I am using Laravel (PHP) and MySQL for my backend. I am creating methods for setting and getting information from the database. Those information are being send as a json to the frontend.
I can send table information like:
[
{
"id": 4,
"name": "Max"
},
{
"id": 5,
"name": "Eric"
}
]
For this I am using laravel methods like: DB::table('user')->select('user.id', 'user.name')->get();
However my friend who is doing the frontend want the following json code:
[
{
"id": 4,
"name": "Max"
"specific_user_price":{
"S":5.00,
"M":6.00,
"XL":8.00
}
},
{
"id": 5,
"name": "Eric"
"specific_user_price":{
"S":5.50,
"M":10.00,
"XL":15.00
}
}
]
"specific_user_price is a table and "S", "M", "XL" are columns which have different values depending on the user. I do not know how I can create specific_user_price as an array in a query. I can use group_concat but he needs the json like displayed above.
My idea was to create additional columns in user "size S price", "size M price" and "size XL price". However my friend want those values as an own array group, because some users only habe access to one size, so he would get null values.
Any ideas which method in PHP or Laravel I can use for that? Or is there a MySQL method for creating such thing in a query?
Firstly use Models, way easier to work with out of the box. Define your User model like this, with a relationship for the price.
use Illuminate\Database\Eloquent\Model;
class User extends Model
{
public function specificUserPrice() {
return $this->hasOne(UserPrice::class);
}
}
You also need to have the UserPrice model defined.
use Illuminate\Database\Eloquent\Model;
class SpecificUserPrice extends Model
{
}
Laravel automatically transforms models, you can get away with the following code in the controller.
public function index()
{
return User::with('specificUserPrice')->get();
}
DB::table('user')->select(['id', 4],['name','max'])->get();
DB::table('user')[0]->get()
This gets you the first element in the array
Related
In my emails table, I have a column named To with column-type Json. This is how values are stored:
[
{
"emailAddress": {
"name": "Test",
"address": "test#example.com"
}
},
{
"emailAddress": {
"name": "Test 2",
"address": "test2#example.com"
}
}
]
Now I want a collection of all emails sent to "test#example.com". I tried:
DB::table('emails')->whereJsonContains('to->emailAddress->address', 'test#example.com')->get();
(see https://laravel.com/docs/5.7/queries#json-where-clauses)
but I do not get a match. Is there a better way to search using Laravel (Eloquent)?
In the debugbar, I can see that this query is "translated" as:
select * from `emails` where json_contains(`to`->'$."emailAddress"."address"', '\"test#example.com\"'))
The arrow operator doesn't work in arrays. Use this instead:
DB::table('emails')
->whereJsonContains('to', [['emailAddress' => ['address' => 'test#example.com']]])
->get()
I haven't used the json column but as the documentation refers, the below code should work fine.
DB::table('emails')
->where('to->emailAddresss->address','test#example.com')
->get();
In case to store array in json format. And just have an array list of IDs, I did this.
items is the column name and $item_id is the term I search for
// $item_id = 2
// items = '["2","7","14","1"]'
$menus = Menu::whereJsonContains('items', $item_id)->get();
Checkout the Laravel API docs for the whereJsonContains method
https://laravel.com/api/8.x/Illuminate/Database/Query/Builder.html#method_whereJsonContains
Using Eloquent => Email::where('to->emailAddress->address','test#example.com')->get();
You can use where clause with like condition
DB::table('emails')->where('To','like','%test#example.com%')->get();
Alternatively, if you have Model mapped to emails table names as Email using Eloquent
Email::where('To','like','%test#example.com%')->get();
I am building API. I ran into issue when building responses such as this one:
{
"id": 1,
"name": "Some name",
"my_joined_table": {
"joined_table_id": "10",
"some_joined_table_field": "some value"
}
},
Joining tables as described in https://laravel.com/docs/5.2/queries would yield result such as:
{
"id": 1,
"name": "Some name",
"joined_table_id": "10",
"some_joined_table_field": "some value"
},
Instead of using join I could just run two queries, one for main table, second one for secondary and then just append second array to first one and spit JSON response, but it's a lot of queries and appending if list is big!
Example code which yields second result in pseudo-code:
$data = Model::select('id', 'name', 'my_joined_table.id as joined_table_id', 'my_joined_table.some_value some_value')
->leftJoin('my_joined_table', function($join) { //conditions_callback
})->get();
return response()->json($data);
Please advice.
EDIT2:
It seems that I can use with as follows:
$data = Model::with('my_second_table')->first();
return response()->json($data);
It does what I want, only the problem, that I cannot specify fields for both first and second tables using ->first($fields) and->with(['my_second_table' => function ($query) { $query->select('id', 'some_value'); }]) unless I specify primary key of second table in ->first($fields). How do I work around this?
TL;DR; Issue: http://laravel.io/bin/YyVjd
You can probably use Laravel Eloquent relationship to achieve it.
https://laravel.com/docs/5.2/eloquent-relationships#one-to-many
Or you can remap the returned data to a new response object using $appends.
Try something here,
http://laraveldaily.com/why-use-appends-with-accessors-in-eloquent/
This is just some clues and there is a lots work to do.
FYI, you can set $visible in your model to specify which attributes is visible.
I have Products that have Models. I need to get some fields from all the Products and its Model slug. This is my code:
$query = Product::with('model');
$query->orderBy('order', 'asc')->orderBy('name', 'asc')->skip($skip)->take($take);
return $query->get(['id', 'name', 'front_image', 'back_image', 'slug']),
Its returning something like:
{
"id": 39,
"name": "ASD1",
"front_image": "adasdsa.jpg",
"back_image": "2evewa.jpg",
"slug": "asd1",
"model": null
}
How can I take model.slug?
The get method is used to tell eloquent to execute a query to the database.
To select only specific fields you can use the select method before get.
For example:
Product::with('model')->select('name', 'model.slug as model_slug')->get();
Sorry about the non-explanatory title but I could not come up with a descriptive one.
I've got the following 3 tables:
- games
- platforms
- games_platforms
And I've got 2 Models in Laravel for both Platform and Game.
public function games()
{
return $this->belongsToMany('App\Game', 'games_platforms')->withPivot('release_date');
}
public function platforms()
{
return $this->belongsToMany('App\Platform', 'games_platforms')->withPivot('release_date');
}
Now this works like a charm, I get a JSON string with all the information in the 3 tables, like this.
[{
"id": 1,
"name": "Borderlands",
"short_description": "",
"created_at": null,
"updated_at": null,
"platforms": [{
"id": 4,
"name": "PC",
"pivot": {
"game_id": 1,
"platform_id": 4,
"release_date": "2016-03-03"
}
}]
}]
Now my problem is as follows. I don't want to show the whole 'pivot' information, just the 'release_date', like this:
"platforms": [{
"id": 4,
"name": "PC",
"release_date": "2016-03-03"
Is there an easy way in Laravel to do such a thing? As far as I can see right now, looking at other posts, is to either write a function that turns the json into an array and I can arrange it then. Or I can write my own query instead of letting Laravel do all that.
Hope you guys can help me with this issue. Thanks!
I would modify the data returned from the query via methods on the collection class:
//replace Game::all() with your actual query
return Game::all()->each(function($game){
$game->platforms->map(function($platform){
$platform->release_date = $platform->pivot->release_date;
unset($platform->pivot);
return $platform;
});
});
I know this is already answered but I believe the proper answer would be to add whatever you want to hide to your hidden attribute on the model.
<?php
class Games extends Eloquent
{
protected $hidden = ['pivot.game_id', 'pivot.platform_id'];
}
I am not sure what your keys are becuase they are different in your two examples.
Please see: https://github.com/laravel/framework/issues/745
A better way is to use Laravel resources,
First create Resource (php artisan make:resource)
Rresource GameResource extends Resource
{
public function toArray($request)
{
return [
'release_date' => $this->pivot->release_date,
'name' => $this->name,
/// All other fields or you can merge both here
];
}
}
Now use this resource;
$data = GameResource::collection(Game::all());
I am in a problem of modifying the cake association hierarchy,I am new to cakephp , below is the snippet of code.
Current Code
$this->Cateogory->find('all');
The output of the above expression is of the below form :
[
{
"Category": {
"id": "37",
"title": "Inner Title",
"color": "#ffffff",
"phone-number": ""
},
"CategoryHas": [],
"PhoneNumberhas": []
}
]
Requirement :-
I am creating API for third party consumption , now the third party has given the output format as below. I want the the output to be of below form.
[
{
"Category": {
"id": "37",
"title": "Inner Title",
"color": "#ffffff",
"phone-number": "",
"CategoryHas": [],
"PhoneNumberhas": []
}
}
]
I am searching on this problem for past 1 day and I consider that I need to make a custom PHP function to solve the problem.
Thanks in advance :
Just use a loop
To convert the array of the first format to an array of the second format - just use a loop e.g.:
foreach ($stuff as &$row) {
$row['Category']['CategoryHas'] = $row['CategoryHas'];
$row['Category']['PhoneNumberhas'] = $row['PhoneNumberhas'];
unset($row['CategoryHas'], $row['PhoneNumberhas']);
}
This would be appropriate to put in your controller if it's only needed in one place.
If you always, always want your results in that format, use an afterFind method (or a behavior) to implement the same logic:
// In the category model
public function afterFind($results, $primary = false) {
foreach ($results as &$row) {
if (isset($row['CategoryHas'])) {
$row['Category']['CategoryHas'] = $row['CategoryHas'];
}
if (isset($row['PhoneNumberhas'])) {
$row['Category']['PhoneNumberhas'] = $row['PhoneNumberhas'];
}
unset($row['CategoryHas'], $row['PhoneNumberhas']);
}
return $results;
}
Be wary of choosing this option - it means that your models act unconventionally, potentially meaning plugins don't work and other developers (or you, tomorrow) are confused by the way the models act.
Your association names are pretty strange. Why are they not better named? You can use proper names internally and reformat it when sending it.
If this is just about getting the API response into a silly format that doesn't provide additional benefit use JSON views and change the structure of the array in the view as needed. This section of the book explains it in detail, read it. Example taken from the book:
// Controller code
class CategoriesController extends AppController {
public function index() {
// ... get the categories
$this->set(compact('categories'));
}
}
// View code - app/View/Posts/json/index.ctp
foreach ($categories &$row) {
if (isset($row['CategoryHas'])) {
$row['Category']['CategoryHas'] = $row['CategoryHas'];
}
if (isset($row['PhoneNumberhas'])) {
$row['Category']['PhoneNumberhas'] = $row['PhoneNumberhas'];
}
unset($row['CategoryHas'], $row['PhoneNumberhas']);
}
echo json_encode(compact('categories'));
I would convince the client and bring some arguments for a proper response format and structure. Most clients usually don't know what they want. Sometimes you have smart asses who think they need something specific but can explain why. Check JSend or http://jsonapi.org/ on how to implement a proper API response format. Understand the reasons for these formats and bring these arguments to the client.