Output Laravel SQL Column as a link? - sql

I have a Javascript table pulling data into it from a Laravel Json response, I have the column 'name' I wish to turn into a link when it is in the table? Is there anyway to do it like this:
"SELECT CONCAT('<a href=computer/?id=',id,'>',name,'</a>') as name,
I would rather use eloquent rather than straight SQL.

I really wouldn't advise building up HTML as part of your SQL query. Say you want to use that JSON feed somewhere else that doesn't support HTML in the field or you want to append a class to link you've got to change your code.
However there is nothing to wrong with building the link and passing that through you could consider using the Appending Values To JSON functionality of eloquent model.
So you could do something like
class Product extends Model
{
/**
* The accessors to append to the model's array form.
*
* #var array
*/
protected $appends = ['link'];
public function getLinkAttribute()
{
return route('routeName', ['id' => $this->id]);
}
}
If you really must pass back an anchor tag then you could use the link_to_route method which is part of the laravelcollective/html package.

Related

Compare two database fields in extbase repository

I am using TYPO3 8. In my extension I have a database table "company" in which I store for each company the total number of places (number_places) and the number of occupied places (occupied_places).
Now I want to limit the search to companies which have available places left.
In MySQL it would be like this:
SELECT * FROM company WHERE number_places > occupied_places;
How can I create this query in the extbase repository?
I tried to introduce the virtual property placesLeft in my model but it did not work.
I don't want to use a raw SQL statement as mentioned below, because I already have implemented a filter which uses plenty of different constraints.
Extbase query to compare two fields in same table
You can do it like this in your repository class, please note the comments inside the code:
class CompanyRepository extends \TYPO3\CMS\Extbase\Persistence\Repository
{
public function findWithAvailablePlaces(bool $returnRawQueryResult = false)
{
// Create a QueryBuilder instance
$queryBuilder = $this->objectManager->get(\TYPO3\CMS\Core\Database\ConnectionPool::class)
->getConnectionForTable('company')->createQueryBuilder();
// Create the query
$queryBuilder
->select('*')
->from('company')
->where(
// Note: this string concatenation is needed, because TYPO3's
// QueryBuilder always escapes the value in the ExpressionBuilder's
// methods (eq(), lt(), gt(), ...) and thus render it impossible to
// compare against an identifier.
$queryBuilder->quoteIdentifier('number_places')
. \TYPO3\CMS\Core\Database\Query\Expression\ExpressionBuilder::GT
. $queryBuilder->quoteIdentifier('occupied_places')
);
// Execute the query
$result = $queryBuilder->execute()->fetchAll();
// Note: this switch is not needed in fact. I just put it here, if you
// like to get the Company model objects instead of an array.
if ($returnRawQueryResult) {
$dataMapper = $this->objectManager->get(\TYPO3\CMS\Extbase\Persistence\Generic\Mapper\DataMapper::class);
return $dataMapper->map($this->objectType, $result);
}
return $result;
}
}
Notes:
If you have lots of records to deal with, I would - for performance reasons - not use the data mapping feature and work with arrays.
If you want to use the fluid pagination widget, be sure you don't and build your own pagination. Because of the way this works (extbase-internally), you'd get a huge system load overhead when the table grows. Better add the support for limited db queries to the repository method, for example:
class CompanyRepository extends \TYPO3\CMS\Extbase\Persistence\Repository
{
public function findWithAvailablePlaces(
int $limit = 10,
int $offset = 0,
bool $returnRawQueryResult = false
) {
// ...
$queryBuilder
->setMaxResults($limit)
->setFirstResult($offset);
$result = $queryBuilder->execute()->fetchAll();
// ...
}
}
I think you cant do this using the default Extbase Query methods like equals() and so on. You may use the function $query->statement() for your specific queries like this.
You also can use the QueryBuilder since TYPO3 8 which has functions to compare fields to each other:
https://docs.typo3.org/typo3cms/CoreApiReference/latest/ApiOverview/Database/QueryBuilder/Index.html#quoteidentifier-and-quoteidentifiers
It's fine to use this QueryBuilder inside Extbase repositories. After this you can use the DataMapper to map the query results to Extbase models.
In case of using "statement()" be aware of escaping every value which may cause any kind of SQL injections.
Based on the current architecture of TYPO3, the data structure is such that comparing of two tables or, mixing results from two tables ought to be done from within the controller, by injecting the two repositories. Optionally, you can construct a Domain Service that can work on the data from the two repositories from within the action itself, in the case of a routine. The service will also have to be injected.
Note:
If you have a foreign relation defined in your table configuration, the results of that foreign relation will show in your defined table repository. So, there's that too.

Laravel dynamic route from MySQL

I am trying to generate routes from one template "tuning.blade.php"
I have a DB with 250 rows I would like to dynamically create 250 routes with one route or one controller.
I want to be able to use these URLs
laravel.dev/tuning/(field from DB row 1)
laravel.dev/tuning/(field from DB row 2 and so on)
I want to put the DB requests in tuning.blade.php so that this template can display all 250 rows using 250 different URLS
I have tried to use the first example from Laravel Docs
class UserController extends BaseController {
/**
* Show the profile for the given user.
*/
public function showProfile($id)
{
$user = User::find($id);
return View::make('user.profile', array('user' => $user));
}
}
Route::get('user/{id}', 'UserController#showProfile');
I also got some interesting search results from http://forumsarchive.laravel.io/viewtopic.php?id=9010
But alsas I always end up with a notfound exception
But I am unsure what to put in my tuning template to display anything at all. my Tuning template reside in app/views/home/tuning.blade.php
Currently I have got the error "Symfony \ Component \ HttpKernel \ Exception \ NotFoundHttpException"
Can anyone put me in the right direction of where I can find a resource to help me understand?
You said you want to be able to use these URLs:
laravel.dev/tuning/(field from DB row 1)
laravel.dev/tuning/(field from DB row 2 and so on)
You can do this by declaring a route like this:
Route::any('/tuning/{field}', 'TuningController#someMethod'); you may get/post
You shouldn't run sql queries from your view and if you want to really declare some dynamic routes for each fields from your database then you may do id right from your routes.php file, for example, assume that you have a table named tunings and that table contains some fields including id, name and some others. Now, to declare routes individually for each routes dynamically using the tuning field of table tunings you may create a method in your TuningController, something like this:
class TuningController extends baseController {
// other methods...
public function registerTuningRoutes()
{
$tunings = Tuning::all(); // Assume that you have a model Tuning
// Or you may use this instead
$tunings = DB::table('tuning')->get();
// Now loop all tunings and declare routes
foreach($tunings as $tuning) {
$url = '/tuning/' . $tuning->name;
$route_name = 'tuning.' . $tuning->name;
Route::any($url, $route_name); // You may use get/post
}
}
public function TuningMethod($tuning = null)
{
// $tuning will contain the current tuning name, check
dd($tuning);
}
}
Now in your routes.php file use something like this:
Registers route for each tuning name in database
App::make('TuningController')->registerTuningRoutes();
From your terminal/command prompt check the routes by running following command:
php artisan routes
But, I think, you don't need to do this, only one route is enough as I mentioned earlier in my answer.

Need a concept on fetching data with HQL while three or more tables are in use

A small briefing on what I am trying to do.
I have three tables Content(contentId, body, timeofcreation), ContentAttachmentMap(contentId, attachmentId) and Attachment(attachmentId, resourceLocation).
The reason I adopted to create the mapping table because in future application the attachment can also be shared with different content.
Now I am using HQL to get data. My objectives is as follows:
Get All contents with/without Attachments
I have seen some examples in the internet like you can create an objective specific class (not POJO) and put the attribute name from the select statement within its constructor and the List of that Class object is returned.
For e.g. the HQL will be SELECT new com.mydomain.myclass(cont.id, cont.body) ..... and so on.
In my case I am looking for the following SELECT new com.mydomain.contentClass(cont.id, cont.body, List<Attachment>) FROM ...`. Yes, I want to have the resultList contain contentid, contentbody and List of its Attachments as a single result List item. If there are no attachments then it will return (cont.id, contentbody, null).
Is this possible? Also tell me how to write the SQL statements.
Thanks in advance.
I feel you are using Hibernate in a fundamentally wrong way. You should use Hibernate to view your domain entity, not to use it as exposing the underlying table.
You don't need to have that contentClass special value object for all these. Simply selecting the Content entity serves what you need.
I think it will be easier to have actual example.
In your application, you are not seeing it as "3 tables", you should see it as 2 entities, which is something look like:
#Entity
public class Content {
#Id
Long id;
#Column(...)
String content;
#ManyToMany
#JoinTable(name="ContentAttachmentMap")
List<Attachment> attachments;
}
#Entity
public class Attachment {
#Id
Long id;
#Column(...)
String resourceLocation
}
And, the result you are looking for is simply the result of HQL of something like
from Content where attachments IS EMPTY
I believe you can join fetch too in order to save DB access:
from Content c left join fetch c.attachments where c.attachments IS EMPTY

Mapping two tables to one entity in Doctrine2

I'm looking at using doctrine for an application I'm working on - but after reading the documentation I'm having trouble conceptualizing how to represent the database structure we have in terms of entities.
I have many tables which have partner tables which hold translation data like the following....
Where I would like to have one Entity (Navigation Element) which had access to the 'label' field depending on what Language I set in my application. The following from the Doctrine documentation seems to suggest that you need to define one (single) table which is used to persist an entity
http://www.doctrine-project.org/docs/orm/2.0/en/reference/basic-mapping.html
By default, the entity will be
persisted to a table with the same
name as the class name. In order to
change that, you can use the #Table
annotation as follows:
Or do I need to define two entities and link them (or allow the translation table to inherit from the element table).
And what strategy would I use to always insert a language_id clause to the Join (to ensure I'm pulling the right label for the currently set language). Is this something I would define in the entity itself, or elsewhere?
This seems to suit a One-To-Many Bidirectional association. This is the scenario from that page translated to your situation:
/** #Entity */
class NavigationElement
{
// ...
/**
* #OneToMany(targetEntity="NavigationElementTranslation", mappedBy="navigationElement")
*/
private $translations;
// ...
public function __construct() {
$this->translations = new \Doctrine\Common\Collections\ArrayCollection();
}
}
/** #Entity */
class NavigationElementTranslation
{
// ...
/**
* #ManyToOne(targetEntity="NavigationElement", inversedBy="translations")
* #JoinColumn(name="navigation_element_id", referencedColumnName="id")
*/
private $navigationElement;
// ...
}
You could add a getLabel($languageId) method to the NavigationElement entity that searches through the translations to get the correct label:
public function getLabel($languageId) {
foreach($this->translations as $trans) {
if($trans->languageId == $languageId)
return $trans->label;
}
throw new InvalidArgumentException();
}
And you could use the following DQL to ensure you only load the translation you want into the $translations property:
$query = $em->createQuery(
"SELECT ne, net
FROM Entity\NavigationElement ne
JOIN ne.translations net WITH net.languageId = :langId"
);
$query->setParameter('langId', $languageId);
$navigationElements = $query->execute();
This situation sounds like one where you would want to cache aggressively. Make sure you look into Doctrine 2's caching mechanisms too.
Also, internationalization can be handled reasonably well in PHP with gettext if you find join tables for translations start to become unmanageable.
I would also direct anyone who has to tackle this same problem to take a look at the following doctrine extension.
http://www.gediminasm.org/article/translatable-behavior-extension-for-doctrine-2

kohana ORM question

i am using kohana ORM in order to get some results from the database. My problem is: even though i have consulted the documentation, i can't find a way to select only the column i am interested in. To be more explicit, i have:
$sale_stock = Model::factory('product_type')
->where('product_type_id','=', $id )
-> find_all();
var dumping it, it selects me all the "SELECT product_type.* from product_type where etc".
But i want to select only the 'stock' field from the salestock table. doing find('stock') instead find_all() returns a weired object... Where am i wrong, and how can i actually select only the column 'stock' using kohana orm?
thank you!
ORM methods find() and find_all() always select all table columns, so there is two ways to get specified fields:
Load full table rows and get columns
from it:
$sale_stock = Model::factory('product_type')
->where('product_type_id','=', $id )
-> find_all();
// get array of id=>stock values
$columns = $sale_stock->as_array('id', 'stock');
Create special method in model using
Query Builder:
// model Model_Product_Type
public function get_stocks($product_type_id)
{
return DB::select(array('stock'))
->from($this->_table_name)
->where('product_type_id', '=', $product_type_id)
->execute($this->_db);
}
I realise this isn't exactly what you're looking for, but I've pulled the following from the Kohana documentation ...
$articles = ORM::factory('article')->select_list('id', 'title');
foreach ($articles as $id => $title)
{
// Display a list of links
echo html::anchor('articles/'.$id, $title);
}
// Display a dropdown list
echo form::dropdown('articles', $articles);
You could think of it as a discount, two fields for the price of one.
It's common practice for ORMs to return a 'non-standard' object when partial model or merged model fields are requested. This prevents confusing operations using the original object (ie. how do you save an object when it contains only 2 of 8 fields, plus maybe some fields from another model?).
If you print_r the object, and give me an indication of how that looks ... it might be just what you want.
I know this is an old question, but i found maybe easier solution:
$sale_stock = ORM::factory('product_type')
->where( 'product_type_id','=', $id )
->find_all();
die($sale_stock->stock);