the data to) you have to avoid this by doing the following:
Extract the inputs of the form you want to nest into a new view, e.g. model1/_formInputs would look like
...
<?php echo $form->labelEx($model,'name'); ?>
<?php echo $form->textField($model,'name');
<?php echo $form->error($model,'name');
...
alter the model1/create and the other views and get a reference to the form created there, by assigning the return of $this->beginWidget to the variable $form (if not already done):
<?php $form = $this->beginWidget('CActiveForm', array(
'id'=>'foo',
)); ?>
replace the former input fields with
<?php $this->renderPartial('model1/_formInputs', array('form' => $form); ?>
Now, for example the old model1/create-view should work as expected
To get your multi-model-form working you just have to get the reference to the form created in modelMaster/create and use it to renderPartial all */_formInputs you require. Please also remember to include the models for the inputs into the renderPartial-call. So modelMaster/create would look something like:
<?php $form = $this->beginWidget('CActiveForm', array(
'id'=>'foo',
)); ?>
/* Master Inputs here */
// Rendering other models' inputs
<?php $this->renderPartial('model1/_formInputs', array('form' => $form, 'model' => $model1); ?>
<?php $this->renderPartial('model2/_formInputs', array('form' => $form, 'model' => $model2); ?>
/* Render Form Buttons here */
<?php $this->endWidget(); ?>
Submit with Ajax, in Yii it is easy to do and it will keep things easy to understand in the controllers, each controller will have a save and respond with json to confirm the save. There is already ajax validation.
/**
* Performs the AJAX validation.
* #param CModel the model to be validated
*/
protected function performAjaxValidation($model)
{
if(isset($_POST['ajax']) && $_POST['ajax']==='employee-form')
{
$valid = CActiveForm::validate($model);
if(strlen($valid) > 2) {
echo $valid;
Yii::app()->end();
}
}
}
As you can see I have modified it to return the error if there is one (validate returns [] if it is valid, I should probably check for that instead of strlen >2 ), otherwise let the script continue, in this case it will go to the save function.
been looking for a solution to add a feature for "Custom Columns"... Meaning, I present a list of columns that I can show the user and he selects the ones he wants to see and after the selection the table is updated and add/removes the needed columns.
Didn't find anything on Google (perhaps it has a different name than what I was looking for...)
Anyone has an Idea on how it can be accomplished?
Thanks in advance!
This is not a complete sample, but can give you some clues on how to implement it. You've to define some kind of form to collect the data about how your grid has to be rendered. I recommend you to create a CFormModel class if there are more than 3 input fields. Create a view file with the form and a div or renderPartial of a file containing a grid:
$form = $this->beginWidget('CActiveFormExt');
echo $form->errorSummary($model);
echo $form->labelEx($model,'column1');
echo $form->dropDownList($model
echo $form->error($model,'column1');
echo CHtml::ajaxSubmitButton('UpdateGrid',array('controller/grid'),
array('update'=>'#grid'),
$this->endWidget();
// you can render the 'default options' before any ajax update
$this->renderPartial('_grid',array($customColumns=>array('id','name'),'dataProvider'=>$dataProvider));
In the _grid.php view file:
$this->widget('zii.widgets.grid.CGridView', array(
'id' => 'grid',
'dataProvider'=>$dataProvider,
'columns' => $customColumns;
));
In the controller:
function actionGrid(){
// recover the form data, and build the custom columns array
$customColumns = array();
$customColumns[] = '.....';
$dataProvider = ...;
$this->renderPartial('_formTrabajo', array('customColumns' => $idSiniestro, 'dataProvider' => $dataProvider'), false);
}
When you click the ajaxSubmitButton, the form is sent to the url specified through ajax, and the reply from the controller must contain the renderPartial of the view containing the grid, so the jQuery call can replace the html correctly. You must pass an array from your controller to the partial view of the grid, with the custom list of columns you want to display.