sangria | Sangria is a collection of Guice extensions

by tavianator Java Version: 1.3 License: Apache-2.0

X-Ray Key Features Code Snippets Community Discussions(6)Vulnerabilities Install Support

kandi X-RAY | sangria Summary

sangria is a Java library. sangria has no vulnerabilities, it has build file available, it has a Permissive License and it has high support. However sangria has 1 bugs. You can download it from GitHub.

Sangria is a collection of Guice extensions

Support

Quality

Security

License

Reuse

Support

sangria has a highly active ecosystem.

It has 7 star(s) with 0 fork(s). There are 2 watchers for this library.

It had no major release in the last 12 months.

sangria has no issues reported. There are no pull requests.

It has a positive sentiment in the developer community.

The latest version of sangria is 1.3

Quality

sangria has 1 bugs (0 blocker, 0 critical, 0 major, 1 minor) and 18 code smells.

Security

sangria has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

sangria code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

sangria is licensed under the Apache-2.0 License. This license is Permissive.

Permissive licenses have the least restrictions, and you can use them in most projects.

Reuse

sangria releases are available to install and integrate.

Build file is available. You can build the component from source.

sangria saves you 1505 person hours of effort in developing the same functionality from scratch.

It has 3354 lines of code, 318 functions and 60 files.

It has low code complexity. Code complexity directly impacts maintainability of the code.

Top functions reviewed by kandi - BETA

kandi has reviewed sangria and discovered the below as its top functions. This is intended to give you an instant insight into sangria implemented functionality, and help decide if they suit your requirements.

Creates a typeLiteral for the given key and value type
Returns a TypeLiteral for the given key and value pair
Build a MapLiteral for the given key and value pair
Adds an entry to the list
Returns the next priority
Creates a typeLiteral from the given type
Wraps a typeLiteral
Converts a type to a ProviderLiteral
Converts a TypeLiteral into a TypeLiteral
Returns a TypeLiteral for the given type
Converts a TypeLiteral to a SetLiteral
Convert ListBinderErrors into TypeLiteral
Creates a new LazyBinder instance
Returns a string representation of this class
Compares with this priority
This method returns a hashCode of this builder
Creates a new priority
Gets the cached value
Converts a type to a ListElement of type
Returns the logger for the specified injection point
Binds this logger to the Slf4J logger
Override this method to bind a context sensitive provider
Returns the logger
Get the logger for the injection point
Compares by priority
Returns a string representation of this list binder

Get all kandi verified functions for this library.

sangria Key Features

No Key Features are available at this moment for sangria.

sangria Examples and Code Snippets

No Code Snippets are available at this moment for sangria.

Community Discussions

Trending Discussions on sangria

Fandom-py: Unable to get image by using page.images[0]

test for membership in Scala type class

Assigning value to arg with val object(arg) = object

Configure c++ debugger in VSCode

All fields and parameters marked as non-null in Graphiql : Sangria, scala

Graphql Sangria Many-to-Many example

QUESTION

Fandom-py: Unable to get image by using page.images[0]

Asked 2022-Feb-01 at 09:46

I'm using fandom module in python for my discord bot. I get this error when I request data by using page.images[0]. It should be an image url. This is the page I want to get.

...

ANSWER

Answered 2022-Feb-01 at 09:46

I fixed it by using pymediawiki module.

This code that use mediawiki can replace fandom module by changing the api of wiki.

Source https://stackoverflow.com/questions/70863772

QUESTION

test for membership in Scala type class

Asked 2021-Oct-20 at 11:48

Is there a way to test whether a type is a member of a type class? E.g.:

...

ANSWER

Answered 2021-Oct-20 at 11:48

(1) Try to introduce type class IsFoo

Source https://stackoverflow.com/questions/69487499

QUESTION

Assigning value to arg with val object(arg) = object

Asked 2021-Jul-15 at 08:24

I am following this tutorial on GraphQL with Sangria. I am wondering about the following line

...

ANSWER

Answered 2021-Jul-15 at 07:42

It is the Extractor pattern, you can reach the same result implementing the unapply method on your arbitrary object (like shown in the example). When you create a case class the compiler produces an unapply method for you, so you can do:

Source https://stackoverflow.com/questions/68389733

QUESTION

Configure c++ debugger in VSCode

Asked 2021-Jul-02 at 22:30

I recently moved to VSCode and im a little be lost.

If i compile my program with this console command

...

ANSWER

Answered 2021-Jul-02 at 22:17

If you already have a compiled version with debug information (-g) then you do not need to include the header files again.

Just remove the line "preLaunchTask": "C/C++: g++.exe compilar archivo activo" from the configuration since your program is already compiled.

Source https://stackoverflow.com/questions/68231069

QUESTION

All fields and parameters marked as non-null in Graphiql : Sangria, scala

Asked 2020-May-08 at 18:22

I am not able to use postman to query a graphql server written in scala using its sangria library. I get a request malformed error in postman if I use the content-type as "application-json" and if I use plain text I get content type not supported. The postman version I am using is Version 6.2.4. The code is pretty straightforward in which I read data from hbase. I use the sangria macro function deriveObjectTypeUnit, myCaseClass to define an object. I am however able to use the graphiql console by concatenating a graphiql.html file from the resources to the route created. The issue with the graphiql console is that it marks all the arguments and fields as non-nullable which it should not as by default the fields are all nullable in graphql. I checked this from the documentation tab of the graphiql console where I can see all my fields and the parameters are marked as non-nullable(Suffixed with an exclamation mark !). Sample query is as follows:

...

ANSWER

Answered 2020-May-08 at 18:22

Field might by nullable by default in GraphQL schemas but in Scala nulls are almost always programming errors (and surely always if you pass or return them as parameters) - you want nullable, you explicitly model them with Option. If you want to stick to GraphQL convention just write all fields in your case classes as Optional.

Source https://stackoverflow.com/questions/61683745

QUESTION

Graphql Sangria Many-to-Many example

Asked 2020-Feb-14 at 08:39

I'm working on a Many-to-Many relation with Sangria and Slick. I have 3 objects: Role, Permission and RolePermission (a mapping table)

...

ANSWER

Answered 2020-Feb-14 at 08:39

You have to define Relation type for both related entities. It's not easy to answer in short comment on StackOverflow, but I've explained such kind of relation in my blog post: https://scalac.io/akka-http-sangria-graphql-backend#implementing-many-to-many-relation I hope it helps..

Source https://stackoverflow.com/questions/60197953

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install sangria

You can download it from GitHub.
You can use sangria like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the sangria component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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