Convert a JSON of different type of objects into java List object

Question

I have a JSON object stored in db in the form of string. I am using it to create dynamic form in the UI. Now the problem is I want to change some values in it based on other changes happening on the application. So suppose I updated label for the field, then I have to get this JSON and change that here. This would be easy If I have stored same type of objects in this json, but my JSON is like follows:

[{
    "name": "someName",
    "xtype": "keyvaluecombo",
    "fieldLabel": "Some Title",
    "refType": "YES_NO",
    "multiSelect": false,
    "helpText": ""
  },
  {
    "name": "someName2",
    "xtype": "keyvaluecombo",
    "fieldLabel": "Some Title2",
    "refType": "YES_NO",
    "multiSelect": false,
    "helpText": ""
  },
  {
    "xtype": "datefield",
    "fieldLabel": "Joining Date",
    "name": "joiningDate",
    "submitFormat": "Y-m-d"
  },
  {
    "xtype": "userselectioncombo",
    "fieldLabel": "Selection",
    "name": "selections",
    "filterBy": {
      "functions": [
        "select"
      ]
    }
  }]

Now this is stored as String in db, what is efficient way of changing fieldLabel based on name. I could have tried working on it as string only and use regular expression, but that didn't feel right.

` have a JSON object stored in db in the form of string` That is absolutely horrible design. If you MUST do this, you should use a database that has json support like postgres — Falmarri
– Falmarri, Commented May 14, 2015 at 6:53
Extract the JSON, use a parser to get the dictionary from it, change the value, serialize the dictionary back to a JSON string, save it. Or use a DB than known JSON. — user1907906
– user1907906, Commented May 14, 2015 at 6:57
Elaborate on which part? My main advice is to use a DB than knows JSON as a data type, PostgreSQL does, AFAIK. If you are not tied to use a relational DB, use a NoSQL DB like MongoDB. — user1907906
– user1907906, Commented May 14, 2015 at 7:06

Saurabh Jhunjhunwala · Accepted Answer · 2015-05-14 07:00:50Z

2

You should write a bean class, which should be mapping to you Json object like,

public class abc {

    private String name;
    private String xtype;
    private String fieldLabel;
    ........
}

Then you should use

import java.lang.reflect.Type;

GsonBuilder gsonBuilder = new GsonBuilder();
Gson gson = gsonBuilder.create();

Type type = new TypeToken<List<abc>>() {
            }.getType();

List<abc> abcList = gson.fromJson(confTemplate,
                    type); // confTemplate is your Json object you get from DB

this will get the list of beans. 

for (abc abcData : abcList ) {
 // you can do your stuff
}

answered May 14, 2015 at 7:00

Saurabh Jhunjhunwala

2,9223 gold badges32 silver badges60 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

user1907906 Over a year ago

That is just creating the object. How would you manipulate the JSON string that is stored in the DB?

kavinder Over a year ago

i too thought like that, but after my changes are done I have to store it back to db in the same format, this will add all fields to all individual jsons. I mean after this they all will be of same object i.e abc.

Saurabh Jhunjhunwala Over a year ago

@Tichodroma, once you get individual bean objects, you can do your manipulation and finally convert the list back to JSon before DB save

Saurabh Jhunjhunwala Over a year ago

@kavinder, once you are through with your manipulation, you can convert your list back to json using gson.toJson(abcList ) and save it into your DB.

Sercan Ozdemir · Accepted Answer · 2015-05-14 07:08:20Z

First of all the best way, changing your database design with a new one that suits your model. Not keeping json in your database as a column. But if you can't do that, because sometimes it's impossible to change old db designs you can trace the following way.

And of course you should read your json from db before start and save it again after the below process.

Create a custom object that suits your model.

public class MyObject{
private String name;
private String fieldLabel;

public String getFieldLabel(){
    return fieldLabel;
}

public void setFieldLabel( String fieldLabel ){
    this.fieldLabel = fieldLabel;
}

public String getName(){
    return name;
}

public void setName( String name ){
    this.name = name;
}

// bla bla other fields

Convert your json into your object, and vice versa see the code example below:

 public static void main( String[] args ){
        Gson gson = new Gson();
        String yourJson = "[{'name':'someName','xtype':'keyvaluecombo','fieldLabel':'Some Title','refType':'YES_NO','multiSelect':false,'helpText':''},{'name':'someName2','xtype':'keyvaluecombo','fieldLabel':'Some Title2','refType':'YES_NO','multiSelect':false,'helpText':''},{'xtype':'datefield','fieldLabel':'Joining Date','name':'joiningDate','submitFormat':'Y-m-d'},{'xtype':'userselectioncombo','fieldLabel':'Selection','name':'selections','filterBy':{'functions':['select']}}]";

        // changing single quotes with double ones.
        yourJson = yourJson.replaceAll( "'", "\"" );

        JsonArray jsonArray = new JsonParser().parse( yourJson ).getAsJsonArray();

        List<MyObject> result = new ArrayList<MyObject>();

        for( JsonElement jsonElement : jsonArray ){
            MyObject myObject = gson.fromJson( jsonElement, MyObject.class );

            // change fields as you wish
            if( myObject.getName().equals( "someName" ) ){
                myObject.setFieldLabel( "TEST" );
            }

            // add it to another list
            result.add( myObject );
        }

        // convert into another json again..
        System.out.println( gson.toJson( result ) );

    }

codeaholicguy · Accepted Answer · 2015-05-14 07:08:32Z

0

Easy to realize that Object in your list have attributes below:

"name"
"xtype"
"fieldLabel"
"refType"
"multiSelect"
"helpText"
"submitFormat"
"filterBy"

So you can create an Object which has over attributes. Using ObjectMapper for deserialize the list:

ObjectMapper mapper = new ObjectMapper();
mapper.readValue(json, new TypeReference<ArrayList<T>>() {})

After have a list object you can loop for change any attribute or change attribute of an specific item you want.

answered May 14, 2015 at 7:08

codeaholicguy

1,6811 gold badge11 silver badges18 bronze badges

1 Comment

kavinder Over a year ago

the problem with your solution was that the fields are dynamic, there is no surety that these will be the only attributes. However, you were close

Collectives™ on Stack Overflow

Convert a JSON of different type of objects into java List object

3 Answers 3

4 Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

4 Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related