Level 73
Something like this:
User::with(['listings' => function ($query)
{
$query->withCount('responses');
}])
->get();
now each of your listing will have responses_count property.
1 like
Feb 23, 2022
5
Level 5
What will be the query ?
One USER can have multiple LISTINGS and a LISTING can have multiple RESPONSES.
I need to calculate total number of responses of listings by a USER.
What will be the query ?
Models
class Listing extends Model
{
use HasFactory;
protected $guarded = [];
public function messages(){
return $this->hasMany(Message::class,'listing_id','id');
}
public function responses(){
return $this->hasMany(ListingResponse::class,'listing_id','id');
}
class ListingResponse extends Model
{
use HasFactory;
protected $guarded = [];
public function user(){
return $this->belongsTo(User::class,'user_id','id');
}
public function listing(){
return $this->belongsTo(Listing::class,'listing_id','id');
}
Level 5
@Nakov I have the user id. So how I will calculate total responses of all the USER listings ?
Level 73
$user = User::with(['listings' => function ($query)
{
$query->withCount('responses');
}])
->find(USER_ID);
$totalResponses = $user->listings->sum('responses_count');
1 like
Level 11
@nakov added a working query, but I would argue that it is not efficient enough - you're loading listings although you probably don't need them, as you need only responses_count, right?
So I would suggest a hasManyThrough relationship.
User.php:
public function responses()
{
return $this->hasManyThrough(ListingResponse::class, Listing::class);
}
And then you do:
$user = User::withCount('responses')->find($userId);
And then, in the Blade, you have $user->responses_count as the result.
3 likes
Level 5
@PovilasKorop Firstly welcome to Laracast. I am already following you on youtube. Thanks for the most efficient way to query.
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