So I want to show id from order table where user id is the same as currently login user id, then later used to show orders have been made by the user
$orderId = Order::select('id')->firstWhere('user_id', auth()->id())->id;
$orders = SubOrder::where('order_id', $orderId)->orderBy('created_at', 'desc')->get();
it works but it only shows the first record, after some digging later I found out that the problem is on the $orderId, it only shows the first record. but I want it to be all the records. if I change the id to get(), it shows nothing since it give the result like "id = 1" instead of the number only. also have tried to change the firstWhere into where and got error like "Property [id] does not exist on the Eloquent builder instance."
please help, thanks
If you are going to use the other Orders associated with the User soon after you get the first Order, return all the relevant Orders and then just grab the first one when you need it.
$orders = Order::where('user_id', auth()->user()->id)->get();
$firstOrder = $orders->first();
$subOrders = SubOrder::whereIn('order_id', $orders->pluck('id'))->get();
Alternatively, you could use a subOrders relationship defined on your Order model.
class Order extends Model
{
public function subOrders()
{
return $this->hasMany(SubOrder::class);
}
}
$orders = Order::where('user_id', auth()->user()->id)->get();
$firstOrder = $orders->first();
$firstOrderSubOrders = $firstOrder->subOrders;
If you're confident you're going to be working with SubOrder records, you can use eager loading on your Order to improve performance.
$orders = Order::where('user_id', auth()->user()->id)
->with('subOrders')
->get();
$firstOrder = $orders->first();
$firstOrderSubOrders = $firstOrder->subOrders;
first() will return the first id queried and stop execution.
$firstOrder = $orders->first();
Related
I'm new to Laravel. I am developing a project. and in this project I have 4 tables related to each other
-Users
-Orders
-OrderParcels
-Situations
When listing the parcels of an order, I want to get the information of that order only once, the user information of that order once again, and list the parcels as a table under it. so far everything ok. but I also want to display the status of the parcels listed in the table as names. I couldn't add the 4th table to the query. do you have a suggestion? I'm putting pictures that explain the structure below.
My current working code is
$orderParcels = Orders::whereId($id)
->with('parcels')
->with('users:id,name')
->first();
and my 'orders' model has method
public function parcels(){
return $this->hasMany(OrderParcels::class);
}
public function users(){
return $this->hasOne(User::class,'id','affixer_id');
}
Note[edit]: I already know how to connect like this
$orderParcels = DB::table('order_parcels as op')
->leftjoin('orders as o','op.orders_id','o.id')
->leftjoin('users as u','o.affixer_id','u.id')
->leftjoin('situations as s','op.status','s.id')
->select('op.*','o.*','u.name','s.situations_name')
->where('op.orders_id',$id)->get();
but this is not working for me, for each parcels record it returns me orders and user info. I want once orders info and once user info.
Laravel provides an elegant way to manage relations between models. In your situation, the first step is to create all relations described in your schema :
1. Model Order
class User extends Model {
public function parcels()
{
return $this->hasMany(OrderParcels::class);
}
public function users()
{
return $this->hasOne(User::class,'id','affixer_id');
}
}
2. Model Parcel
class Parcel extends Model {
public function situations()
{
return $this->hasOne(Situation::class, ...);
}
}
Then, you can retrieve all desired informations simply like this :
// Retrieve all users of an order
$users = $order->users; // You get a Collection of User instances
// Retrieve all parcels of an order
$parcels = $order->parcels; // You get a Collection of User instances
// Retrieve the situation for a parcel
$situations = $parcel->situations // You get Situation instance
How it works ?
When you add a relation on your model, you can retrieve the result of this relation by using the property with the same name of the method. Laravel will automatically provide you those properties ! (e.g: parcels() method in your Order Model will generate $order->parcels property.
To finish, in this situation where you have nested relations (as describe in your schema), you should use with() method of your model to eager load all the nested relation of order model like this :
$orders = Orders::with(['users', 'parcels', 'parcels.situations'])->find($id)
I encourage you to read those stubs of Laravel documentation :
Define model relations
Eager loading
Laravel Collection
Good luck !
Use join to make a perfect relations between tables.
$output = Orders::join('users', 'users.id', '=', 'orders.user_id')
->join('order_parcels', 'order_parcels.id', '=', 'orders.parcel_id')
->join('situations', 'situation.id', '=', 'order_parcels.situation_id')
->select([
'orders.id AS order_id',
'users.id AS user_id',
'order.parcels.id AS parcel_id',
'and so on'
])
->where('some row', '=', 'some row or variable')->get();
I have two tables, products and product_images, now I want to show product which has at least one image.
In controller I have a simple function in ProductController to fetch all the products:
public function products(){
$allProducts = $this->product->paginate(15);
return view('frontend.pages.products',compact('allProducts'));
}
But, I want to send the products which has at least one image of each product.
What should I do to achieve that?
Edit:
I have created relationship between tables, now how can I get my desired answer?
I have written this in the Controller:
$allProducts = $this->product->whereHas('product_images', function ($query){
$query->where();
})->get();
Assuming table schema
product
-id
-name
product_image
-id
-product_id
-url //any columns you needed
$product_ids = DB::table('product')
->join('product_image','product_image.product_id','=','product.id')
->select('product.id','roduct.name')
->groupBy('product.id')
->get();
$product_count = count($product_ids);
Eloquent has this built in.
Example
// Retrieve all posts that have three or more comments...
$posts = App\Post::has('comments', '>=', 3)->get();
In your case
In your case you could change the $allProducts line to be
$allProducts = $this->product()->has('product_image', '>=', 1)->paginate(15);
I didn't test the code above.
See documentation for more information on this topic.
https://laravel.com/docs/6.x/eloquent-relationships#querying-relationship-existence
You can use selectRaw:
$products = Product::leftJoin('product_images',function ($join){
$join->on('products.product_id','=','product_images.product_id');
})->selectRaw("products.product_id i, count(product_images.id) c")
->groupBy('products.product_id')
->where('c','>=',3)
->get();
Getting answer querying realtionship:
The query will look like following:
$allProducts = $this->product->has('Images')->paginate(15);
But, make you sure you have created relation in Model like the following:
public function Images(){
return $this->hasMany('\App\Models\ProductImages');
}
I have used hasMany relationship because one product can have multiple images.
I have a table with user data (users) and a table with prices (prices).
My prices table can contain multiple prices pr. user since I want to keep historical data.
I've defined my relation as a one-to-one
$this->hasOne("App\Model\Price","userid","id")->orderBy("id","desc")->take(1);
to allow me to see the users current price.
What I want to do now, is to select every user that has a current price of 100, but how do I do this? I know I could go for a left join, but as I read the documentation, it should be possible without a left join.
I've built a pseudo-query to explain what I'm after;
User::with("price")->where("prices.price","100")->get();
I've read through the documentation (Eloquent: Querying relationships), but that doesn't seem to be useful to my question.
I've also read several questions here on SO but unfortunately to no avail.
You may try this:
$currentPrice = 100;
$users = User::whereHas('price', function($query) use ($currentPrice) {
$query->where('price', $currentPrice); // price is the field name
})
->with("price")->get();
Since you have more than a single price for per user then you may also declare another relationship method to get all the price models instead of one and you may do it using something like this:
// In User model
public function prices()
{
return $this->hasMany("App\Model\Price", "userid", "id");
}
In this case, with::price will give you the last single record and with::prices will give you all the related prices. So, if you want then you may write something like the following to get all users with their all related prices who has the (latest/current) price of 100:
$currentPrice = 100;
$users = User::whereHas('price', function($query) use($currentPrice) {
$query->where('price', $currentPrice); // price is the field name
})
->with("prices") // with all prices
->get();
You can use the combination of whereHas() and with() as:
$users = User::whereHas("price", function($q) use ($currentPrice) {
$q->where("price", $currentPrice);
})
->with(["price" => function ($q) {
$query->where("price", $currentPrice);
})
->get();
I have table with many to many relationship.
User many to many Permission
I already define many to many relationship on both model, and create the pivot table also.
What I want is get all user which contain permission name
What I have done so far is
User::all()->permissions->contains('name', 'access.backend.admin')->get();
But it give me
Undefined property: Illuminate\Database\Eloquent\Collection::$permissions on line 1
What wrong with my code?
User::All() returns a collection not model object. You have iterate over the collection to get the model object and use ->permissions().
For exapmle:
$users = User::all();
foreach ($users as $user) {
$user->permissions->contains('name', 'access.backend.admin'); // returns boolean
}
Or you can get a single model from DB as:
$user = User::first();
$user->permissions->contains('name', 'access.backend.admin'); // returns boolean
Update 1
To get users which contain desired permission use filter() method as:
$filtered_users = $users->filter(function ($user) {
if ($user->permissions->contains('name', 'access.backend.admin')) {
return $user;
}
});
Update 2
You can also write a query which returns the desired result as:
$filtered_users = User::whereHas('permissions', function($q) {
$q->where('name', 'access.backend.admin');
})->get()
I have a similar case of questions and tags, they have many to many relationship.
So when i have to fetch all question with a particular tag then i do this
$tag = Tag::where('name','laravel')->get()->first();
I first retrieved the Tag model with name laravel.
and then retrieved all questions having tag laravel.
$questions = $tag->questions;
Similarly you can do this
$permission = Permission::where('name','access.backend.admin')->get()->first();
$users = $permission->users;
I have an Eloquent ORM relationship defined as follows:
ProductConfiguration:
public function product()
{
return $this->belongsTo('Excel\Products\Product');
}
public function currency()
{
return $this->belongsTo('Excel\Currencies\Currency');
}
Product
public function productConfigurations()
{
return $this->hasMany('Excel\Products\ProductConfiguration');
}
public function productType()
{
return $this->belongsTo('Excel\Products\ProductType');
}
I expect that if I do the following that I will load all product configurations of a specified product type, with the related products, nested product type details and the product configuration currency
$results = ProductConfiguration::with(
array(
'product' => function($query) use ($product_type) {
$query->where('product_type_id' , $product_type);
},
'product.productType',
'currency'
)
)
->get();
however the returned collection has 'product' set to NULL. the Currency Relationship is there, but the product relationship is not. I can see the outputted SQL queries and the query that selects the products retrieves the correct products if I paste it directly into my sql editor
select * from `products`
where `products`.`id` in ('12', '13')
and `product_type_id` = '1'
Am I correct to think that the results from this query should be included in my collection, or is there some obvious flaw in my thinking?
I think you don't want to achieve that. Now what you get is getting all ProductConfiguration with products that are only of certain_type.
So in case you have some configuration that has other type for product you will get null because you limited results from product to only the one that has certain product type.
I might be wrong, but you probably wanted to get those ProductConfiguration that belongs to Product that is type of certain_type. In this case you should use whereHas:
$results = ProductConfiguration::
with('product', 'product.productType', 'currency')->
whereHas('product', function($q) use ($product_type)
{
$q->where('product_type_id', '=', $product_type);
})->get();
I hate to post this as an answer but since i don't have enough rep to comment so try this first:
$results = ProductConfiguration::with('product')->get();
dd($results->toArray());
See what you get, if you get some data, try this
$results = ProductConfiguartion::with(array('products' => function($query){
$query->where('product_type_id' , $product_type);
})->get();
dd($results);
See what you get, if you get null: your $product_type variable may be something you didnt expect, so try dd($product_type) to make sure its what your expecting.