Good morning,
I've been trying for quite a lot of time to translate this query(which returns an array of stdClass) into query builder so I could get objects back as Eloquent models.
This is how the query looks like untranslated:
$anketa = DB::select( DB::raw("SELECT *
FROM v_anketa a
WHERE not exists (select 1 from user_poeni where anketa_id=a.id and user_id = :lv_id_user)
Order by redni_broj limit 1"
), array( 'lv_id_user' => $id_user,
));
I have tried this, but it gives a syntax error near the inner from in the subquery:
$anketa = V_anketa::selectRaw("WHERE not exists (select 1 from user_poeni where anketa_id=a.id and user_id = :lv_id_user)", array('lv_id_user' => $id_user,)
)->orderBy('redni_broj')->take(1)->first();
The problem is this exists and a subquery in it. I couldn't find anything regarding this special case.
Assume each table has an appropriate Eloquent model.
V_anketa is a view. The db is postgresql.
As far as the query goes I believe this should work:
$anketa = V_anketa::whereNotExists(function ($query) use ($id_user) {
$query->select(DB::raw(1))
->from('user_poeni')
->where('anketa.id', '=', 'a.id')
->where('user_id', '=', $id_user);
})
->orderBy('redni_broj')
->first();
but I'm not clear on what do you mean by "assuming every table has an Eloquent model" and "V_anketa" is a view...
Assuming the SQL query is correct, this should work:
$anketa = DB::select(sprintf('SELECT * FROM v_anketa a WHERE NOT EXISTS (SELECT 1 FROM user_poeni WHERE anketa_id = a.id AND user_id = %s) ORDER BY redni_broj LIMIT 1', $id_user));
If you want to get back an Builder instance you need to specify the table:
$anketa = DB::table('')->select('');
If you however, want to get an Eloquent Model instance, for example to use relations, you need to use Eloquent.
Related
I need a following code to convert to Laravel query can any one help me with these.
SELECT id, `leave_name`, `total_leave_days`, leave_id, leave_taken_days FROM `leaves` AS t1 INNER JOIN ( SELECT leave_id, SUM(`leave_taken_days`) AS leave_taken_days FROM `leave_applications` WHERE user_id = 2 AND statuses_id = 2 GROUP BY leave_id ) AS t2 ON t1.id = t2.leave_id
I even tried but the output is not showing atall.
$user_leaves = DB::table('leaves')
->select('id', 'leave_name', 'total_leave_days', 'leave_id', 'leave_taken_days')
->join('leave_application', 'leave_application.leave_id', '=', 'leave.id')
->select('leave_application.leave_id', DB::raw("SUM(leave_taken_days) as leave_application.leave_taken_days"))
->where('user_id','=', 2)
->where('statuses_id','=', 2)
->get();
How can I solve this issue?
UPDATE
Relations between two models.
Leave Model
public function leave_application()
{
return $this->belongsTo(LeaveApplication::class, 'id' , 'leave_id');
}
Leave Application Model
public function leave()
{
return $this->belongsTo(Leave::class, 'leave_id', 'id');
}
Try this :
$user_leaves = Leave::select('leaves.id', 'leaves.leave_name', 'leaves.total_leave_days', 'leave_applications.leave_id', DB::raw('SUM(leave_applications.leave_taken_days) as leave_taken_days'))
->with('leave_application')
->whereHas('leave_application', function($q) {
$q->where('user_id', 2)
->where('statuses_id', 2);
})
->groupBy('leaves.id')
->get();
On this topic I would like to give my recommendations for some tools to help you out in the future.
SQL Statement to Laravel Eloquent to convert SQL to Laravel query builder. This does a decent job at low level queries. It also saves time when converting old code.
The other tool I use to view the query that is being run is Clock Work
I keep this open in a tab and monitor slow nasty queries or, also gives me perspective on how the query builder is writing SQL. If you have not use this extension I highly recommend getting and using it.
Actually I found my answer,
$user_leaves = DB::table('leaves as t1')
->select('t1.id', 't1.leave_name', 't1.total_leave_days', 't2.leave_id', 't2.leave_taken_days')
->join(DB::raw('(SELECT leave_id, SUM(leave_taken_days) AS leave_taken_days FROM leave_applications WHERE user_id = ' . $user_id . ' AND statuses_id = 2 GROUP BY leave_id) AS t2'), function ($join) {
$join->on('t1.id', '=', 't2.leave_id');
})
->get();
You can use DB:select("your query", params) and put your query and params (as an array (optional)
As below sample:
$result = DB:select("
SELECT id, `leave_name`, `total_leave_days`, leave_id, leave_taken_days
FROM `leaves` AS t1
INNER JOIN (
SELECT leave_id, SUM(`leave_taken_days`) AS leave_taken_days
FROM `leave_applications`
WHERE user_id = 2
AND statuses_id = 2
GROUP BY leave_id
) AS t2 ON t1.id = t2.leave_id" , $params
);
return response()->json($result);
How to order laravel eloquent query using parent model?
I mean I have an eloquent query where I want to order the query by its parent without using join relationship?
I used whereHas and order by on it, but did not work.
Here is a sample of my code:
$query = Post::whereHas('users')->orderBy('users.created_at')->get();
If you want to order Post by a column in user you have to do a join in some way unless you sort after you retrieve the result so either:
$query = Post::select('posts.*')
->join('users', 'users.id', 'posts.user_id')
->orderBy('users.created_at')->get();
Note that whereHas is not needed anymore because the join (which is an inner join by default) will only result in posts that have a user.
Alternatively you can do:
$query = Post::has('users')
->with('users')
->get()
->sortBy(function ($post) { return $post->users->created_at; });
The reason is that eloquent relationships are queried in a separate query from the one that gets the parent model so you can't use relationship columns during that query.
I have no clue why you wanted to order Posts based on their User's created_at field. Perhaps, a different angle to the problem is needed - like accessing the Post from User instead.
That being said, an orderBy() can accept a closure as parameter which will create a subquery then, you can pair it with whereRaw() to somewhat circumvent Eloquent and QueryBuilder limitation*.
Post::orderBy(function($q) {
return $q->from('users')
->whereRaw('`users`.id = `posts`.id')
->select('created_at');
})
->get();
It should generate the following query:
select *
from `posts`
order by (
select `created_at`
from `users`
where `users`.id = `posts`.id
) asc
A join might serve you better, but there are many ways to build queries.
*As far as I know, the subquery can't be made to be aware of the parent query fields
You can simply orderBy in your Post model.
public function users(){
return $this->belongsTo(User::class, "user_id")->orderByDesc('created_at');
}
I hope this helps you.
You can try
Post::query()
->has('users')
->orderBy(
User::select('created_at')
->whereColumn('id', 'posts.user_id')
->orderBy('created_at')
)
->get();
The sql generated would be like
select * from `posts`
where exists (select * from `users` where `posts`.`user_id` = `users`.`id`)
order by (select `created_at` from `users` where `id` = `posts`.`user_id` order by `created_at` asc) asc
But I guess join would be a simpler approach for this use case.
Laravel Docs - Eloquent - Subquery Ordering
I use laravel 5.3
My sql query is like this :
SELECT *
FROM (
SELECT *
FROM products
WHERE `status` = 1 AND `stock` > 0 AND category_id = 5
ORDER BY updated_at DESC
LIMIT 4
) AS product
GROUP BY store_id
I want to change it to be laravel eloquent
But I'm still confused
How can I do it?
In cases when your query is to complex you can laravel RAW query syntax like:
$data = DB::select(DB::raw('your query here'));
It will fire your raw query on the specified table and returns the result set, if any.
Reference
If you have Product model, you can run
$products = Product::where('status', 1)
->where('stock', '>', 0)
->where('category_id', '=', 5)
->groupBy('store_id')
->orderBy('updated_at', 'desc')
->take(4)
->get();
I think this should give you the same result since you pull everything from your derived table.
How to make this MySQL query in Query Builder?
SELECT traction_instances.*,
(SELECT COUNT(station_id)
FROM traction_stations
WHERE traction_instances.instance_id = traction_stations.monitoring_instance) as `count stations`
FROM traction_instances
What I've tried up to now is the following:
$instances = DB::connection('monitors')->table(DB::raw('traction_instances as ti, traction_stations as ts'))
->select(array('ti.instance_id', 'ti.status', 'ti.cpu_usage', DB::raw('count(ts.station_id) as station_count')))
->where(DB::raw('ti.instance_id'), '=', DB::raw('ts.monitoring_instance'))
->get();
With this, I get the single 'Instance' row (and there are 4 of them) with count of 'Stations' the instance is monitoring, but I need all the instances, and if there are no stations that that instance is monitoring, the station_count should be 0.
Found a solution:
$instances = DB::connection('monitors')->table('traction_instances as ti')
->where('ti.instance_type', 'Audio Stream Processor')
->select(DB::raw('ti.*, (SELECT COUNT(*) FROM traction_stations AS ts WHERE ti.instance_id = ts.monitoring_instance) AS station_count'))
->get();
I want to get the user that wrote the most articles. I do so fine in two ways with ActiveRecord like the following:
$table = Articles::find()
->select('articles.*, COUNT(*) AS cnt')
->with('user','userDetails')
->groupBy('articles.user_id')
->orderBy(('cnt DESC'))
->limit(10)
->offset($offset)
->all();
and with a query like the following:
$query = (new Query())
->select('articles.user_id, COUNT(*) AS num_articles')
->from('articles')
->join('LEFT JOIN', 'user_details', 'user_details.user_id = articles.user_id')
->groupBy('articles.user_id')
->orderBy('num_articles DESC')
->limit(10)
->offset($offset)
->all();
The problem is that the ActiveRecord gives me further needed informations userDetails that I need. But I do not get the amount of articles of user that should be on cnt
With the Query I get the user_id and the amount of articles. But I do not get it working by joining with userDetails. All of these does not work: LEFT JOIN, RIGHT JOIN, INNER JOIN.
I am interested in resolving both for learning, but for concrete I need help with the ActiveRecord problem.
Okay well I solved it for the ActiveRecord. The ActiveRecords needs a public $var; in the Model. So to get the amount you have to add the mentioned public... to your Model so that in my case:
public $cnt; extends the ActiveRecord of Articles
now I can access it with the given Request in my Question. But this just solves the first point. I am still interested in the second way for more complex Queries.
I dont have much idea about active record but I think the below is something what you are looking for
select * from user_details where user_id in
(select A.user from
(select user_id as user, COUNT(*) AS num_articles
from articles group by user_id order by num_articles desc LIMIT 10)A
);
for second point you should include required column from joined table to select statement:
$query = (new Query())
->select('articles.user_id, COUNT(*) AS num_articles, user_details.username as username')
->from('articles')
->join('LEFT JOIN', 'user_details', 'user_details.user_id = articles.user_id')
->groupBy('articles.user_id')
->orderBy('num_articles DESC')
->limit(10)
->offset($offset)
->all();