This is my query, I tried this query it works.
SELECT *
FROM conference_venue
WHERE id_venue NOT IN (SELECT id_venue FROM submission_data WHERE id_submission = 1);
i want to display data in conference_venue. but I don't want to display data whose id_venue is the same as the submission_data table (same as id_venue whose id_submission is mentioned).
I'm trying to make a query for the laravel version, but it's a blank white screen with no errors.
DB::table('conference_venue')
->whereNotIn('id_venue', function($q){
$q->select('id_venue')
->from('submission_data')
->where('id_submission', '=', 1);
})->select('*')->get();
This query works when I try it in sql query console but fails when I try it with Laravel query builder.
You can try this:
DB::table('conference_venue')
->select('*')
->whereRaw(
'conference_venue.id_venue NOT IN (SELECT submission_data.id_venue FROM submission_data WHERE id_submission = 1)'
);
Or better yet, create a Model for conference_venue and submission_data (ie: ConferenceVenue, SubmissionData) and you can add Eloquent relationships for ConferenceVenue and SubmissionData.
Eloquent relationships, which supports a variety of common
relationships (One To One, One To Many, Many To Many, etc.), are
defined as methods on your Eloquent model classes. Since relationships
also serve as powerful query builders, defining relationships as
methods provides powerful method chaining and querying capabilities.
Eloquent: Relationships
On you ConferenceVenue Class, you can add a method something similar to the following:
public function available() {
return this->hasMany(SubmissionData, 'id_venue')
->select('*') // You can also specify relevant columns ONLY
->whereRaw(
'conference_venue.id_venue NOT IN (SELECT submission_data.id_venue FROM submission_data WHERE id_submission = 1)'
);
}
Where you can use the relationship method as follows:
$available = ConferenceVenue::with('available')->get();
Good morning,
Apologies if this has been ask before but we are unable to find any answer to an issue we are having.
We are working with a legacy database that is not owned by us (read-only) and are attempting to use Eloquent (Models) in Laravel to solve some common issues.
Is it possible to setup Eloquent's Eager-loading to create additional ON clauses to the HasMany / HasOne relationship builder?
Please see below of what we are trying to achieve without raw queries.
public function policy()
{
return $this->hasMany(Policy::class, 'Group', 'Group')
// This breaks as `on` isn't defined on Eloquent\Builder. Is this concept possible? Multiple on clauses
->on('Reference', 'Reference');
}
In our controller we have attempted the following which also fails.
Vehicle::with([
'policy' => function ($query) {
// Model isn't instantiated yet, but we need an additional on clause here
$query->on('Reference', 'Reference');
}
]);
Can the above be achieved or do we have to revert back to using raw queries?
Thank you in advance for any help.
You can use the Compoships package:
class Vehicle extends Model
{
use \Awobaz\Compoships\Compoships;
public function policy()
{
return $this->hasMany(Policy::class, ['Group', 'Reference'], ['Group', 'Reference']);
}
}
I am assuming the query already exists as raw SQL and what you are trying to achieve is to convert it to using eloquent. If that is the case you may find it quicker an deasier to use the built in raw query.
I have a Objective model which has many Action and every Action has one ActionYear. already defined in model.
How to use orderby to sort action in objective through action_year's specific column.
$query = Objective::with('actions.actionYear')
->orderBy('action_year.created_at')
->get();
this through error Undefined table: 7 ERROR: missing FROM-clause entry for table "action_year".
How to solve this. Thank you.
You can order the eager loaded models with a closure:
$query = Objective::with(['actions.actionYear' => function($q){
$q->orderBy('action_year.created_at');
})->get();
If you use function with() then this only makes sure the named relationship is loaded also to avoid extra need for SQL queries later. Othewise it does not introduce any additional changes compared to Objective::all()
One way how to achieve sorted collection is to use sortBy() function after loading the data like this
$query = $query->sortBy(function($objective){
return $objective->actionYear->created_at;
});
I have read the documentation but I can't quite figure out how to run the following query in Laravel 4
SELECT
COUNT(*)
FROM
acl a,
routes r
WHERE
(a.user_id = 1 OR
a.group_id IN(SELECT group_id FROM user_group_junction WHERE user_id = 1)) AND
r.route = 'protected' AND
a.routes_id = r.id;
So how would I run the query in Laravel 4 using eloquent?
Yes each table has a model and relationships are defined
Based on my selected answer the following is what I came up with (And works)
Acls::join('routes','routes.id','=','acl.routes_id')
->where('routes.route','=','protected')
->Where(function($in_parenthesis) use($user_id){
$in_parenthesis->whereIn('acl.group_id',function($where_in) use($user_id){
$where_in->select('group_id')
->from('user_group_junction')
->where('user_id','=',$user_id);
})
->orWhere('acl.user_id','=',$user_id);
})
->count();
Methods called on Eloquent models pass through to the Illuminate\Database\Eloquent\Builder class, which itself extends from the Illuminate\Database\Query\Builder class. This means that all the things you can do with the query builder, you can also do with Eloquent models. The exception being that you don't need to define the table.
So for example if you wanted to perform a join and a where like you've done above, you would just do:
$query = Acl::join('routes', 'acl.routes_id', '=', 'routes.id')
->where('routes.route', '=', 'protected');
$results = $query->get();
Obviously this isn't your whole query, but you can figure out the rest.
I have got 2 joined tables in Eloquent namely themes and users.
theme model:
public function user() {
return $this->belongs_to('User');
}
user model:
public function themes() {
return $this->has_many('Theme');
}
My Eloquent api call looks as below:
return Response::eloquent(Theme::with('user')->get());
Which returns all columns from theme (that's fine), and all columns from user (not fine). I only need the 'username' column from the user model, how can I limit the query to that?
Change your model to specify what columns you want selected:
public function user() {
return $this->belongs_to('User')->select(array('id', 'username'));
}
And don't forget to include the column you're joining on.
For Laravel >= 5.2
Use the ->pluck() method
$roles = DB::table('roles')->pluck('title');
If you would like to retrieve an array containing the values of a single column, you may use the pluck method
For Laravel <= 5.1
Use the ->lists() method
$roles = DB::table('roles')->lists('title');
This method will return an array of role titles. You may also specify a custom key column for the returned array:
You can supply an array of fields in the get parameter like so:
return Response::eloquent(Theme::with('user')->get(array('user.username'));
UPDATE (for Laravel 5.2)
From the docs, you can do this:
$response = DB::table('themes')
->select('themes.*', 'users.username')
->join('users', 'users.id', '=', 'themes.user_id')
->get();
I know, you ask for Eloquent but you can do it with Fluent Query Builder
$data = DB::table('themes')
->join('users', 'users.id', '=', 'themes.user_id')
->get(array('themes.*', 'users.username'));
This is how i do it
$posts = Post::with(['category' => function($query){
$query->select('id', 'name');
}])->get();
First answer by user2317976 did not work for me, i am using laravel 5.1
Using with pagination
$data = DB::table('themes')
->join('users', 'users.id', '=', 'themes.user_id')
->select('themes.*', 'users.username')
->paginate(6);
Another option is to make use of the $hidden property on the model to hide the columns you don't want to display. You can define this property on the fly or set defaults on your model.
public static $hidden = array('password');
Now the users password will be hidden when you return the JSON response.
You can also set it on the fly in a similar manner.
User::$hidden = array('password');
user2317976 has introduced a great static way of selecting related tables' columns.
Here is a dynamic trick I've found so you can get whatever you want when using the model:
return Response::eloquent(Theme::with(array('user' => function ($q) {
$q->addSelect(array('id','username'))
}))->get();
I just found this trick also works well with load() too. This is very convenient.
$queriedTheme->load(array('user'=>function($q){$q->addSelect(..)});
Make sure you also include target table's key otherwise it won't be able to find it.
This Way:
Post::with(array('user'=>function($query){
$query->select('id','username');
}))->get();
I know that this is an old question, but if you are building an API, as the author of the question does, use output transformers to perform such tasks.
Transofrmer is a layer between your actual database query result and a controller. It allows to easily control and modify what is going to be output to a user or an API consumer.
I recommend Fractal as a solid foundation of your output transformation layer. You can read the documentation here.
In Laravel 4 you can hide certain fields from being returned by adding the following in your model.
protected $hidden = array('password','secret_field');
http://laravel.com/docs/eloquent#converting-to-arrays-or-json
On Laravel 5.5, the cleanest way to do this is:
Theme::with('user:userid,name,address')->get()
You add a colon and the fields you wish to select separated by a comma and without a space between them.
Using Model:
Model::where('column','value')->get(['column1','column2','column3',...]);
Using Query Builder:
DB::table('table_name')->where('column','value')->get(['column1','column2','column3',...]);
If I good understood this what is returned is fine except you want to see only one column. If so this below should be much simpler:
return Response::eloquent(Theme::with('user')->get(['username']));
#You can get selected columns from two or three different tables
$users= DB::Table('profiles')->select('users.name','users.status','users.avatar','users.phone','profiles.user_id','profiles.full_name','profiles.email','profiles.experience','profiles.gender','profiles.profession','profiles.dob',)->join('users','profiles.user_id','=','users.id')
->paginate(10);
Check out, http://laravel.com/docs/database/eloquent#to-array
You should be able to define which columns you do not want displayed in your api.