I am trying to validate a simple form in Yii2 framework.
validation method on model is -
/**
* {#inheritdoc}
*/
public function rules()
{
return array(
array('name', 'required'),
// ... other rules
);
}
When i try to make request it says -
array(1) { ["name"]=> array(1) { [0]=> string(21) "Name cannot be blank." } }
But the thing is i am sending name field in request ( POST ). But after providing name field , it gives an error.
If i do var_dump for request It shows me that name field is in there the request.
Here is the var_dump of request data and the validation error. -
array(1) { ["name"]=> string(6) "distro" } array(1) { ["name"]=> array(1) { [0]=> string(21) "Name cannot be blank." } }
Thank you in advance.
If I correct understanding your question, your problem is in your name of input.
In Yii2 ActiveRecord all data is in array of model name. for example:
Your models name is Users, when you using ActiveRecord your data must send like this to rules work on it:
{
'Users' => {
'username' => 'john_doe',
'fullname' => 'John Doe',
}
}
so change your html form input name to: model_name[field_name]
like this :
<input type="text" name="Users[username]">
Related
I have problem for action view, update, and delete in index.php page, it always show bad request (#400) Missing required parameters: id_kategori and the address always go to localhost/training/frontend/web/index.php?r=kategori%2F(view/update/delete)&id=1, but when i change the address manually to localhost/training/frontend/web/index.php?r=kategori%2Fview&id_kategori=1 it's no problem, also i can create action but then it will redirect page to localhost/training/frontend/web/index.php?r=kategori%2Fview&id=1. Here's the code, its generate from Gii CRUD:
public function actionView($id_kategori)
{
return $this->render('view', [
'model' => $this->findModel($id_kategori),
]);
}
public function actionUpdate($id_kategori)
{
$model = $this->findModel($id_kategori);
if ($this->request->isPost && $model->load($this->request->post()) && $model->save()) {
return $this->redirect(['view', 'id_kategori' => $model->id_kategori]);
}
return $this->render('update', [
'model' => $model,
]);
}
public function actionDelete($id_kategori)
{
$this->findModel($id_kategori)->delete();
return $this->redirect(['index']);
}
Should i rename id_kategori column to id and other id_column just to id?
Version: Yii 2 (2.0.43)
Template: Advanced Template
define $id_kategori=null in function
public function actionView($id_kategori=null)
{
return $this->render('view', [
'model' => $id_kategori ? $this->findModel($id_kategori) : null,
]);
}
You need to do like this
public function actionView($id) {
return $this->render('view', [
'model' => $this->findModel($id),
]);
}
I have been working through the getting started guide for Yii. After generating the CRUD code I receive an error when clicking on the "View/Edit/Delete" buttons. It appears that the generated links are incorrect in that they reference "id" rather than "code" (which is the primary key).
E.g the View link is
http://localhost/yii/demo/basic/web/index.php?r=country%2Fview&id=AU
which generates the error
Bad Request (#400)
Missing required parameters: code
when I manually change the link to
http://localhost/yii/demo/basic/web/index.php?r=country%2Fview&code=AU
it works correctly. Why is it referencing "id"?
I also had to modify the CountryController.php file, findModel function which was again referencing "id"
if (($model = Country::findOne($id)) !== null) {
when changed to "code" this worked.
protected function findModel($code)
{
if (($model = Country::findOne($code)) !== null) {
return $model;
}
throw new NotFoundHttpException('The requested page does not exist.');
}
Does anyone have any ideas, the db is MySQL and the Code column is defined as the Primary Key.
This is a bug in the current yii code.
It expects ID to be the primary key always
A way of fixing it would be to remove the action column from the gridview in index.php and add
[
'class' => 'yii\grid\ActionColumn',
'header' => 'Actions',
'headerOptions' => ['style' => 'color:#337ab7'],
'template' => '{view}{update}{delete}',
'urlCreator' => function ($action, $model, $key, $index) {
if ($action === 'view') {
$url ='index.php?r=country/view&code='.$model->code;
return $url;
}
if ($action === 'update') {
$url ='index.php?r=country/update&code='.$model->code;
return $url;
}
if ($action === 'delete') {
$url ='index.php?r=country/delete&code='.$model->code;
return $url;
}
}
],
Just got done going through old answers regarding how to change gridview to get to this.
I developed a custom multi-banner module to display on the header of my prestashop page. When i save my changes on form i would like to display it on header.
My module form:
My table to see all banner inserted:
I tried to use smarty and orm to retrieve all information but i don't know how to display on my header.
Any issues? I hope i can understand my trobles :)
Thanks
Here's the code of some of my files:
//in main file:
public function hookDisplayHeaderBanner($params) {
$banner = Banner::getBannerToDisplay();
// If there is a banner to display
if ($banner) {
$this->context->smarty->assign([
'banner' => $banner
]);
return $this->display(__FILE__, 'views/templates/hook/display-custom-banner.tpl');
}
// Nothing to display
return false;
}
//In classes/Banner.php
class Banner extends ObjectModel
{
public $id;
public $color;
public $background_color;
public $content;
public static $definition = [
'table' => 'custom_banner',
'primary' => 'id_custom_banner',
'multilang' => false,
'fields' => [
'color' => [
'type' => self::TYPE_STRING,
],
'background_color' => [
'type' => self::TYPE_STRING,
],
'content' => [
'type' => self::TYPE_STRING
]
]
];
/*
*
* Return the banner to display
*
* Here we put the logic to select the right banner
*/
public static function getBannerToDisplay()
{
$sql = 'SELECT *
FROM `' . _DB_PREFIX_ . self::$definition['table'] . '`
WHERE active = 1';
return Db::getInstance()->getRow($sql);
}
}
//In display-custom-banner.tpl
<div id="banner" style="background-color:{$banner.background_color}!important;">
<p style="color:{$banner.color}!important;">
{$banner.content}
</p>
</div>
Using:
return Db::getInstance()->getRow($sql);
only retrieve the first row of the result, if you would like to return more than one banner you would like to use Db::getInstance()->executeS($sql); to get all results.
Also check your DB table structure and a var_dump() of the involved variables, this would help understanding if there are more issues.
I have create API of update my project details, I test it in POSTMAN app it shows the success message but there no effect in the database.
Here are my code:
ProjectsController.php
public function UpdateProject($id)
{
$data = Input::all();
$q = Project::where('id',$id)->update($data);
return response()->json([
'code' => SUCCESS,
'message' => 'Project data update successfully'
]);
}
api.php
Route::post('UpdateProject/{id}','ProjectsController#UpdateProject');
Postman - see image.
output in postman:
{
"code": "200",
"message": "Project data update successfylly"
}
Can anyone help me out?
Thank you
I think you need to check all input details closely , it also comes with token when you submit the form so you need to save all details except token
Change this
$data = Input::all();
to this
$data = Input::except('_token');
I hope this resolves the issue.
in your model add fillable :
protected $fillable = ['name', 'project_group_id','number','ROM','address','city','state','zip','start_date','end_date','duration','description','timeline_type','project_type_id','project_category_id','office_id'];
You have forgotten to run the ->save() method after updating the data:
public function UpdateProject($id)
{
$data = Input::all();
$q = Project::find($id)
$q = $q->fill($data);
$q->save();
return response()->json([
'code' => SUCCESS,
'message' => 'Project data update successfully'
]);
}
You can use this method it will reduce your code
Route (api)
Route::post('UpdateProject/{project}','ProjectsController#UpdateProject');
ProjectsController.php
public function UpdateProject(Request $request, Project $project)
{
$data = $request->all();
$project->update($data);
return response()->json([
'code' => SUCCESS,
'message' => 'Project data update successfully'
]);
}
I want to UPDATE record without losing the File that i have uploaded. Below is my create action.
public function actionCreate()
{
$model = new Page;
if (isset($_POST['Page']))
{
$model->attributes = $_POST['Page'];
$model->filename = CUploadedFile::getInstance($model, 'filename');
if ($model->save())
{
if ($model->filename !== null)
{
$dest = Yii::getPathOfAlias('application.uploads');
$model->filename->saveAs($dest . '/' . $model->filename->name);
$model->save();
}
$this->redirect(array('view', 'id' => $model->id));
}
}
$this->render('create', array(
'model' => $model,
));
}
In yii, update method is designed to update the values that you are posting from your form. It will not update all Model properties.
For suppose, your Page model has 4 model properties, assume all are mandatory.
title
description
keywords
image
Since you don't want to update image field, you can make it non mandatory field by setting scenario for image field in your Page model rules.
class Page extends CActiveRecord
{
/*
Coding
*/
public function rules()
{
return array(
array('title, description,keywords', 'required'),
array('image', 'file', 'types'=>'jpg, png'),
array('image', 'required', 'on' => 'insert'),
// => You need image field to be entered on Create, not on Update
.................
}
}
So, Show file input in your form on Create, Hide on Update action so that image field won't submit.
//Your page form
<?php if($model->isNewRecord):?>
echo $form->labelEx($model, 'image');
echo $form->fileField($model, 'image');
echo $form->error($model, 'image');
<?php endif;?>
Since your are using Yii-2 you can use skippOnEmpty validator for the same.