I am using cakephp and I want to insert form data in my database, but I am getting the following error:
Table posts for model Post was not found in datasource default.
How can I save my data to database. And in cakephp how can we send data from controller to model?
// this is my view code
echo $this->Form->create('posts', array('action' => 'Add'));
echo $this->Form->input('title', array('label' => 'Enter your email address:'));
echo $this->Form->end('Add');
// this is my controller code
public function Add() {
$this->request->data['posts']['title'] = $this->request->data["posts"]["title"];
$this->Post->save($this->request->data);
}
Error says that model "Post" is not created. You need to create file PostsTable.php and put it in src/Model/Table folder.
Also just try CakePHP tutorial from CakeBook.
Related
I am trying to make column row editable using x editable. I'm using codeigniter framework. When I post url to controller using Ajax post, value didn't post into array in controller.
this is my script
this is my view
<td class="numeric"><center><a href="#" class="status_order" name="status" data-type="select"
data-pk="<?=$data['id_faktur'];?>" data-value="<?=$data['status'];?>"
data-title="STATUS PEMBELIAN">
<?=$data['status'];?></a></center></td>
this is my controller
public function edit() {
$data = array (
'status' => $this->input->post('value')
);
$this->admin_order_model->update_status($value, $data);
redirect('admin/produk/penjualan');
}
i don't know why array is empty... help me please
I edit a row on the Post table or add a new row to it from edit() function in PostsController. The function looks like this:
public function edit($id = null) {
// Has any form data been POSTed?
if ($this->request->is('post')) { //Replaced 'post' by 'get' in this line
// If the form data can be validated and saved...
if ($this->Post->save($this->request->data)) {
// Set a session flash message and redirect.
$this->Session->setFlash('Updated the Post!');
return $this->redirect('/posts');
}
}
// If no form data, find the post to be edited
// and hand it to the view.
$this->set('post', $this->Post->findById($id));
}
I simply replaced 'post' by 'get' to see what would happen and it went on creating new rows without even taking me to the form. I still get the flash message 'Updated the Post!', but without taking any form data.
If the code in edit.ctp is required, here it is:
<?php
echo $this->Form->Create('Post');
echo $this->Form->input('id', array('type' => 'hidden','default'=>$post['Post' ['id']));
echo $this->Form->input('title',array('default'=>$post['Post']['title']));
echo $this->Form->input('body',array('default'=>$post['Post']['body']));
echo $this->Form->end('Update');
?>
Any thoughts on why this might be happening?
Edit: Added CakePHP Version
I am using CakePHP 2.4.5
What you are doing makes no sense.
Why would you want to switch the "post" by "get" here?
Of course it will then generate new rows, as you effectively trigger a save on each page load (GET).
Don't do that.
The code you had there was just fine - IF you also took PUT into consideration.
For edit forms, it is not a post, but:
if ($this->request->is('put')) {}
PS: If you want to make sure it always works for both add/edit, use
if ($this->request->is(array('post', 'put')) {}
But NEVER replace it with "get".
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' => $dataProvider'), false);
}
When you click the ajaxSubmitButton, the form is sent to the url specified through ajax, and the reply from the controller must contain the renderPartial of the view containing the grid, so the jQuery call can replace the html correctly. You must pass an array from your controller to the partial view of the grid, with the custom list of columns you want to display.
How to o retrieve value from a text box in a form (view) to controller in cake php?
Here's an example from the CakePHP book on saving your Model data:
function edit($id) {
//Has any form data been POSTed?
if(!empty($this->data)) {
//If the form data can be validated and saved...
if($this->Recipe->save($this->data)) {
//Set a session flash message and redirect.
$this->Session->setFlash("Recipe Saved!");
$this->redirect('/recipes');
}
}
//If no form data, find the recipe to be edited
//and hand it to the view.
$this->set('recipe', $this->Recipe->findById($id));
}
If your model was Recipe, and your input was named "title", then the value would be in $this->data['Recipe']['title'], if you setup your view like so:
echo $this->Form->create('Recipe');
echo $this->Form->hidden('id');
echo $this->Form->input('title');
echo $this->Form->end('Save Recipe');
So, look here: http://book.cakephp.org/view/1031/Saving-Your-Data
And try to do the Blog tutorial, it might help you get started: http://book.cakephp.org/view/1528/Blog