How To Display The Data In View Browser Using Checkboxlist......

Yii 1.1 General Discussion
1

i have a form which is made with the help of 2 checkboxlist and textfield.i have no database table so i use CFormModel.in my form i also use checkall for the select of all checkbox ,this is work.but when i submit my form then i found

Error 400

Your request is invalid.

when i submit the form data reach to the controller in "actioncreate" using function

if(isset($_POST['Search'])


$model->attributes=$_POST['Search'];

here i have not the table so "id" is absent.then here how to select form data and sent to the "actionview and loadmodel" for the view the data in browser.

please anyone help me. I’m still new to Yii and struggling up the learning curve.

thanks

2

please post your form and controller action.

3

check out the rules first, in your controller

4
5

please anyone????????

6

the id will get only after saving the model, in this u r not saving the model,

u r passing the POST value as id to view, how it works??


$this->redirect(array('view','id'=>$value));

look at the action view




public function actionView($id)

	{

		$this->render('view',array(

			'model'=>$this->loadModel($id),

		));

	}

here needs the id, in ur case the &id is getting the whole post value, so the loadmodel($id) didnt works

dit u got the points??

7

a lot of thanks Rajith…

i get all the points as you told me.actually i have no id because i have no table so i use CFormModel. here i use id for the checking purpose data is pass or not ,if id use here i found that data is pass to the url.

now i want to pass this data to the view and action load without id.Is any way to do that if yes give me some suggestion.How to "actioncreate" pass the data to the "actionview" and "actionload" without using id…

8

yes…

if u want to view the data

remove the loadmodel

pass the post value to the view as model




public function actionView($id)

        {

                $this->render('view',array(

                        'model'=>$id,

                ));

        }

like this

i didnt get what ur actually purpose!!

9

if u just want to see the post values in the view

u just renderpartial a view as u need

then u can change this




$this->redirect(array('view','id'=>$value));

to

example


$this->renderPartial('view', array('model'=>$value,'brand'=>$brand));

then u will get $model,$brand etc at the view…

10

it is also not working .i want to call u if u r free rajith sir.give me reply pls…

11

what is the error?

u will get $model at the view

change the view according to ur use

12

if u r using like


$this->redirect(array('view','id'=>$value));

u r passing a big array through the url…did u notice that, u may use renderpartial

what is your purpose?

u just want to post some values from a form and want see the value in the view? is it?

want to implement a search?

13

ya i want to implement search.

i notice that when i pass $this->redirect(array(‘view’,‘id’=>$value)); then all data who i select go to the url.i want to that data go to the view.

14

so use renderpartial

u can renderpartial a view

check out the codes for create and update pages, u will get the idea

15 



<div class="row">

		<?php echo $form->labelEx($model,'brand'); ?>

		<?php echo $form->checkboxList($model, 'brand',$model->getBrandOptions(), array( 'checkAll' => 'SELECT ALL')); ?>

		<?php echo $form->error($model,'brand'); ?>

	</div>

	

<div class="row">

		<?php echo $form->labelEx($model,'OS'); ?>

		<?php echo $form->checkBoxList($model,'OS',$model->getOsOptions(),array( 'checkAll' => 'SELECT ALL')); ?>

		<?php echo $form->error($model,'OS'); ?>

	</div>

	

<h3> Price </h3>


 <div class="row">

    <?php echo $form->labelEx($model,'range'); ?>

    <?php echo $form->textField($model,'range',array('size'=>20,'maxlength'=>20)); ?>

    <?php echo $form->error($model,'range'); ?>

</div><div class="row">

    <?php echo $form->labelEx($model,'to'); ?>

    <?php echo $form->textField($model,'to',array('size'=>20,'maxlength'=>20)); ?>

    <?php echo $form->error($model,'to'); ?>

</div>

u may change all these

use CHtml text fields and checkboxList, then u can avoid the model, to implement a simple search the model is not necessary thing

just need a form

create an action in the search controller




public function actionSearch()

	{

$model=new Search;


if(isset($_POST['Search']))

{

//set ur search condition here


//then render the same form or another page here with the search values


}


$this->render('ur-form-page',array('model'=>$model));





}