Ascending Form numbers in Yii - php

In Yii I am doing a form in which the form has input fields for user details.I have made necessary input fields for all those fields.Where a user can submit all the values.Now I have a field where it will show the form number which will not be entered by user.It will be generated randomly with ascending order like
for the 1st form it will be like this FORM:001, For the second form it will be like this FORM:002 and it will go on.
Now I want that the form number will be like Form:001 so how to do that?HAny help and suggestions will be highly appriciable.
[UPDATED]
<div class="row">
<?php echo $form->labelEx($model,'id'); ?>
<?php echo Yii::app()->db->getLastInsertId();?>
<?php echo $form->error($model,'id'); ?>
</div>
This is the code for view > _form.php file.
and the result is ID 0

What you need is either of the following:
$maxFormId= Yii::app()->db->createCommand()
->select('max(id) as max')
->from('tbl_yourtable')
->queryScalar();
$yourFormId = "Form:".($maxFormId+ 1);
or alternatively run the following after the insert, and only display the form id then:
$yourFormId = "Form:".Yii::app()->db->getLastInsertId();
UPDATE:
public function actionCreate()
{
$model=new YourModel;
if(isset($_POST['YourForm']))
{
$model->attributes=$_POST['YourForm'];
if($model->save())
$this->redirect(array('view','id'=>$model->id));
}
$this->render('create',array(
'model'=>$model,
));
}
On the above:
$this->redirect(array('view','id'=>$model->id));
automatically gives you the last ID inserted so you can just put the following in your view:
echo "Form:".$id;

Related

How to get selected from jmultiselect2side in yii and display in a new page

I'm new to yii and I don't understand the extensions much
but I used this extension called jmultiselect2side because I'm trying to make a site where users could reserve stuff like apparatuses in the lab
Anyway, I need a code that would get the Selected Items and then display them in another page for viewing purposes
I haven't put anything in the controller but the name of my controller and model is Apparatus
Here is my view:
<?php
$model= Apparatus::model()->findByAttributes(array('ApparatusCode'=>'1'));
// complete user list to be shown at multiselect order by ApparatusCode
$Apparatus= Apparatus::model()->findAll(
array('order' => 'ApparatusCode'));
?>
<center>
<?php
$this- >widget('application.extensions.jmultiselect2side.Jmultiselect2side',array(
'model'=>$model,
'attribute'=>'ApparatusName', //selected items
'labelsx'=>'Available',
'labeldx'=>'Selected',
'moveOptions'=>false,
'autoSort'=>'true',
'search'=>'Search:',
'list'=>CHtml::listData( // available items
$Apparatus,
'ApparatusCode',
'ApparatusName'),
));
?>
please help as soon as possible :/
put all above elements in a form. Set action for the form. Submit the form then the action in which you are handling this submit request, you can write there
if(isset($_POST))
{
foreach($_POST['Apparatus']['ApparatusName'] as $name)
{
do what ever you want
}
}
$name will represent the each selected item

CGridView search not functioning when merged with create form

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.

How to send the request through POST and how to access attributes of the model?

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; ?>

Yii framework CJUIautocomplete showing error

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
),
);
}

Yii framework - Different validation rules depending on selected options

Is it possible to create model rules that are dependent from selection?
I have a model "Deposit" which is used to enter the money transfer details..I have drop down list with two possible choices "Cash Transfer","Cheque Transfer", and i have fields cash_deposit_date,bank_name.. which is required only for cash transfer and cheque_date, cheque_no and in_favour_of.. which is required only at the time of cheque transfer.. how can i do that..?
I know i can use scenarios like this,
$model=new Deposit("cash_transfer");
or
$model=new Deposit("cheque_transfer");
but how can I change scenario depending on the value selected in dropdown list?
Couldn't you just do this:
Generated HTML:
<form>
....
<select name="scenario">
<option value="cash_transfer">Cash Transfer</option>
<option value="cheque_transfer">Checque Transfer</option>
</select>
....
</form>
Code:
// allow only lowercase letters and underscore
$scenario = preg_replace('/[^a-z_]/', '', $_POST['scenario']);
if (!empty($scenario)) {
$model = new Deposit($scenario);
} else {
die('Missing scenario!');
}
Extending from jupaju answer,
That dropdown list also a model field(transfer_type), so I have done something like this..
After checking POST values are set, I use $model->scenario = $model->transfer_type==1 ? 'cash_transfer':'cheque_transfer' to change the scenario.
My code is
$model=new Deposit;
// Uncomment the following line if AJAX validation is needed
// $this->performAjaxValidation($model);
if(isset($_POST['Deposit']))
{
$model->attributes=$_POST['Deposit'];
$model->scenario = $model->transfer_type==1 ? 'cash_transfer':'cheque_transfer';
if($model->save())
$this->redirect(array('admin'));
}
$this->render('create',array('model'=>$model));
Now it is working..
I think the best way would be to use a personalized validation rule in your model and then check if the model's "scenario" property is set to one or other option.
You can read more about custom validation rules in here
Hope this helps. Good luck!
This is more simple form me
In your model put:
public $pay_type;
/**
* #return array validation rules for model attributes.
*/
public function rules()
{
// NOTE: you should only define rules for those attributes that
// will receive user inputs.
return array(
...
array('pay_type', 'required'),
array('cash_deposit_date', 'validationTransfer'),
...
);
}
public function validationTransfer($attribute,$params)
{
// this a sample, put all your need
if($this->pay_type=='cash_transfer' and $this->cash_deposit_date==='')
$this->addError('cash_deposit_date','If you will pay to Cash Transfer, enter your cahs deposit date');
// this a sample, put all your need
if($this->pay_type=='cheque_transfer' and $this->cheque_date==='')
$this->addError('cheque_date','If you will pay to Cheque Transfer, enter your cheque date');
// this a sample, put all your need
if($this->pay_type=='cheque_transfer' and $this->cheque_no==='')
$this->addError('cheque_no','If you will pay to Cheque Transfer, enter your Cheque No');
}
In the view into your widget form
<?php echo $form->labelEx($model,'pay_type',array('class'=>'control-label')); ?>
<?php echo $form->dropDownList($model,'pay_type',array("cash_transfer"=>"Cash Transfer","cheque_transfer"=>"Cheque Transfer"),array('class'=>'form-control','empty'=>'Pay Type...')); ?>
<?php echo $form->error($model,'pay_type',array('class'=>'help-block')); ?>
Note: css class like "help-block", "control-label", "form-control" on witget form are optionals, maybe you use Bootstrap 3 and it will looks good

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