fastjson | FASTJSON 2.0.x has been released faster and more secure | JSON Processing library

by alibaba Java Version: 1.1.77.android_noneautotype License: Apache-2.0

X-Ray Key Features Code Snippets Community Discussions(10)Vulnerabilities Install Support

kandi X-RAY | fastjson Summary

fastjson is a Java library typically used in Utilities, JSON Processing applications. fastjson has no bugs, it has no vulnerabilities, it has build file available, it has a Permissive License and it has high support. You can download it from GitHub, Maven.

FASTJSON 2.0.x has been released, faster and more secure, recommend you upgrade.

Support

Quality

Security

License

Reuse

Support

fastjson has a highly active ecosystem.

It has 25359 star(s) with 6526 fork(s). There are 1316 watchers for this library.

It had no major release in the last 12 months.

There are 1840 open issues and 1669 have been closed. On average issues are closed in 272 days. There are 174 open pull requests and 0 closed requests.

It has a positive sentiment in the developer community.

The latest version of fastjson is 1.1.77.android_noneautotype

Quality

fastjson has 0 bugs and 0 code smells.

Security

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

fastjson code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

fastjson 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

fastjson releases are available to install and integrate.

Deployable package is available in Maven.

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

Installation instructions are not available. Examples and code snippets are available.

fastjson saves you 177099 person hours of effort in developing the same functionality from scratch.

It has 181091 lines of code, 19450 functions and 3015 files.

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

Top functions reviewed by kandi - BETA

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

Build JavaBeanInfo from given class .
Deserialize field information .
Scan a date in ISO8601 format .
Gets an object writer .
Write a string with double quotes .
Generate write as array .
tokenize object
Scan a field name .
Get the value of a property
Writes the value of the property .

Get all kandi verified functions for this library.

fastjson Key Features

No Key Features are available at this moment for fastjson.

fastjson Examples and Code Snippets

No Code Snippets are available at this moment for fastjson.

Community Discussions

Trending Discussions on fastjson

SpringBoot: Validation without @Valid annotation

How to use lambda expression in android?

how to generate signed apk in android studio

How to don't serialize the class field in FastJson?

@JSONField it did not work in spring boot

Error when starting spring-boot app. How to find out which dependency is causing AbstractMethodError

Can't understand/find error in JSON config file referenced by Unity logs

Rails - How to add pagination to Fastjson api?

java.lang.StackOverflowError happens when connect to Neo4j, in Springboot Mybatis

Error creating bean with name 'projectController': Unsatisfied dependency expressed through field 'service';

QUESTION

SpringBoot: Validation without @Valid annotation

Asked 2021-Nov-23 at 04:53

I have a web socket handler inherited from AbstractWebSocketHandler that handles text messages. My DTOs use javax.validation.constraints for validation. So, in my REST endpoints, I simply can use the @Valid annotation to invoke the validator. However, as far as I know, this annotation is not usable in my web socket handler. How can I invoke the SpringBoot validator programmatically without this annotation?

Besides, is it possible to use the SpringBoot de-serializer for messages instead of JSON.parseObject?

Example:

...

ANSWER

Answered 2021-Nov-23 at 04:53

You will use a Validator to fill a list of ConstraintViolation. An example could looks like this :

Source https://stackoverflow.com/questions/70029336

QUESTION

How to use lambda expression in android?

Asked 2021-Nov-02 at 06:49

I have tried everything I could find from StackOverflow to make this problem right but none of them succeeded.

The strange thing is : I can use lambda such as

...

ANSWER

Answered 2021-Nov-02 at 05:21

go to App level build.gradle>

then put this code in android {} block

Source https://stackoverflow.com/questions/69805214

QUESTION

how to generate signed apk in android studio

Asked 2021-Oct-28 at 02:25

In my case, I have my signature key and I have generated the apk with function menubuild--->generate signed Bundle/Apk

Here are the steps how I generate this apk:

step 1:

step 2:

finally:

When I put this generated release apk into my real device, it will work well.

But when I tried to send this apk to my application center, it said that I don't have any signature file in this apk.

According to many issues found on google, I have noticed that there was actually no signature file in my apk at all. It seems that Android studio's build--->generate signed Bundle/Apk didn't work at all, there was only a release with no signature that generated.

I'm quite new in android developing. I wonder if there are some mistakes in my gradle settings.

My application has 2 gradle.build file like this picture:

the gradle.build out side the app directory is :

...

ANSWER

Answered 2021-Oct-28 at 02:25

Thanks for @Raj Suvariya and @Nitish

The problem was solved according to their kindness instructions. And how I solved this problem is showing below:

First of all I'm using AS of version

Source https://stackoverflow.com/questions/69738029

QUESTION

How to don't serialize the class field in FastJson?

Asked 2021-Jun-10 at 03:04

Today I'm using FastJson(https://github.com/alibaba/fastjson), the following is my demo code.

...

ANSWER

Answered 2021-Jun-10 at 03:04

@JSONField(serialze=false)
public String getPassword() {
  return password;
}

Source https://stackoverflow.com/questions/67835239

QUESTION

@JSONField it did not work in spring boot

Asked 2021-May-24 at 06:06

I use @JSONField annotation for pointX and pointY, I want to change pointX to point_x! I want to pointX show like point_x! But it did not work! It always is pointX and pointY! but other field is ok!

My java bean like this!

...

ANSWER

Answered 2021-May-24 at 06:06

I resolved this, it cause by org.springframework.beans.BeanUtils.copyProperties
I used org.springframework.beans.BeanUtils.copyProperties change DAO java bean to View java bean.
My Dao java bean

Source https://stackoverflow.com/questions/67635422

QUESTION

Error when starting spring-boot app. How to find out which dependency is causing AbstractMethodError

Asked 2021-Apr-27 at 06:37

I'm having an error as the below when I'm starting my program.

...

ANSWER

Answered 2021-Mar-25 at 06:00

seems it's related to your config file

use -X flag with your maven command to see the full stack trace

mvn -X spring-boot:run

Source https://stackoverflow.com/questions/66792625

QUESTION

Can't understand/find error in JSON config file referenced by Unity logs

Asked 2021-Apr-05 at 14:19

I have customized this JSON config file for a mod from the game Valheim. The mods make use of a Unity program in order to load the mods up and it shows a console log so you can find errors while loading the game with said mods. I am getting this constant error in the Unity console regarding the file, but cannot understand the error or find any errors in the files JSON formatting.

The error is as follows

...

ANSWER

Answered 2021-Apr-05 at 14:19

What the error message is saying is that 14451th character is wrong (from the json parser's point of view).

This character index is not very helpful to me. So, to explore this further, I've removed the comments with a regex (in the regex101 link, click on "substitution" on the left panel to see the version without comments). [*]

Then I pasted the comment-less version in an online JSON parser.

The output is a bit more precise:

Source https://stackoverflow.com/questions/66933866

QUESTION

Rails - How to add pagination to Fastjson api?

Asked 2021-Feb-27 at 07:55

The default result of rendering FastJsonApi gem serialized_json like below:

...

ANSWER

Answered 2021-Feb-27 at 07:55

The desired document would be invalid according to the JSON API specification. You would need to include next and previous links in a link section. The current and total_count would belong in the meta section.

Source https://stackoverflow.com/questions/66387607

QUESTION

java.lang.StackOverflowError happens when connect to Neo4j, in Springboot Mybatis

Asked 2020-Dec-31 at 15:28

As the title suggests, I met the StackOverFlow problem when I connect to Neo4j in Springboot with Mybatis. The CQL runs well in Neo4j desktop, but the api returns a 500 result with the same CQL in mapper.

The error message goes like:

...

ANSWER

Answered 2020-Dec-31 at 15:28

As far as I know, there is no MyBatis integration in:

neither Spring Data Neo4j 5 (added via spring-boot-starter-data-neo4j in your POM, included up until Spring Boot 2.3)
nor Neo4j OGM (which is the library Spring Data Neo4j 5 is built upon).

If you want to get started, you should probably get started with Spring Data Neo4j 6 (aka SDN 6), i.e. the latest version of Spring Data Neo4j at the time of writing. If you upgrade Spring Boot to the latest version (2.4 at the time of writing), you will automatically get it.

You should probably start with the reference documentation of SDN 6. You also can find an example here.

As you will learn, you can get rid of MyBatis, SDN 6 (and 5 via Neo4j OGM) takes care of the mapping already.

Source https://stackoverflow.com/questions/65521129

QUESTION

Error creating bean with name 'projectController': Unsatisfied dependency expressed through field 'service';

Asked 2020-Nov-20 at 07:59

I use the following configuration and code to create a spring boot application.

pom.xml:

...

ANSWER

Answered 2020-Nov-18 at 06:08

Spring's complaint is that you haven't defined a Bean of type ProjectMapper, which it is saying that it needs to satisfy an Autowire request. Looking at your code, I don't see a definition for such a Bean in what you've shown us here.

The exact point of the problem is hidden, I'm guessing, because you aren't showing us the definition of the ServiceImpl class. My guess is that that's where the actual reference is that Spring is trying to wire up. Apparently, the field name is baseMapper.

Source https://stackoverflow.com/questions/64887470

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

Vulnerabilities

No vulnerabilities reported

Install fastjson

You can download it from GitHub, Maven.
You can use fastjson 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 fastjson 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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