So I am having some trouble figuring out how to do a feed style mysql call, and I don't know if its an eloquent issue or a mysql issue. I am sure it is possible in both and I am just in need of some help.
So I have a user and they go to their feed page, on this page it shows stuff from their friends (friends votes, friends comments, friends status updates). So say I have tom, tim and taylor as my friends and I need to get all of their votes, comments, status updates. How do I go about this? I have a list of all the friends by Id number, and I have tables for each of the events (votes, comments, status updates) that have the Id stored in it to link back to the user. So how can I get all of that information at once so that I can display it in a feed in the form of.
Tim commented "Cool"
Taylor Said "Woot first status update~!"
Taylor Voted on "Best competition ever"
Edit #damiani
So after doing the model changes I have code like this, and it does return the correct rows
$friends_votes = $user->friends()->join('votes', 'votes.userId', '=', 'friend.friendId')->orderBy('created_at', 'DESC')->get(['votes.*']);
$friends_comments = $user->friends()->join('comments', 'comments.userId', '=', 'friend.friendId')->orderBy('created_at', 'DESC')->get(['comments.*']);
$friends_status = $user->friends()->join('status', 'status.userId', '=', 'friend.friendId')->orderBy('created_at', 'DESC')->get(['status.*']);
But I would like them all to happen at once, this is because mysql sorting thousands of records in order is 100x faster then php taking 3 lists, merging them and then doing it. Any ideas?
I'm sure there are other ways to accomplish this, but one solution would be to use join through the Query Builder.
If you have tables set up something like this:
users
id
...
friends
id
user_id
friend_id
...
votes, comments and status_updates (3 tables)
id
user_id
....
In your User model:
class User extends Eloquent {
public function friends()
{
return $this->hasMany('Friend');
}
}
In your Friend model:
class Friend extends Eloquent {
public function user()
{
return $this->belongsTo('User');
}
}
Then, to gather all the votes for the friends of the user with the id of 1, you could run this query:
$user = User::find(1);
$friends_votes = $user->friends()
->with('user') // bring along details of the friend
->join('votes', 'votes.user_id', '=', 'friends.friend_id')
->get(['votes.*']); // exclude extra details from friends table
Run the same join for the comments and status_updates tables. If you would like votes, comments, and status_updates to be in one chronological list, you can merge the resulting three collections into one and then sort the merged collection.
Edit
To get votes, comments, and status updates in one query, you could build up each query and then union the results. Unfortunately, this doesn't seem to work if we use the Eloquent hasMany relationship (see comments for this question for a discussion of that problem) so we have to modify to queries to use where instead:
$friends_votes =
DB::table('friends')->where('friends.user_id','1')
->join('votes', 'votes.user_id', '=', 'friends.friend_id');
$friends_comments =
DB::table('friends')->where('friends.user_id','1')
->join('comments', 'comments.user_id', '=', 'friends.friend_id');
$friends_status_updates =
DB::table('status_updates')->where('status_updates.user_id','1')
->join('friends', 'status_updates.user_id', '=', 'friends.friend_id');
$friends_events =
$friends_votes
->union($friends_comments)
->union($friends_status_updates)
->get();
At this point, though, our query is getting a bit hairy, so a polymorphic relationship with and an extra table (like DefiniteIntegral suggests below) might be a better idea.
Probably not what you want to hear, but a "feeds" table would be a great middleman for this sort of transaction, giving you a denormalized way of pivoting to all these data with a polymorphic relationship.
You could build it like this:
<?php
Schema::create('feeds', function($table) {
$table->increments('id');
$table->timestamps();
$table->unsignedInteger('user_id');
$table->foreign('user_id')->references('id')->on('users')->onDelete('cascade');
$table->morphs('target');
});
Build the feed model like so:
<?php
class Feed extends Eloquent
{
protected $fillable = ['user_id', 'target_type', 'target_id'];
public function user()
{
return $this->belongsTo('User');
}
public function target()
{
return $this->morphTo();
}
}
Then keep it up to date with something like:
<?php
Vote::created(function(Vote $vote) {
$target_type = 'Vote';
$target_id = $vote->id;
$user_id = $vote->user_id;
Feed::create(compact('target_type', 'target_id', 'user_id'));
});
You could make the above much more generic/robust—this is just for demonstration purposes.
At this point, your feed items are really easy to retrieve all at once:
<?php
Feed::whereIn('user_id', $my_friend_ids)
->with('user', 'target')
->orderBy('created_at', 'desc')
->get();
Related
Very long time I searching for solution for this problem:
Lets say we have 2 tables one table is Clients table and second table is ClientAssignment table:
the ClientAssignment table is related to Clients table:
public function assignment()
{
return $this->hasOne(ClientAssignment::class, 'client_id');
}
now when I want to count how many ClientAssignment has Clients and i do it like that:
$users =[1,2,3,4 .....]
$userAssignments = array();
foreach ($users as $user) {
$user_assignments = Client::whereHas('assignment', function ($query) use ($user) {
$query->where('assigned_id', $user);
});
$ua['user_id'] = $user;
$ua['count'] = $user_assignments->count();
array_push($userAssignments, $ua);
}
The code works well but it hits the performance and query execution time ~20 + seconds on a relatively small table with 80k Clients,
My question if can be another way to do the same thing but with minimum performance and query execution time hit ?
According to your post, I think
Client --- hasOne ----> ClientAssignment <----- hasMany ---- User
[client_id, assigned_id]
So User can hasMany Client through ClientAssignment,
However, your client_id and assigned_id are all in client_assignment table. So you can not use hasManyThrough, this just like a pivot table;
Fortunately, you can directly count the client_id and get the assigned_id as user_id just use this pivot table.
The query is like this (Use distinct(client_id) for preventing the dirty records):
ClientAssignment::whereIn('assigned_id', $users)
->groupBy('assigned_id')
->select('assigned_id AS user_id',
DB::raw('COUNT(DISTINCT(client_id)) AS count'))
->get()->toArray();
And add assigned_id index to improve the performance.
$table->index('assigned_id');
Use the count()
Client::first()->assignments->count();
// Or
Client::find(ID)->assignments()->count();
// or ...
I am learning relationships in Laravel php framework and I am trying to build this query
SELECT * FROM users u INNER JOIN link_to_stores lts ON u.id=lts.user_id INNER JOIN stores s ON lts.store_id=s.store_id WHERE lts.privilege = 'Owner'
I built this in Model
Link_to_store.php
public function store()
{
return $this->belongsTo('App\Store');
}
public function user()
{
return $this->belongsTo('App\User');
}
User.php
public function store_links()
{
return $this->hasMany('App\Link_to_store');
}
Store.php
public function user_links()
{
return $this->hasMany('App\Link_to_store');
}
I tried this query but this only joins user and link_to_store table
$personal_stores = Auth::user()->store_links->where('privilege','=','Owner');
Now I am confused how to join store table too. Can anyone help with this?
Schema is like this
Stores Table
store_id store_name
Users Table
id name
Link_to_stores Table
id store_id user_id privilege
I suppose store_links is actually a pivot table. In this case, you can use belongsToMany(), this will automatically take care of the pivot table.
To do this, in your User model you change the store function to this:
function stores() {
return $this->belongsToMany('App\Store', 'store_links', 'user_id', 'store_id')->withPivot('privilege');
}
Because the primary key of stores is not id, you will have to define this in you Store model with the following line:
protected $primaryKey = 'store_id';
Now to get the stores for a user, you simply call
$stores = Auth::user->stores()->wherePivot('privilege', 'Owner')->get();
I am learning relationships in Laravel php framework and I am trying to build this query
SELECT * FROM users u INNER JOIN link_to_stores lts ON u.id=lts.user_id INNER JOIN stores s ON lts.store_id=s.store_id WHERE lts.privilege = 'Owner'
You are trying to do a join here. You can do a join like this:
$stores = User::join('link_to_stores as lts', 'users.id', '=', 'lts.user_id')->join('stores as s', 'lts.store_id', '=', 's.id')->where('lts.privilege', 'Owner')->get();
But like Jerodev pointed out, it seems like Many to Many relationship might make more sense in your case. The difference is that relationship will actually execute 2 queries (1 for original model, 1 for relationship). It will then attach the related models to the original model (which is extremely handy).
I have two tables, say "users" and "users_actions", where "users_actions" has an hasMany relation with users:
users
id | name | surname | email...
actions
id | id_action | id_user | log | created_at
Model Users.php
class Users {
public function action()
{
return $this->hasMany('Action', 'user_id')->orderBy('created_at', 'desc');
}
}
Now, I want to retrieve a list of all users with their LAST action.
I saw that doing Users::with('action')->get();
can easily give me the last action by simply fetching only the first result of the relation:
foreach ($users as $user) {
echo $user->action[0]->description;
}
but I wanted to avoid this of course, and just pick ONLY THE LAST action for EACH user.
I tried using a constraint, like
Users::with(['action' => function ($query) {
$query->orderBy('created_at', 'desc')
->limit(1);
}])
->get();
but that gives me an incorrect result since Laravel executes this query:
SELECT * FROM users_actions WHERE user_id IN (1,2,3,4,5)
ORDER BY created_at
LIMIT 1
which is of course wrong. Is there any possibility to get this without executing a query for each record using Eloquent?
Am I making some obvious mistake I'm not seeing? I'm quite new to using Eloquent and sometimes relationship troubles me.
Edit:
A part from the representational purpose, I also need this feature for searching inside a relation, say for example I want to search users where LAST ACTION = 'something'
I tried using
$actions->whereHas('action', function($query) {
$query->where('id_action', 1);
});
but this gives me ALL the users which had had an action = 1, and since it's a log everyone passed that step.
Edit 2:
Thanks to #berkayk looks like I solved the first part of my problem, but still I can't search within the relation.
Actions::whereHas('latestAction', function($query) {
$query->where('id_action', 1);
});
still doesn't perform the right query, it generates something like:
select * from `users` where
(select count(*)
from `users_action`
where `users_action`.`user_id` = `users`.`id`
and `id_action` in ('1')
) >= 1
order by `created_at` desc
I need to get the record where the latest action is 1
I think the solution you are asking for is explained here http://softonsofa.com/tweaking-eloquent-relations-how-to-get-latest-related-model/
Define this relation in User model,
public function latestAction()
{
return $this->hasOne('Action')->latest();
}
And get the results with
User::with('latestAction')->get();
I created a package for this: https://github.com/staudenmeir/eloquent-eager-limit
Use the HasEagerLimit trait in both the parent and the related model.
class User extends Model {
use \Staudenmeir\EloquentEagerLimit\HasEagerLimit;
}
class Action extends Model {
use \Staudenmeir\EloquentEagerLimit\HasEagerLimit;
}
Then simply chain ->limit(1) call in your eager-load query (which seems you already do), and you will get the latest action per user.
My solution linked by #berbayk is cool if you want to easily get latest hasMany related model.
However, it couldn't solve the other part of what you're asking for, since querying this relation with where clause would result in pretty much the same what you already experienced - all rows would be returned, only latest wouldn't be latest in fact (but latest matching the where constraint).
So here you go:
the easy way - get all and filter collection:
User::has('actions')->with('latestAction')->get()->filter(function ($user) {
return $user->latestAction->id_action == 1;
});
or the hard way - do it in sql (assuming MySQL):
User::whereHas('actions', function ($q) {
// where id = (..subquery..)
$q->where('id', function ($q) {
$q->from('actions as sub')
->selectRaw('max(id)')
->whereRaw('actions.user_id = sub.user_id');
})->where('id_action', 1);
})->with('latestAction')->get();
Choose one of these solutions by comparing performance - the first will return all rows and filter possibly big collection.
The latter will run subquery (whereHas) with nested subquery (where('id', function () {..}), so both ways might be potentially slow on big table.
Let change a bit the #berkayk's code.
Define this relation in Users model,
public function latestAction()
{
return $this->hasOne('Action')->latest();
}
And
Users::with(['latestAction' => function ($query) {
$query->where('id_action', 1);
}])->get();
To load latest related data for each user you could get it using self join approach on actions table something like
select u.*, a.*
from users u
join actions a on u.id = a.user_id
left join actions a1 on a.user_id = a1.user_id
and a.created_at < a1.created_at
where a1.user_id is null
a.id_action = 1 // id_action filter on related latest record
To do it via query builder way you can write it as
DB::table('users as u')
->select('u.*', 'a.*')
->join('actions as a', 'u.id', '=', 'a.user_id')
->leftJoin('actions as a1', function ($join) {
$join->on('a.user_id', '=', 'a1.user_id')
->whereRaw(DB::raw('a.created_at < a1.created_at'));
})
->whereNull('a1.user_id')
->where('aid_action', 1) // id_action filter on related latest record
->get();
To eager to the latest relation for a user you can define it as a hasOne relation on your model like
namespace App\Models;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Support\Facades\DB;
class User extends Model
{
public function latest_action()
{
return $this->hasOne(\App\Models\Action::class, 'user_id')
->leftJoin('actions as a1', function ($join) {
$join->on('actions.user_id', '=', 'a1.user_id')
->whereRaw(DB::raw('actions.created_at < a1.created_at'));
})->whereNull('a1.user_id')
->select('actions.*');
}
}
There is no need for dependent sub query just apply regular filter inside whereHas
User::with('latest_action')
->whereHas('latest_action', function ($query) {
$query->where('id_action', 1);
})
->get();
Migrating Raw SQL to Eloquent
Laravel Eloquent select all rows with max created_at
Laravel - Get the last entry of each UID type
Laravel Eloquent group by most recent record
Laravel Uses take() function not Limit
Try the below Code i hope it's working fine for u
Users::with(['action' => function ($query) {
$query->orderBy('created_at', 'desc')->take(1);
}])->get();
or simply add a take method to your relationship like below
return $this->hasMany('Action', 'user_id')->orderBy('created_at', 'desc')->take(1);
Good morning everyone!
I'm using Laravel 5.0 and Eloquent to build a displaying page for results of some replies on the database.
Replies are in reservations table, which "belongs" to users table since every reservation is linked to a person.
class Reservation extends Model {
public function user()
{
return $this->belongsTo('App\User');
}
}
I would like to display results, the reservations, ordered by the last name column of the user. So like:
$reservations = Reservation::orderBy( /*users.last_name*/ )->get();
But I dont'know how. Thank you in advance for your time.
You'll need to join the tables to order by a foreign column.
$reservations = Reservation::join('users', 'users.id', '=', 'users.reservation_id')
->orderBy('users.last_name', 'asc')->get();
I have a Pivot table thats used to join two other tables that have many relations per hotel_id. Is there a way I can eagerload the relationship that pulls the results for both tables in one relationship? The raw SQL query, works correctly but when using belongsToMany the order is off.
Amenities Pivot Table
id
hotel_id
distance_id
type_id
Distance Table
id
name
Type Table
id
name
RAW Query (This works fine)
SELECT * FROM amenities a
LEFT JOIN distance d ON a.distance_id = d.id
LEFT JOIN type t ON a.type_id = t.id WHERE a.hotel_id = ?
My "Hotels" Model is using belongsToMany like so
public function distance() {
return $this->belongsToMany('Distance', 'amenities', 'hotel_id', 'distance_id');
}
public function type() {
return $this->belongsToMany('Type', 'amenities', 'hotel_id', 'type_id');
}
This outputs the collection, but they are not grouped correctly. I need to loop these into select fields side by side as entered in the pivot table, so a user can select a "type" and the "distance", but the order is off when using the collection. The raw query above outputs correctly.
Hotels::where('id','=','200')->with('distance', 'type')->take(5)->get();
Ok Solved it. So apparently you can use orderBy on your pivot table. Incase anyone else has this issue this is what I did on both relationships.
public function distance() {
return $this->belongsToMany('Distance', 'amenities', 'hotel_id', 'distance_id')->withPivot('id')->orderBy('pivot_id','desc');
}
public function type() {
return $this->belongsToMany('Type', 'amenities', 'hotel_id', 'type_id')->withPivot('id')->orderBy('pivot_id','desc');
}
It's not really a great practice to include other query building steps in the relationship methods on your models. The relationship method should just define the relationship, nothing else. A cleaner method is to apply eager load constraints. (scroll down a bit) Consider the following.
Hotels::where('id', 200)->with(array(
'distance' => function ($query)
{
$query->withPivot('id')->orderBy('pivot_id','desc');
},
'type' => function ($query)
{
$query->withPivot('id')->orderBy('pivot_id','desc');
},
))->take(5)->get();
If you find that you are eagerly loading this relationship in this way often, consider using scopes to keep things DRY. The end result will allow you to do something like this.
Hotels::where('id', 200)->withOrderedDistance()->withOrderedType()->take(5)->get();
P.S. Your models should be singular. Hotel, not Hotels. The model represents a single record.
Solved by using ->withPivot('id')->orderBy('pivot_id','desc');
Posted answer in the question.