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Is there any API in Java to work with JSON?

D
derk last edited by

What standard api Working with json? For everything to come out of the box, I mean, from normal. java jdk 1.7
♪ php Right in the box, there's a json job in Java, isn't there?
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inna last edited by
Unfortunately, there is currently no built JSON API in Java. The developers wanted to add a standardized JSON API in Java 9, but then decided to add other important phishes and remove it from the list. We don't know if he'll be in Java 10.
In the meantime, we have to use separate libraries.
Like https://docs.oracle.com/javaee/7/api/javax/json/package-summary.html which can be taken from http://mvnrepository.com/artifact/javax.json/javax.json-api/1.0 :
```
<dependency>
    <groupId>javax.json</groupId>
    <artifactId>javax.json-api</artifactId>
    <version>1.0</version>
</dependency>
```
or http://www.json.org/java/index.html which can be taken from http://mvnrepository.com/artifact/org.json/json/20151123
```
<dependency>
    <groupId>org.json</groupId>
    <artifactId>json</artifactId>
    <version>20151123</version>
</dependency>
```
and other libraries.
https://stackoverflow.com/questions/2591098/how-to-parse-json-in-java
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I was able to do this using track by, that in this case the identifiers of a course are the sigla and periodo: <select name="curso" ng-options="curso.cursoId.sigla + ' - ' + curso.cursoId.periodo for curso in cursos track by curso.cursoId.sigla + curso.cursoId.periodo" class="form-control" ng-model="disciplina.curso" required disabled> <option value="">Selecione o curso</option> </select> From what I read about, used the object id with track by the Angular uses this id as being the identifier of the object in the DOM, so it can validate whether the object in disciplina.curso is equal to any of the objects in the list cursos, and if that is the case, the dropdown will display the object of ng-model as the selected from the list.
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In Java 1.4, there was no feature https://docs.oracle.com/javase/tutorial/java/generics/index.html . This feature was implemented from Java 1.5 (which officially became called Java 5).Although you are not realizing, the compiler is doing as follows:Instand an ArrayList from Object and assign the variable list.Add a String na list At this time, the cast does not need why String is an instance of Object also (Every class that does not extend another class declaratively through the word extends, will automatically extend Object]Take the Object from list to be in her position 0 and send print.Although you are imagining that you are calling System.out.println(String), you are actually calling System.out.println(Object), which internally will make System.out.println(Object.toString())Here, you are trying to get an object from list of Object and trying to assign to one String directly.He cannot, for though all String be it Object, not always all Object will be String. In order for you to do this assignment, you will need to talk to the compiler who takes the risk, making a http://www.javabeginner.com/java-object-typecasting/ :String s = (String) list.get(0); // eu assumo o risco If you want to make a String List, from Java 5, you can use genérics to solve your problem ( http://docs.oracle.com/javase/1.5.0/docs/api/java/util/List.html List<String> strings = new ArrayList<String>(); strings.add("hello"); String hello = strings.get(0); // ok, strings é uma List<String>. In java 5, if you try to declare a List without assigning a generic parameter, it will assume that it is a List to maintain compatibility with Java 1.4.Useful links: http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/Object.html

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