I've set up a log in process where a verification code is generated, and when successful, is then removed. However, i want to make sure that if there's multiple verification codes for the same user, upon log in success, delete all records for that user.
Here's my code
if ($model->validate() && $model->login()) {
//delete this verification code
$verificationCode->delete();
//delete all existing codes for user_id
VerificationCode::model()->deleteAll('user_id',$user->id);
Yii::app()->user->setReturnUrl(array('/system/admin/'));
$this->redirect(Yii::app()->user->returnUrl);
}
However, this seems to just delete all the records, regardless on different user_id's in table. Can anyone see where I'm going wrong?
If you want to delete record with specified attributes, the cleanest way for this is to use deleteAllByAttributes():
VerificationCode::model()->deleteAllByAttributes(['user_id' => $user->id]);
Seems you call the function delete() in wrong way ... try passing value this way
VerificationCode::model()->deleteAll('user_id = :user_id', array(':user_id' => $user->id));
For Yii2, the documented way is to use the function deleteAll().
I normally pass the arguments as an array, like so:
VerificationCode::deleteAll(['user_id' => $user->id]);
Also, you can use the afterDelete method, to make sure that everytime or everywhere someone deletes one verificationCode, your application will also delete every userVerificationCode. Put this in your verificationCode model class:
protected function afterDelete()
{
parent::afterDelete();
VerificationCode::model()->deleteAll('user_id = :user:id',[':user_id' =>$this->user_id]);
//... any other logic here
}
You can use below method for deleting all user_id entry from database:
$criteria = new CDbCriteria;
// secure way for add a new condition
$criteria->condition = "user_id = :user_id ";
$criteria->params[":user_id"] = $user->id;
// remove user related all entry from database
$model = VerificationCode::model()->deleteAll($criteria);
or you can use another method directly in controller action
VerificationCode::model()->deleteAll("user_id= :user_id", [":user_id"
=>$user->id]);
use below method for redirecting a URL
$this->c()->redirect(Yii::app()->createUrl('/system/admin/'));
Related
I wanted to let the system to show error message when detect duplicated entry of full_name column without applying unique in the full_name column from public function rules() in model.
My code is like this :
if ($model->load(Yii::$app->request->post()) ) {
$model->full_name = $model->first_name .'' . $model->last_name ;
$name = StudentInfo::find()->select('full_name')->where(['full_name'=> $model->full_name]);
if($name == $model->full_name ){
echo "<script type='text/javascript'>alert('Same student name is detected');</script>";
}
else{
$model->status ="Active";
$model->call_format = Countries::find()->select('phonecode')->where(['name'=> $model->country]);
$model->date_created = new Expression('NOW()');
$user->user_type ='student';
$user->user_name = $model->full_name;
$user->user_status = $model->status;
$user->authKey = Yii::$app->security->generateRandomString(10);
$user->accessToken = Yii::$app->security->generateRandomString(10);
$user->save();
$model->save();
return $this->redirect(['view', 'id' => $model->id]);
}
}
But it shows error like :missing required parameters: id. When i apply model->save(false) ,it seems that the sql statement wont run because of duplicate entry in full_name column. How do i fix it?
Well, there is a construct exists() for such a purposes (see Yii2: check exist ActiveRecord model in database ).
if(StudentInfo::find()->where(['full_name'=> $model->full_name])->exists()){
echo "<script type='text/javascript'>alert('Same student name is detected');</script>";
}
else{...}
it generates the EXISTS query, which is faster and you don't have to load all the data from DB.
If you don't have such a column in your table, then check it by the first/last name.
change it:
$name = StudentInfo::find()->select('full_name')->where(['full_name'=> $model->full_name]);
To:
$name = StudentInfo::find()->select('full_name')->where(['full_name'=> $model->full_name])->one();
Also, if you use the select() method, to use the update() and save() or updateCounters() ... methods, you need the row ID in the same query.
Example:
->select('id') or ->select(['id', 'full_name'])
info: Multi-parameter is an array in select()
:missing required parameters: id
could mean that it couldn't find id, not by duplicate entry in full_name column. please check again
There are two problems with your code.
$name = StudentInfo::find()->select('full_name')->where(['full_name'=> $model->full_name]);
When this line is executed the $name variable will contain instance of yii\db\ActiveQuery. You want to call some method, that will actually execute your query and return result.
You can use scalar() to get the selected value. In that case the $name will contain the content of full_name column from result.
$name = StudentInfo::find()
->select('full_name')
->where(['full_name'=> $model->full_name])
->scalar();
Or you can use count() to get the number of rows that match condition. In that case you may leave out the select() method call but you will need to modify your condition
$count = StudentInfo::find()
->where(['full_name'=> $model->full_name])
->count();
if ($count > 0) {
echo "<script type='text/javascript'>alert('Same student name is detected');</script>";
} else {
// ...
}
The other problem is that you are not checking whether your $model->save() was successful. If your $model is new instance and the id attribute is auto-generated then when $model->save fails the $model->id is empty and then you are trying to redirect user to view with empty id.
Your code should look like this:
if ($user->save() && $model->save()) {
return $this->redirect(['view', 'id' => $model->id]);
}
If the save fails because of validation the validation errors will be stored in models and if you are using ActiveForm widget the errors will be displayed. If you are not using ActiveForm you should do something to tell user that operation failed.
Since you are saving two different models you might want to consider use of transactions to prevent a situations where $user model is saved but save of $model fails.
I need a little help and I can’t find an answer. I would like to replicate a row from one data table to another. My code is:
public function getClone($id) {
$item = Post::find($id);
$clone = $item->replicate();
unset($clone['name'],$clone['price']);
$data = json_decode($clone, true);
Order::create($data);
$orders = Order::orderBy('price', 'asc')->paginate(5);
return redirect ('/orders')->with('success', 'Success');
}
and i got an error :
"Missing argument 1 for
App\Http\Controllers\OrdersController::getClone()"
.
I have two models: Post and Order. After trying to walk around and write something like this:
public function getClone(Post $id) {
...
}
I got another error
Method replicate does not exist.
Where‘s my mistake? What wrong have i done? Maybe i should use another function? Do i need any additional file or code snippet used for json_decode ?
First of all, make sure your controller gets the $id parameter - you can read more about how routing works in Laravel here: https://laravel.com/docs/5.4/routing
Route::get('getClone/{id}','YourController#getClone');
Then, call the URL that contains the ID, e.g.:
localhost:8000/getClone/5
If you want to create an Order object based on a Post object, the following code will do the trick:
public function getClone($id) {
// find post with given ID
$post = Post::findOrFail($id);
// get all Post attributes
$data = $post->attributesToArray();
// remove name and price attributes
$data = array_except($data, ['name', 'price']);
// create new Order based on Post's data
$order = Order::create($data);
return redirect ('/orders')->with('success', 'Success');
}
By writing
public function getClone(Post $id)
you are telling the script that this function needs a variable $id from class Post, so you can rewrite this code like this :
public function getClone(){
$id = new Post;
}
However, in your case this does not make any sence, because you need and integer, from which you can find the required model.
To make things correct, you should look at your routes, because the url that executes this function is not correct, for example, if you have defined a route like this :
Route::get('getClone/{id}','YourController#getClone');
then the Url you are looking for is something like this :
localhost:8000/getClone/5
So that "5" is the actual ID of the post, and if its correct, then Post::find($id) will return the post and you will be able to replicate it, if not, it will return null and you will not be able to do so.
$item = Post::find($id);
if(!$item){
abort(404)
}
Using this will make a 404 page not found error, meaning that the ID is incorrect.
When a new user is registered and is redirected to profile page. All these $userconatact , $qualification and $profileimage variables are null. Home page is setup with if (condition) to have buttons to fill form or show available data to user. There are three forms and user has to fill these form one by one. All these variables or one of them could be null but I still want to send the null value to view. Is there any way to make this view happen.
class ProfileController extends Controller
{
public function showprofile($type=null, $id='id'){
$contactquery = Usercontact::where('user_id',Auth::user()->id)->select('*')->get();
$usercontact = $contactquery->toArray();
$qualificationquery = userqualification::where('user_id',Auth::user()->id)->select('*')->get();
$qualification = $qualificationquery->toArray();
$profileimagequery = Profileimage::where('user_id',Auth::user()->id)->select('id','user_id','profileimage')->get();
$profileimage = $profileimagequery->toArray();
return view('home')->withUsercontact($usercontact)->withQualification($qualification)->withProfileimage($profileimage);
}
}
I tried using if statements but if didn't work
return view('home')
if(!empty($usercontact)){->withUsercontact($usercontact)}
if(!empty($qualification)){->withQualification($qualification)}
if(!empty($profileimage)){->withProfileimage($profileimage);}
Is there any way to use if statements like this to make it easy.
1) Use compact which will automatically create key value pairs for you.
return view('home')->compact('usercontact','qualification','profileImage')
2) Here are a few other improvements to your code. You can use the Auth facade to access the logged-in user.
$usercontact = Auth::user();
3) Define relationships on your user model and access the qualifications by this. This will return a Collection.
$qualification = $usercontact->qualifications;
You can use compact to send data from your controllers to views.
return view('home')->compact('usercontact', 'qualification', profileimage');
Your code can be refactored as below.
public function showprofile($type=null, $id='id')
{
$usercontact = Usercontact::where('user_id',Auth::user()->id)->first();
$qualification = Userqualification::where('user_id',Auth::user()->id)->first();
$profileimage = Profileimage::where('user_id',Auth::user()->id)->select('id','user_id','profileimage')->first();
return view('home')->compact('usercontact', 'qualification', 'profileimage');
}
Note the Userqualification model name in the code according to the convention. Try to add relationships to your models to make your query easier.
Try this style to send data to view
return view('home',['Usercontact'=>$usercontact,'Qualification'=>$qualification,'Profileimage'=>$profileimage]);
In view you will be able to get data by array key
I'm using a yii\filters\PageCache filter for my action and I want to define a cache dependency based on action parameter.
E.g. I want to use yii\caching\DbDependency to select updated_at column of the row with id from the request. How can I reference the id parameter of the action?
Yii::$app->request->get('id');
If you want to add dependency in your cache then try this
$db = Yii::$app->db; // or Category::getDb()
$dep = new DbDependency();
$dep->sql = 'SELECT max(update_at) FROM table';
$model = $db->cache(function ($db) {
return model::find()->asArray()->orderBy('id ASC')->all();
},expirytime, $dep);
The code will check your data in cache if it exist will load from cache otherwise create it and cached it.it will also check your update_at filed if its value is changed then it will update your cache,you don't need to worry about that
and if you want to access current action in yii then try this code
Yii::$app->controller->id; //will return current controller//
Yii::$app->controller->action->id; //will return current action//
Yii::$app->controller->module->id; //will return current module//
i hope this will help you
Is there any way to update a record in Laravel using eloquent models just if a change has been made to that record? I don't want any user requesting the database for no good reason over and over, just hitting the button to save changes. I have a javascript function that enables and disables the save button according with whether something has changed in the page, but I would like to know if it's possible to make sure to do this kind of feature on the server side too. I know I can accomplish it by myself (meaning: without appealing to an internal functionality of the framework) just by checking if the record has change, but before doing it that way, I would like to know if Laravel eloquent model already takes care of that, so I don't need to re-invent the wheel.
This is the way I use to update a record:
$product = Product::find($data["id"]);
$product->title = $data["title"];
$product->description = $data["description"];
$product->price = $data["price"];
//etc (string values were previously sanitized for xss attacks)
$product->save();
You're already doing it!
save() will check if something in the model has changed. If it hasn't it won't run a db query.
Here's the relevant part of code in Illuminate\Database\Eloquent\Model#performUpdate:
protected function performUpdate(Builder $query, array $options = [])
{
$dirty = $this->getDirty();
if (count($dirty) > 0)
{
// runs update query
}
return true;
}
The getDirty() method simply compares the current attributes with a copy saved in original when the model is created. This is done in the syncOriginal() method:
public function __construct(array $attributes = array())
{
$this->bootIfNotBooted();
$this->syncOriginal();
$this->fill($attributes);
}
public function syncOriginal()
{
$this->original = $this->attributes;
return $this;
}
If you want to check if the model is dirty just call isDirty():
if($product->isDirty()){
// changes have been made
}
Or if you want to check a certain attribute:
if($product->isDirty('price')){
// price has changed
}
You can use $product->getChanges() on Eloquent model even after persisting. Check docs here
I like to add this method, if you are using an edit form, you can use this code to save the changes in your update(Request $request, $id) function:
$post = Post::find($id);
$post->fill($request->input())->save();
keep in mind that you have to name your inputs with the same column name. The fill() function will do all the work for you :)
use only this:
Product::where('id', $id)->update($request->except(['_token', '_method']));
At times you need to compare the newly changed value with the previous one and if you are looking for that here is the solution.
if (
$obj->isDirty('some_field_name') &&
$obj->some_field_name != $obj->getOriginal('some_field_name')
) {
// Make required changes...
}
});
}
The reference of the derived solution is here.
Maybe Laravel has updated since, but wasChanged is working for me better than isDirty in all of these previous answers.
For example:
if($post->wasChanged('status') && $post->status == 'Ready') // Do thing