How to return a single object with json-simple in java?

Question

I've this json object:

public static final String JSON_TEXT = "{" +
"\"Subjects\": [{" +
            "\"primaryKey\": \"1\"," +
            "\"subjectName\": \"English\"" +
        "}," +
        "{" +
            "\"primaryKey\": \"2\"," +
            "\"subjectName\": \"Spanish\"" +
        "}" +
    "]," +
    "\"Exams\": [{" +
            "\"primaryKey\": \"1\"," +
            "\"grade\": \"10\"" +
        "}," +
        "{" +
            "\"primaryKey\": \"2\"," +
            "\"grade\": \"20\"" +
        "}" +
    "]" +
"}";

and need to build some methods for a json dao implementation.

I've built

public Collection<Subject> getSubjects()

successfully but I got totally stuck with

public Subject findSubjectById(Integer subjectId)

This was my first idea so far:

public Subject findSubjectById(Integer subjectId) {
    JSONObject obj = (JSONObject)JSON_PARSER.parse(JSON_TEXT);
    if (obj.get(subjectId) != null)
        try {
            JSONArray subjectsArray = (JSONArray) obj.get("Subjects");

            for (int i = 0; i < subjectsArray.length(); i++){   

            }
                } catch (ParseException e) {
                    e.printStackTrace();
                }
        return subject;

}

Any ideas/ examples are highly appreciated to solve this. Thanks in advance!

What do you mean you got stuck? Can you show us what you have so far? So that we can guide you further from your current position. — Adelin, Dec 22 '17 at 12:31
I edited my question. But my idea is super incomplete because I got stuck. — hagep, Dec 22 '17 at 13:06

score 0 · Answer 1 · answered Dec 22 '17 at 13:19

0

Try this in your for loop

JSONObject subDtl=subjectsArray.getJSONObject(i);
int subKey=subDtl.getInt("primaryKey"); 
String subName=subDtl.getString("subjectName");

answered Dec 22 '17 at 13:19

Sreejesh K Nair

563
1
6
16

There has been a problem with subjectsArray.length() already and now with getJSONObject(i), getInt("primaryKey") and getString("subjectName") as well... – hagep Dec 22 '17 at 13:33
use this JSONArray subjectsArray = obj.optJSONArray("Subjects"); – Sreejesh K Nair Dec 22 '17 at 13:53
Still the same errors plus obj.optJSONArray as well now. – hagep Dec 22 '17 at 13:58

score 0 · Answer 2 · answered Dec 22 '17 at 19:46

I know you don't use Gson but it would resolve all your issues. Also I recommend to change field names to "subjects" and "exams" instead of "Subjects" and "Exams".

Here is the complete solution that is at least worth to be considered.

public class Data {

    private List<Subject> subjects;

    private List<Exam> exams;

    public Data(String json) {
        Data data = new Gson().fromJson(json, Data.class);
        this.subjects = data.subjects;
        this.exams = data.exams;
        Reporter.log(new Gson().toJson(this), true);
    }

    public Subject findSubjectById(Integer id) {
        return subjects.stream().filter(subject -> subject.primaryKey.equals(id)).findAny().orElse(null);
    }

    public Exam findExamById(Integer id) {
        return exams.stream().filter(exam -> exam.primaryKey.equals(id)).findAny().orElse(null);
    }

    public List<Subject> getSubjects() {
        return subjects;
    }

    public void setSubjects(List<Subject> subjects) {
        this.subjects = subjects;
    }

    public List<Exam> getExams() {
        return exams;
    }

    public void setExams(List<Exam> exams) {
        this.exams = exams;
    }

    public class Subject {
        private Integer primaryKey;
        private String subjectName;

        public Integer getPrimaryKey() {
            return primaryKey;
        }

        public void setPrimaryKey(Integer primaryKey) {
            this.primaryKey = primaryKey;
        }

        public String getSubjectName() {
            return subjectName;
        }

        public void setSubjectName(String subjectName) {
            this.subjectName = subjectName;
        }
    }

    public class Exam {
        private Integer primaryKey;
        private String grade;

        public Integer getPrimaryKey() {
            return primaryKey;
        }

        public void setPrimaryKey(Integer primaryKey) {
            this.primaryKey = primaryKey;
        }

        public String getGrade() {
            return grade;
        }

        public void setGrade(String grade) {
            this.grade = grade;
        }
    }

}

Here are tests that verify our solution

@SpringBootTest(classes = TestNGWithSpringApplication.class)
public class JsonDataIT extends AbstractTestNGSpringContextTests {

    String json = "{  \n"
            + "   \"subjects\":[  \n"
            + "      {  \n"
            + "         \"primaryKey\":\"1\",\n"
            + "         \"subjectName\":\"English\"\n"
            + "      },\n"
            + "      {  \n"
            + "         \"primaryKey\":\"2\",\n"
            + "         \"subjectName\":\"Spanish\"\n"
            + "      }\n"
            + "   ],\n"
            + "   \"exams\":[  \n"
            + "      {  \n"
            + "         \"primaryKey\":\"1\",\n"
            + "         \"grade\":\"10\"\n"
            + "      },\n"
            + "      {  \n"
            + "         \"primaryKey\":\"2\",\n"
            + "         \"grade\":\"20\"\n"
            + "      }\n"
            + "   ]\n"
            + "}";

    @Test
    public void findSubjectTest() {
        Data.Subject subject = new Data(json).findSubjectById(new Integer(2));
        Assert.assertEquals(subject.getSubjectName(), "Spanish");
    }

    @Test
    public void findExamTest() {
        Data.Exam exam = new Data(json).findExamById(new Integer(2));
        Assert.assertEquals(exam.getGrade(), "20");
    }

}

That should work

How to return a single object with json-simple in java?

2 Answers2