I would like to display the posts of everyone the current user follows, ordered by date desc.
I have a many to many relationship supplying all the people the user is following.
$users = User::find(Auth::user()->id)->follow()->get();
I have a one to many relationship displaying the posts for any user.
$updates = App\User::find(?????)->updates()->orderBy('created_at', 'desc')->get();
The question mark's shows where the followers ID's need to be placed.
I can put the above query inside the for each loop but that obviously works its way through each follower rather than all posts in date order.
I suspect I may need to set a new relationship and work from the beginning. Can anyone advise.
User Model
public function updates()
{
return $this->hasMany('App\update');
}
/**
* User following relationship
*/
// Get all users we are following
public function follow()
{
return $this->belongsToMany('App\User', 'user_follows', 'user_id', 'follow_id')->withTimestamps()->withPivot('id');;;
}
// This function allows us to get a list of users following us
public function followers()
{
return $this->belongsToMany('App\User', 'user_follows', 'follow_id', 'user_id')->withTimestamps();;
}
}
Update Model
public function user_update()
{
return $this->belongsTo('App\User');
}
Thank you.
Since you want the posts, it is probably going to be easier starting a query on the Post model, and then filter the posts based on their relationships.
Assuming your Post model has an author relationship to the User that created the post, and the User has a follower relationship to all the Users that are following it, you could do:
$userId = Auth::user()->id;
$posts = \App\Post::whereHas('author.follower', function ($q) use ($userId) {
return $q->where('id', $userId);
})
->latest() // built in helper method for orderBy('created_at', 'desc')
->get();
Now, $posts will be a collection of your Post models that were authored by a user that is being followed by your authenticated user.
Related
Good day. I have this relationship problem in laravel
I have three models
User
State
School
I have established the following relationships between them
a. School - User (A user can enroll in many schools)
In User Model
public function schools(){
return $this->belongsToMany('App\Models\School', 'school_user', 'user_id', 'school_id');
}
b. A school can have many users
In School Model
public function users(){
return $this->belongsToMany('App\Models\User', 'school_user', 'school_id', 'user_id');
}
c. A school belongs to a state (i.e c an be found in a state)
In School Model
public function states(){
return $this->belongsTo('App\Models\State');
}
d. A state has many schools
In State Model
public function schools(){
return $this->hasMany('App\Models\School');
}
Now, I know how to query models with immediate relationship. For example, I can get the users associated with a school and vice-versa. I can also get the schools in a state.
My Question
How do I get all users associated with schools in a given state (using a query)? Assuming all we have is just the name of a state.
And probably get all schools the users are associated with and the date(time) in which there were associated with the schools?
This is what I have done. The second query is not giving me an answer
public function details(){
//Get all schools associated with a state
$schools = State::with('schools')->where('id',1)->first();
foreach($schools->schools as $data){
//Get all users associated with the schools
$users = School::with('users')->where('id',$data->id);
dd($users);
}
}
Apart from the fact that this approach is not probably right, I am not getting any answer. Is there a query or method that can solve this?
Thanks.
You should see the related models in the relations array attribute
$school= App\Models\School::with('User', 'State')->get();
Using a whereHas you can get all users that belong to schools with the given state_id.
$state = State::find(1);
$users = User::whereHas('schools', function ($query) use ($state) {
$query->where('state_id', $state->id);
})->get();
Querying Relationship Existence
If you need even more power, you may use the whereHas and orWhereHas
methods to define additional query constraints on your has queries,
such as inspecting the content of a comment:
use Illuminate\Database\Eloquent\Builder;
// Retrieve posts with at least one comment containing words like code%...
$posts = Post::whereHas('comments', function (Builder $query) {
$query->where('content', 'like', 'code%');
})->get();
// Retrieve posts with at least ten comments containing words like code%...
$posts = Post::whereHas('comments', function (Builder $query) {
$query->where('content', 'like', 'code%');
}, '>=', 10)->get();
I have an user. User create many ads.I want to see users details with ads where ads shown by paginate. For this, i make two model(User & Ads)
public function user(){
return $this->hasOne(User::class, 'id', 'user_id');
}
public function ads(){
return $this->hasMany(Ads::class, 'user_id', 'id');
}
In controller i call like this:
$users = Ads::with(['user'=> function ($q) use ($id){
$q->where('id',$id);
}])->paginate(2);
But here user's details are shown when forelse loop are called. But i don't want this.
So, How can i get user's details with ads's pagination?
I think you're overcomplicating.
You has two models.
In User, you can put this relationship:
public function ads(){
return $this->hasMany(App\Ad, 'user_id', 'id');
}
In Ads model, you put this relationship:
public function user(){
return $this->belongsTo(App\User, 'id', 'user_id');
}
In your controller, you simple call like this:
//If you want a list of User's Ads
$user = User::find(1);
$userAds = $user->ads()->paginate(2); //you paginate because can be many
//If you want the details from a user who made the Ad #1
$ad = Ad::find(1);
$user = $ad->user; //Don't need paginate because is only one.
I have 3 models
User
Channel
Reply
model relations
user have belongsToMany('App\Channel');
channel have hasMany('App\Reply', 'channel_id', 'id')->oldest();
let's say i have 2 channels
- channel-1
- channel-2
channel-2 has latest replies than channel-1
now, i want to order the user's channel by its channel's current reply.
just like some chat application.
how can i order the user's channel just like this?
channel-2
channel-1
i already tried some codes. but nothing happen
// User Model
public function channels()
{
return $this->belongsToMany('App\Channel', 'channel_user')
->withPivot('is_approved')
->with(['replies'])
->orderBy('replies.created_at'); // error
}
// also
public function channels()
{
return $this->belongsToMany('App\Channel', 'channel_user')
->withPivot('is_approved')
->with(['replies' => function($qry) {
$qry->latest();
}]);
}
// but i did not get the expected result
EDIT
also, i tried this. yes i did get the expected result but it would not load all channel if there's no reply.
public function channels()
{
return $this->belongsToMany('App\Channel')
->withPivot('is_approved')
->join('replies', 'replies.channel_id', '=', 'channels.id')
->groupBy('replies.channel_id')
->orderBy('replies.created_at', 'ASC');
}
EDIT:
According to my knowledge, eager load with method run 2nd query. That's why you can't achieve what you want with eager loading with method.
I think use join method in combination with relationship method is the solution. The following solution is fully tested and work well.
// In User Model
public function channels()
{
return $this->belongsToMany('App\Channel', 'channel_user')
->withPivot('is_approved');
}
public function sortedChannels($orderBy)
{
return $this->channels()
->join('replies', 'replies.channel_id', '=', 'channel.id')
->orderBy('replies.created_at', $orderBy)
->get();
}
Then you can call $user->sortedChannels('desc') to get the list of channels order by replies created_at attribute.
For condition like channels (which may or may not have replies), just use leftJoin method.
public function sortedChannels($orderBy)
{
return $this->channels()
->leftJoin('replies', 'channel.id', '=', 'replies.channel_id')
->orderBy('replies.created_at', $orderBy)
->get();
}
Edit:
If you want to add groupBy method to the query, you have to pay special attention to your orderBy clause. Because in Sql nature, Group By clause run first before Order By clause. See detail this problem at this stackoverflow question.
So if you add groupBy method, you have to use orderByRaw method and should be implemented like the following.
return $this->channels()
->leftJoin('replies', 'channels.id', '=', 'replies.channel_id')
->groupBy(['channels.id'])
->orderByRaw('max(replies.created_at) desc')
->get();
Inside your channel class you need to create this hasOne relation (you channel hasMany replies, but it hasOne latest reply):
public function latestReply()
{
return $this->hasOne(\App\Reply)->latest();
}
You can now get all channels ordered by latest reply like this:
Channel::with('latestReply')->get()->sortByDesc('latestReply.created_at');
To get all channels from the user ordered by latest reply you would need that method:
public function getChannelsOrderdByLatestReply()
{
return $this->channels()->with('latestReply')->get()->sortByDesc('latestReply.created_at');
}
where channels() is given by:
public function channels()
{
return $this->belongsToMany('App\Channel');
}
Firstly, you don't have to specify the name of the pivot table if you follow Laravel's naming convention so your code looks a bit cleaner:
public function channels()
{
return $this->belongsToMany('App\Channel') ...
Secondly, you'd have to call join explicitly to achieve the result in one query:
public function channels()
{
return $this->belongsToMany(Channel::class) // a bit more clean
->withPivot('is_approved')
->leftJoin('replies', 'replies.channel_id', '=', 'channels.id') // channels.id
->groupBy('replies.channel_id')
->orderBy('replies.created_at', 'desc');
}
If you have a hasOne() relationship, you can sort all the records by doing:
$results = Channel::with('reply')
->join('replies', 'channels.replay_id', '=', 'replies.id')
->orderBy('replies.created_at', 'desc')
->paginate(10);
This sorts all the channels records by the newest replies (assuming you have only one reply per channel.) This is not your case, but someone may be looking for something like this (as I was.)
In my application, I have setup a User model that can have subscribers and subscriptions through a pivot table called subscriptions.
public function subscribers()
{
return $this->belongsToMany('Forum\User', 'subscriptions', 'subscription_id', 'subscriber_id');
}
public function subscriptions()
{
return $this->belongsToMany('Forum\User', 'subscriptions', 'subscriber_id', 'subscription_id');
}
My question is, what relationship should I use to get a list of paginated Post models (belong to a User) from the User's subscriptions?
You can use the whereHas method to filter based on relationships. Assuming your Post model has a user relationship defined, your code would look something like:
// target user
$user = \App\User::first();
$userId = $user->id;
// get all of the posts that belong to users that have your target user as a subscriber
\App\Post::whereHas('user.subscribers', function ($query) use ($userId) {
return $query->where('id', $userId);
})->paginate(10);
You can read more about querying relationship existence in the documentation.
You can do something like this
\App\Post::with(['subscriptions' => function ($query) {
$query->where('date', 'like', '%date%');
}])->paginate(15);
Or without any conditions
\App\Post::with('subscriptions')->paginate(15);
In the documentation of Eloquent it is said that I can pass the keys of a desired relationship to hasManyThrough.
Lets say I have Models named Country, User, Post. A Country model might have many Posts through a Users model. That said I simply could call:
$this->hasManyThrough('Post', 'User', 'country_id', 'user_id');
This is fine so far! But how can I get these posts only for the user with the id of 3 ?
Can anybody help here?
So here it goes:
models: Country has many User has many Post
This allows us to use hasManyThrough like in your question:
// Country model
public function posts()
{
return $this->hasManyThrough('Post', 'User', 'country_id', 'user_id');
}
You want to get posts of a given user for this relation, so:
$country = Country::first();
$country->load(['posts' => function ($q) {
$q->where('user_id', '=', 3);
}]);
// or
$country->load(['posts' => function ($q) {
$q->has('user', function ($q) {
$q->where('users.id', '=', 3);
});
})
$country->posts; // collection of posts related to user with id 3
BUT it will be easier, more readable and more eloquent if you use this instead:
(since it has nothing to do with country when you are looking for the posts of user with id 3)
// User model
public function posts()
{
return $this->hasMany('Post');
}
// then
$user = User::find(3);
// lazy load
$user->load('posts');
// or use dynamic property
$user->posts; // it will load the posts automatically
// or eager load
$user = User::with('posts')->find(3);
$user->posts; // collection of posts for given user
To sum up: hasManyThrough is a way to get nested relation directly, ie. all the posts for given country, but rather not to search for specific through model.
$user_id = 3;
$country = Country::find($country_id);
$country->posts()->where('users.id', '=', $user_id)->get();
$this->hasManyThrough('Post', 'User', 'country_id', 'user_id')->where(column,x);
What happen here is you get the collection in return you can put any condition you want at the end.