gson.fromJson return null values

Question

This is My JSON String : "{'userName' : 'Bachooo'}"

Converting JSON String to LoginVO logic is:

Gson gson = new GsonBuilder().excludeFieldsWithoutExposeAnnotation().create();
LoginVO loginFrom  = gson.fromJson(jsonInString, LoginVO.class);
System.out.println("userName " + loginFrom.getUserName()); // output null

My LoginVO.class is:

public class LoginVO {

 private String userName;
 private String password;

 public String getUserName()
 {
    return userName;
 }
 public void setUserName(String userName)
 {
    this.userName = userName;
 }
 public String getPassword()
 {
    return password;
 }
 public void setPassword(String password)
 {
    this.password = password;
 }

}

Note I am using jdk 1.8.0_92

Output of loginForm.getUserName() is NULL instead of "Bachooo" any idea about this issue?

if i only use Gson gson = new Gson() Infinity loop at recursive calling com.google.gson.internal.$Gson$Types.resolve($Gson$Types.java:375) — PAncho, May 26 '16 at 11:36
what `excludeFieldsWithoutExposeAnnotation` says? "only expose fields that are annotated and ignore the rest " — Braj, May 26 '16 at 11:38
Possible duplicate of [How should I escape strings in JSON?](http://stackoverflow.com/questions/3020094/how-should-i-escape-strings-in-json) — Siguza, May 26 '16 at 11:38

score 8 · Accepted Answer · edited Sep 02 '20 at 23:26

8

Since you are setting excludeFieldsWithoutExposeAnnotation() configuration on the GsonBuilder you must put @Expose annotation on those fields you want to serialize/deserialize.

So in order for excludeFieldsWithoutExposeAnnotation() to serialize/deserialize your fields you must add that annotation:

@Expose
private String userName;
@Expose
private String password;

Or, you could remove excludeFieldsWithoutExposeAnnotation() from the GsonBuilder.

edited Sep 02 '20 at 23:26

ricardopereira

11,118
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answered May 26 '16 at 11:36

Vikas Madhusudana

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Thanks man its work 100% all I need @Expose To my variable i need to initialized. – PAncho May 26 '16 at 11:42
1

Or the other option is to just say new GsonBuilder().create() if you dont have access to edit the class – Vikas Madhusudana May 26 '16 at 11:44

Divyanshu Singh · Answer 2 · 2023-01-09T05:24:53.623

0

Adding what resolved this for me. So in my API following gson implementation was getting used:

Gson gson = new GsonBuilder().setFieldNamingPolicy(FieldNamingPolicy.LOWER_CASE_WITH_UNDERSCORES).create();

I had to use the same implementation in my test, before which gson was failing to parse attributes.

So in essence check how your gson is configured in api/handler, and use same configuration in your test.

edited Jan 09 '23 at 05:24

answered Jan 09 '23 at 05:23

Divyanshu Singh

35
1
8

score -2 · Answer 3 · edited Nov 18 '19 at 05:05

-2

Try like this, please. Here is the example class:

class AngilinaJoile {
    private String name;

    // setter

    // getter  
}

And here is how you deserialize it with Gson:

Gson gson = new Gson();  
String jsonInString = "{'name' : 'kumaresan perumal'}";
AngilinaJoile angel = gson.fromJson(jsonInString, AngilinaJoile.class);

edited Nov 18 '19 at 05:05

zeroDivider

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answered May 26 '16 at 11:49

Kumaresan Perumal

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gson.fromJson return null values

3 Answers3