I have done the query below, and I am able to get the json data from console log, my question is how to make a pagination link using the json data?
$viewData = DB::table('tbl_contents')
->where('record_type', '=', 'Event')
->where('status', '=', 1)
->paginate(3);
return json_encode($viewData);
You may also return the pagination link, see my code below:
$viewData = DB::table('tbl_contents')
->where('record_type', '=', 'Event')
->where('status', '=', 1)
->paginate(3);
return [
'items' => json_encode($viewData),
'pagination_link' => $viewData->links()
];
Related
I am applying search filter in my project so first of all I get data from multiple tables and store in two different variables and then merge these two variable into one so I can filter data from that merged variable. So my code is like that
$data1=Model::query()
->Join('...')
->leftJoin('...')
->where('id',login_user)
->select(...)
->whereRaw('id IN (select MAX(id) FROM table GROUP BY name)')
->groupBy('name')
->get();
$data2=Model2::query()
->leftJoin(...)
->select(...)
->where('id',login_user)
->whereNotIn(..)
->get();
both data1 and data2 return same column with different values so I merge both variable like that
$results = $data1->concat($data2);
No when I already get data so now I need to add filter data from $results so i make post method for that .
so When user request to filter data with name I write query like that
if ($request->name!="") {
$results->when(request('name'), function($q){
$q->Where('name', request('name'));
});
}
$records = $results;
return response()->json(['success'=>true,'message'=>'success', 'data' => $records]);
But that query is not filtering the data and return me all data.I am new in laravel so I don't know what I have done wrong in that any favour will be helpful for me ,thanks.
if (request()->has('name')) {
$results->when(request()->get('name'), function($q){
return $q->where('name', request()->get('name'));
});
}
$records = $results;
return response()->json(['success'=>true,'message'=>'success', 'data' => $records]);
As you use 'when()', you can drop the if expression all together:
$results->when(request()->has('name'), function($q){
return $q->where('name', request()->get('name'));
});
$records = $results;
return response()->json(['success'=>true,'message'=>'success', 'data' => $records]);
request() is a Laravel helper for Request $request
Edit: the where() clause in the ORM is with a small 'w', not 'W' as in orWhere
my problem is about the JSON format I recieve when sending a query to the database.
The output I want:
[{"errors":[{"something":"something"}],"created_at":"2020-10-20 10:10:10","name":"something","id":99}]
I get :
[{"id":99,"name":"something","device_results":[{"devices_id":99,"created_at":"2020-10-20 10:10:10","errors":[{"something":"something"}]}]}]
Which I get with:
public function errors()
{
$errors = DB::table('devices')
->join('device_results', 'devices.id', '=', 'device_results.devices_id')
->select('errors', 'created_at', 'name', 'device_results.id')
->orderBy('created_at', 'desc')
->whereJsonLength('errors', '>', 0)
->get();
return $errors;
}
I would like to get the same results with using my model because if I do this:
public function errors()
{
$devices = Devices::with(['device_results' => function($query) {
$query->select('device_results.devices_id','created_at','errors')
->whereJsonLength('errors', '>', 0)
->orderBy('created_at', 'desc')
->get();
}])->get();
return $devices;
}
Is it always in this format or can I use the model and get the same format as with the DB class? I am using Vue for the front end and would like to avoid nesting.
Thank you
Have you tried to create a Resource ?
The doc : https://laravel.com/docs/8.x/eloquent-resources
Is there any way to get relational data ( one to many relation records ) using joins in laravel without creating different records in collection :
Example:
$orders = DB::table('orders')
->join('users', 'users.id', '=', 'orders.users_id')
->join('order_items', 'order_items.order_id', '=', 'orders.id')
->select('users.*', 'order_items.*')
->get();
So here what's happening is that its creating 6 records if order has 6 items but i want something like single records in which it has array or collection where are order items are listed.
Output I want is generally of this order:
Collection {# ▼
#items: array:[▼
0 => {
+'id':1,
...,
+ 'items: [
//here i want all the records of relation order items
]
}
]
}
Is there any way to achieve this result without using with() or load() and just only with joins or raw queries?
You should try this:
$orders = DB::table('orders')
->select('users.*', 'order_items.*')
->leftJoin('users', 'users.id', '=', 'orders.users_id')
->leftJoin('order_items', 'order_items.order_id', '=', 'orders.id')
->get();
This answer is speculative, but perhaps you want to report a single row for each user, with a CSV list of order items:
$orders = DB::table('orders AS o')
->join('users AS u', 'u.id', '=', 'o.users_id')
->join('order_items AS oi', 'oi.order_id', '=', 'o.id')
->select('u.id', DB::raw('GROUP_CONCAT(oi.name) AS orders'))
->groupBy('u.id')
->get();
If there's no reason stopping you from using the Eloquent, is the best way for working with databases.
The Eloquent ORM included with Laravel provides a beautiful, simple
ActiveRecord implementation for working with your database
Here is the Documentation Eloquent: Getting Started
The model relationship made easy in laravel, you can get you want as following:
The User model relationships:
//user orders
public function orders() {
return $this->hasMany('App\Order');
}
The Order model relationships:
//order items
public function items() {
return $this->hasMany('App\OrderItem');
}
//order owner
public function user() {
return $this->belongsTo('App\User');
}
The OrderItem model relationships:
//order
public function order() {
return $this->belongsTo('App\Order');
}
Here is a quick example of how you may get the order's items
$orders = User::find($id)->orders
foreach($orders as $order) {
$orderItems = $order->items;
}
you can make the collection the way you prefer, but strongly recommend using Eloquent Resources, since most apps these days expecting JSON responses.
I'm trying to get this MySQL query to work in laravel 5.7 query builder.
It works fine in phpmyadmin
SELECT c.Symbol
, s.SectorName
, cprs.strenght
, s.parentid
, ssbpi.Risklevel
, ssbpi.ColumnType
FROM Companies AS c
JOIN Sectors AS s ON s.SectorID = c.SectorID
JOIN Company_PriceRS AS cprs ON cprs.CompanyID = c.CompanyID
JOIN SubSectorsBPIsData AS ssbpi ON ssbpi.subcategoryid = s.parentid
WHERE cprs.PostDate = '2017-05-08'
AND WHERE CompanyPriceRS.strenght = 'strong'
AND WHERE SubSectorsBPIsData.ColumnType = $ColumnType
ColumnType is a variable that comes from a dropdown and it's already being captured and working properly.
I have tried the normal way according to documentation:
$Completequerytry1 = DB::table('Companies')
->join('Sectors', 'Sectors.SectorID', '=', 'Companies.SectorID')
->join('CompanyPriceRS', 'CompanyPriceRS.CompanyID', '=', 'Companies.CompanyID')
->$join('SubSectorsBPIsData ', 'SubSectorsBPIsData.subcategoryid', '=', 'Sectors.parentid')
->where('CompanyPriceRS.strenght', '=', 'strong')
->where('SubSectorsBPIsData.ColumnType', '=', $ColumnType)
->where('CompanyPriceRS.Postdate', '=', '2017-05-08');
->select('Companies.Symbol', 'Sectors.SectorName', 'CompanyPriceRS.strenght', 'Sectors.parentid', 'SubSectorsBPIsData.subcategoryid','SubSectorsBPIsData.ColumnType')
->limit(10);
->select('Companies.Symbol', 'Sectors.SectorName', 'CompanyPriceRS.strenght', 'Sectors.parentid', 'SubSectorsBPIsData.subcategoryid','SubSectorsBPIsData.ColumnType')
->limit(10);
echo '<pre>';
print_r($Completequerytry1);
Error:
Symfony\Component\Debug\Exception\FatalThrowableError thrown with message "syntax error, unexpected '->' (T_OBJECT_OPERATOR)"
Using functions With several nested joins:
$Completequerytry1 = DB::table('Companies')
->join('Sectors', function ($join) use ($ColumnType) {
$join->on('Sectors.SectorID', '=', 'Companies.SectorID')
->join('CompanyPriceRS', function ($join2) {
$join2->on('CompanyPriceRS.CompanyID', '=', 'Companies.CompanyID')
->join('SubSectorsBPIsData', function ($join3) {
$join3->on('SubSectorsBPIsData.subcategoryid', '=', 'Sectors.parentid')
->where(function ($query1) {
$query1->where('CompanyPriceRS.strenght', '=', 'strong') //filter 1
->where('SubSectorsBPIsData.ColumnType', '=', $ColumnType) //filter2
->where('CompanyPriceRS.Postdate', '=', '2017-05-08'); // filter 3
});
});
});
})
->select('Companies.Symbol', 'Sectors.SectorName', 'CompanyPriceRS.strenght', 'Sectors.parentid', 'SubSectorsBPIsData.subcategoryid','SubSectorsBPIsData.ColumnType')
->limit(10);
echo '<pre>';
print_r($Completequerytry1);
Error:
ErrorException (E_NOTICE)
Undefined variable: ColumnType
3: Then tried functions with nested WHERE
$Completequerytry1 = DB::table('Companies')
->join('Sectors', 'Sectors.SectorID', '=', 'Companies.SectorID')
->join('CompanyPriceRS', 'CompanyPriceRS.CompanyID', '=', 'Companies.CompanyID')
->$join('SubSectorsBPIsData ', 'SubSectorsBPIsData.subcategoryid', '=', 'Sectors.parentid') //ERROR IS GIVEN ON THIS LINE
->where(function ($query1) {
$query1->where('CompanyPriceRS.strenght', '=', 'strong')
->where('SubSectorsBPIsData.ColumnType', '=', $ColumnType)
->where('CompanyPriceRS.Postdate', '=', '2017-05-08');
});
->select('Companies.Symbol', 'Sectors.SectorName', 'CompanyPriceRS.strenght', 'Sectors.parentid', 'SubSectorsBPIsData.subcategoryid','SubSectorsBPIsData.ColumnType')
->limit(10);
echo '<pre>';
print_r($Completequerytry1);
Error:
Undefined variable: join
Still don't know what i'm missing.
Should i create functions for the JOINs and the WHEREs ?
Running out of ideas. Thanks in advance for your insights :)
It turns out that the query was being sent to a variable directly from the controller, not to the view.
I was doing this only for testing but it was using too much memory with print_r(); function making impossible to get a result. Even when using dd(); i wasn't getting what i wanted (I think it's because some sintax errors i had).
So i passed the final variable to the view and it works fine since laravel could handle the data differently.
Also i used ->limit(10); to split the results and avoid memory overload.
Here is the final code working:
$Completequerytry1 = DB::table('Companies')
->join('Sectors', 'Sectors.SectorID', '=', 'Companies.SectorID')
->join('Company_PriceRS', 'Company_PriceRS.CompanyID', '=', 'Companies.CompanyID')
->join('SubSectorsBPIsData', 'SubSectorsBPIsData.subcategoryid', '=', 'Sectors.parentid')
->select('Sectors.SectorName', 'Companies.Symbol', 'Company_PriceRS.strenght', 'Sectors.parentid', 'SubSectorsBPIsData.Risklevel','SubSectorsBPIsData.ColumnType')
->where('Company_PriceRS.strenght', '=', 'strong')
->where('SubSectorsBPIsData.ColumnType', '=','X')
->where('Company_PriceRS.Postdate', '=', '2017-05-08')
->limit(10)
->get();
return view('besttrades2', array('Completequerytry1' => $Completequerytry1)); //sending my query variable to the view
Then in the view only used:
<?= $Completequerytry1; ?>
or show it in whatever way you want.
Sometimes it's better to have another project only for testing separately.
I have two models Category and Transaction the table structures are like this
Categories:
id,category_name,..
and
Transactions:
id,category_id,amount..
The relation is
Category hasMany transactions
public function transactions()
{
return $this->hasMany('App\Transaction');
}
Transactions blongsTo Category
public function category()
{
return $this->belongsTo('App\Category', 'category_id');
}
I want to retrieve all data of transaction table which are sorted by category name.
Most importantly I want to get it using the eloquent method.
I have tried eager load which I think doesn't work on the belongsTo relationship.
Here is the code I have used for the eager load.
$transactions = Transaction::with(['category' => function ($query) {
$query->orderBy('category_name', 'asc');
}])->paginate(10);
So far I can achieve this by writing a query like below, but I'd like to use the eloquent method.
$transactions = Transaction::select(DB::raw('transactions.*'))
->leftJoin(DB::raw('(select id,category_name from categories) as categories'), 'categories.id', '=', 'transactions.category_id')
->orderBy('category_name', 'asc')
->paginate(10);
It'd be nice if someone can help me with this. Thank You.
Note: I am using Laravel 5.1
You have to provide the method name that defines the relationship, in your case this is category.
$transactions = Transaction::all()->with('category')->group_by('category.name')->get();
$transactions = Transaction::with('categories')->group_by('category.name')->get();
$cs = Course::where(['courses.active' => 1])
->whereHas('course_dates', function ($join) use ($now) {
$join->where('course_dates.start_date_time', '>', $now);
$join->orderBy('course_dates.start_date_time', 'asc');
})
->whereHas('category', function ($join) use ($cat_slug) {
$join->where('categories.url_slug', '=', $cat_slug);
})
->whereHas('language', function ($join) use ($cat_slug) {
$join->where('languages.string_id', '=', strtoupper(App::getLocale()));
})
->with(['course_dates' => function($q){
$q->orderBy('course_dates.start_date_time', 'desc');
}])
->join('course_dates' ,'course_dates.course_id', '=', 'courses.id')
->orderby('course_dates.start_date_time')
->limit(7)
->get();
To return Eloquent models ordered by related model (hasMany) column, I had to join the tables and then orderBy, still get the models, but correctly ordered by course_date.start_date_time.
Laravel 5.7, I don't think there is a cleaner solution (at least after few hours of tinkering and searching the web).