JsonObject getAsString throwing Exception

Question

I am new to java, and I am trying to create a json string with name and value.

public static String serializeToken(String name,String value){

    JsonObject json = new JsonObject();
    json.addProperty(name, value);

    return json.getAsString();
}

But the call to getAsString throws an exception and message is just:

JsonObject

https://stackoverflow.com/questions/4105795/pretty-print-json-in-java — cahen, Aug 02 '17 at 15:44

score 1 · Accepted Answer · answered Aug 02 '17 at 15:51

Here's the documentation of getAsString method and this is what it says:

convenience method to get this element as a string value.

Throws:

ClassCastException - if the element is of not a JsonPrimitive and is not a valid string value. IllegalStateException - if the element is of the type JsonArray but contains more than a single element.

So, if the element is not a Primitive (which it is not, in this case), it will throw an Exception. If you want to print json String then you need to call toString method, e.g.:

JsonObject json = new JsonObject();
json.addProperty("test", "value");
String jsonString = json.toString(); 
System.out.println(jsonString);

it worked. one more thing please.. when you say element is not primitive. what do you mean by `element`? — Mubashar Abbas, Aug 02 '17 at 16:00
element = object which the method is being called on. `json` in our case.. — Darshan Mehta, Aug 02 '17 at 16:25

JsonObject getAsString throwing Exception

1 Answers1