I have a project of Laravel with two tables
Variants
id | name | data (jsonb)
Devices
id | name | variants_id | data (jsonb)
Some of the values in Devices tables used from variant but it also has it's own values too.
So when we create a new device we use variant data and save it with devices along with it's own data.
Now when I want to update variant and only update those fields in all the devices without replacing non-matching values.
Is there a better way than running a foreach loop on all devices.
This is what I'm doing right now.
foreach($variant->devices as $device)
{
$data = $device->data;
foreach($variant->data as $key => $value)
{
$data[$key] = $value;
}
$device->data = $data;
$device->save();
}
This is the data in Variant
{"depth": 66.9, "width": 58.2, "height": 100.8, "pit_id": "1234", "variable": false, "liters_per_cm": 3.88194, "measurement_unit": "cm"}
This is the data in Devices
{"lat": "16.636192", "lng": "19.238912", "depth": 66.9, "width": 58.2, "height": 100.8, "pit_id": "1234", "total_liters": 391.299552, "liters_per_cm": 3.88194, "measurement_unit": "cm"}
I hope this makes sense and help will be highly appreciated
i believe you can achieve this by where condition
$variant // you already have
Device::where('variants_id',$variant->id)->update([
'data'=> $variant->data
]);
is this you want ?
Related
I need some specific columns from two relations.
In my questions model I have two relations
public function ans_options()
{
return $this->hasMany('App\Models\AnswerChoices', 'ac_quest_id', 'q_id');
}
public function question_category()
{
return $this->hasOne("App\Models\Categories", 'cat_id', 'q_category');
}
I tried
Questions::with(array(
'questionCategory' => function($query) {$query->select('cat_id','cat_value');},
'ans_options' => function($query1) {$query1->select('ac_id','ac_choiceTxt');}
))->get();
am getting only the columns of question_category not in ans_options
{
"q_id": 349,
"q_project_id": 140,
"q_text": "<p>Gender</p>",
"question_category": {
"cat_id": 1,
"cat_value": "normal"
},
"ans_options": []
}
But when I try the below code all columns of ans_options are getting.
Questions::with('questionCategory:cat_id,cat_value','ans_options')->get();
like
{
"q_id": 349,
"q_project_id": 140,
"q_text": "<p>Gender</p>",
"question_category": {
"cat_id": 1,
"cat_value": "normal"
},
"ans_options": [
{
"ac_id": 334,
"ac_quest_id": 349,
"ac_choiceTxt": "Male",
"ac_modifiedOn": "2021-11-24T06:22:00.000000Z",
"ac_status": "active"
},
{
"ac_id": 335,
"ac_quest_id": 349,
"ac_choiceTxt": "Female",
"ac_modifiedOn": "2021-11-24T06:22:00.000000Z",
"ac_status": "active"
}
]
}
I need only ac_id and ac_choiceTxt from ans_options. How can I achieve that?
to make Laravel able to load the relation, you should select the foreign key that responsible for that relation
Questions::with(array(
'questionCategory' => function ($query) {
$query->select('cat_id', 'cat_value');
},
'ans_options' => function ($query1) {
$query1->select(
'ac_id',
'ac_choiceTxt',
'ac_quest_id'
);
}
))->get();
just add 'ac_quest_id' to your select.
You can make model where available only this fields
Extend current model from new (with removing columns from curent)
use "with" from new short model
I don't know another way at now...
Then we can add the primary key here also. It will get the same result and then no need of closures here.
Questions::with('questionCategory:cat_id,cat_value',
'ans_options:ac_id,ac_choiceTxt,ac_quest_id')
->get();
I would like to perform a query in a table where the column is of json type and it contains array of objects, but only if a certain condition is met
This is my current code
$initial_results = DB::table('toys')->select('id','name')->where(['name' => 'sammy', 'email' => 'whateveremail']);
if($sk ==='yes') {
$results = $initial_results->>whereRaw('JSON_CONTAINS(`info`,\'{"sku":"B07V3SSLN11"}\')')
>whereRaw('JSON_CONTAINS(`info`,\'{"asin":"DTI-LALF3-EA18"}\')')
->get();
} else {
$results = $initial_results->get();
}
But I always get 0 result if the condition is met. In database, the info I want to query indeed exist. What is the proper way to query a json column which contains array of objects? See my example data
[
{
"sku": "DTI-LALF3-EA18",
"adId": 244077676726655,
"asin": "B07V3SSLN11",
"cost": 0,
},
{
"sku": "DTI-LALF3-EA18",
"adId": 242968940906362,
"asin": "B07V3SSLN11",
"cost": 10,
.........
................
I even tried
$initial_results = DB::table('toys')->select('id','name')->where(['name' => 'sammy', 'email' => 'whateveremail'])->->whereIn(DB::raw("JSON_EXTRACT(info, '$[*].asin')"),['B07V3SSLN11']);
Thanks in advance
You can query JSON columns by using the -> operator in your clause:
->where('info->asin', 'DTI-LALF3-EA18')
JSON Where Clauses Docs
Assume you have a GraphQL type and it includes many fields.
How to query all the fields without writing down a long query that includes the names of all the fields?
For example, If I have these fields :
public function fields()
{
return [
'id' => [
'type' => Type::nonNull(Type::string()),
'description' => 'The id of the user'
],
'username' => [
'type' => Type::string(),
'description' => 'The email of user'
],
'count' => [
'type' => Type::int(),
'description' => 'login count for the user'
]
];
}
To query all the fields usually the query is something like this:
FetchUsers{users(id:"2"){id,username,count}}
But I want a way to have the same results without writing all the fields, something like this:
FetchUsers{users(id:"2"){*}}
//or
FetchUsers{users(id:"2")}
Is there a way to do this in GraphQL ??
I'm using Folkloreatelier/laravel-graphql library.
Unfortunately what you'd like to do is not possible. GraphQL requires you to be explicit about specifying which fields you would like returned from your query.
Yes, you can do this using introspection. Make a GraphQL query like (for type UserType)
{
__type(name:"UserType") {
fields {
name
description
}
}
}
and you'll get a response like (actual field names will depend on your actual schema/type definition)
{
"data": {
"__type": {
"fields": [
{
"name": "id",
"description": ""
},
{
"name": "username",
"description": "Required. 150 characters or fewer. Letters, digits, and #/./+/-/_ only."
},
{
"name": "firstName",
"description": ""
},
{
"name": "lastName",
"description": ""
},
{
"name": "email",
"description": ""
},
( etc. etc. ...)
]
}
}
}
You can then read this list of fields in your client and dynamically build a second GraphQL query to get the values of these fields.
This relies on you knowing the name of the type that you want to get the fields for -- if you don't know the type, you could get all the types and fields together using introspection like
{
__schema {
types {
name
fields {
name
description
}
}
}
}
NOTE: This is the over-the-wire GraphQL data -- you're on your own to figure out how to read and write with your actual client. Your GraphQL javascript library may already employ introspection in some capacity. For example, the apollo codegen command uses introspection to generate types.
2022 Update
Since this answer was originally written, it is now a recommended security practice to TURN OFF introspection in production. Reference: Why you should disable GraphQL introspection in production.
For an environment where introspection is off in production, you could use it in development as a way to assist in creating a static query that was used in production; you wouldn't actually be able to create a query dynamically in production.
I guess the only way to do this is by utilizing reusable fragments:
fragment UserFragment on Users {
id
username
count
}
FetchUsers {
users(id: "2") {
...UserFragment
}
}
I faced this same issue when I needed to load location data that I had serialized into the database from the google places API. Generally I would want the whole thing so it works with maps but I didn't want to have to specify all of the fields every time.
I was working in Ruby so I can't give you the PHP implementation but the principle should be the same.
I defined a custom scalar type called JSON which just returns a literal JSON object.
The ruby implementation was like so (using graphql-ruby)
module Graph
module Types
JsonType = GraphQL::ScalarType.define do
name "JSON"
coerce_input -> (x) { x }
coerce_result -> (x) { x }
end
end
end
Then I used it for our objects like so
field :location, Types::JsonType
I would use this very sparingly though, using it only where you know you always need the whole JSON object (as I did in my case). Otherwise it is defeating the object of GraphQL more generally speaking.
GraphQL query format was designed in order to allow:
Both query and result shape be exactly the same.
The server knows exactly the requested fields, thus the client downloads only essential data.
However, according to GraphQL documentation, you may create fragments in order to make selection sets more reusable:
# Only most used selection properties
fragment UserDetails on User {
id,
username
}
Then you could query all user details by:
FetchUsers {
users() {
...UserDetails
}
}
You can also add additional fields alongside your fragment:
FetchUserById($id: ID!) {
users(id: $id) {
...UserDetails
count
}
}
Package graphql-type-json supports custom-scalars type JSON.
Use it can show all the field of your json objects.
Here is the link of the example in ApolloGraphql Server.
https://www.apollographql.com/docs/apollo-server/schema/scalars-enums/#custom-scalars
Assume you have a GraphQL type and it includes many fields.
How to query all the fields without writing down a long query that includes the names of all the fields?
For example, If I have these fields :
public function fields()
{
return [
'id' => [
'type' => Type::nonNull(Type::string()),
'description' => 'The id of the user'
],
'username' => [
'type' => Type::string(),
'description' => 'The email of user'
],
'count' => [
'type' => Type::int(),
'description' => 'login count for the user'
]
];
}
To query all the fields usually the query is something like this:
FetchUsers{users(id:"2"){id,username,count}}
But I want a way to have the same results without writing all the fields, something like this:
FetchUsers{users(id:"2"){*}}
//or
FetchUsers{users(id:"2")}
Is there a way to do this in GraphQL ??
I'm using Folkloreatelier/laravel-graphql library.
Unfortunately what you'd like to do is not possible. GraphQL requires you to be explicit about specifying which fields you would like returned from your query.
Yes, you can do this using introspection. Make a GraphQL query like (for type UserType)
{
__type(name:"UserType") {
fields {
name
description
}
}
}
and you'll get a response like (actual field names will depend on your actual schema/type definition)
{
"data": {
"__type": {
"fields": [
{
"name": "id",
"description": ""
},
{
"name": "username",
"description": "Required. 150 characters or fewer. Letters, digits, and #/./+/-/_ only."
},
{
"name": "firstName",
"description": ""
},
{
"name": "lastName",
"description": ""
},
{
"name": "email",
"description": ""
},
( etc. etc. ...)
]
}
}
}
You can then read this list of fields in your client and dynamically build a second GraphQL query to get the values of these fields.
This relies on you knowing the name of the type that you want to get the fields for -- if you don't know the type, you could get all the types and fields together using introspection like
{
__schema {
types {
name
fields {
name
description
}
}
}
}
NOTE: This is the over-the-wire GraphQL data -- you're on your own to figure out how to read and write with your actual client. Your GraphQL javascript library may already employ introspection in some capacity. For example, the apollo codegen command uses introspection to generate types.
2022 Update
Since this answer was originally written, it is now a recommended security practice to TURN OFF introspection in production. Reference: Why you should disable GraphQL introspection in production.
For an environment where introspection is off in production, you could use it in development as a way to assist in creating a static query that was used in production; you wouldn't actually be able to create a query dynamically in production.
I guess the only way to do this is by utilizing reusable fragments:
fragment UserFragment on Users {
id
username
count
}
FetchUsers {
users(id: "2") {
...UserFragment
}
}
I faced this same issue when I needed to load location data that I had serialized into the database from the google places API. Generally I would want the whole thing so it works with maps but I didn't want to have to specify all of the fields every time.
I was working in Ruby so I can't give you the PHP implementation but the principle should be the same.
I defined a custom scalar type called JSON which just returns a literal JSON object.
The ruby implementation was like so (using graphql-ruby)
module Graph
module Types
JsonType = GraphQL::ScalarType.define do
name "JSON"
coerce_input -> (x) { x }
coerce_result -> (x) { x }
end
end
end
Then I used it for our objects like so
field :location, Types::JsonType
I would use this very sparingly though, using it only where you know you always need the whole JSON object (as I did in my case). Otherwise it is defeating the object of GraphQL more generally speaking.
GraphQL query format was designed in order to allow:
Both query and result shape be exactly the same.
The server knows exactly the requested fields, thus the client downloads only essential data.
However, according to GraphQL documentation, you may create fragments in order to make selection sets more reusable:
# Only most used selection properties
fragment UserDetails on User {
id,
username
}
Then you could query all user details by:
FetchUsers {
users() {
...UserDetails
}
}
You can also add additional fields alongside your fragment:
FetchUserById($id: ID!) {
users(id: $id) {
...UserDetails
count
}
}
Package graphql-type-json supports custom-scalars type JSON.
Use it can show all the field of your json objects.
Here is the link of the example in ApolloGraphql Server.
https://www.apollographql.com/docs/apollo-server/schema/scalars-enums/#custom-scalars
I've created this code from laravel:
public function findConfig($id)
{
$config = DB::table('configuration')
->join('model', 'model.configuration_id','=', 'configuration.id')
->select('configuration.id','configuration.description', 'model.name','configuration.price')
->where('configuration.id','=', $id)
->get();
$encode = json_encode($config, JSON_UNESCAPED_SLASHES);
$response = Response::make($encode, 200);
$response->header('Content-Type', 'application/json');
return $response;
}
then the return is somehow like this
[{
"id": "1",
"description": "{\"item\":[{'colours\":[\"red\",\"blue\",\"green\"]},{\"motors\":[ {\"name\":\"450W/48V\",\"price\":\"2,000\"},{\"name\":\"550W/48V\", \"price\":\"3,000\" }] } ]}",
"name": "k5-A",
"price": "300000"
},
{
"id": "1",
"description": "{\"item\":[{'colours\":[\"red\",\"blue\",\"green\"]},{\"motors\":[ {\"name\":\"450W/48V\",\"price\":\"2,000\"},{\"name\":\"550W/48V\", \"price\":\"3,000\" }] } ]}",
"name": "r-A",
"price": "300000"
}
]
How can I remove the slashes and instead of string as return type, it should be in JSON?
As lukasgeiter said, generally it isn't a good idea to store json in a db. It may get difficult to filter by that field.
If you decide to do so, and need to get the decoded data, you can use an accessor in the model. I don't know if it is the best practice. If the description is saved in the db as a json you can do this:
For the "configuration" table you may have a "Configuration" model (The official Laravel website recommends to name the table in plural, and the model in it's singular, like: table -> configurations and the model configuration). In that file you can add this:
public function getDescriptionAttribute($value)
{
return json_decode($value, true);
}
Now, the description field is returned as an array.
You can see more about accessors and mutators here: http://laravel.com/docs/4.2/eloquent#accessors-and-mutators