使用Java生成有效JSON的异常

Question

I have a JSON string :

{
    "_id": -1,
    "create_date": 0,
    "update_date": 0,
    "description": "test",
    "active": true
}

In Java, I try to parse it using org.json.simple.parser.JSONParser :

JSONParser jsonParser = new JSONParser();
org.json.simple.JSONObject jsonObject = (org.json.simple.JSONObject) jsonParser.parse(phoneJSON);

I try to retrieve the _id field value:

String s_id = ((String) jsonObject.get("_id"));

but encounter the following exception:

java.lang.Long cannot be cast to java.lang.String

Furthermore, if I display the field value in the console:

System.out.println("print value -  _id: "+jsonObject.get("_id"));

I get:

print value -  _id: -1

output in the console.

I have seen this post:

java.lang.Long cannot be cast to java.lang.String

But is does not help me.

What am I not understanding?

Answer 1

The problem is not in the parsing process. But in the casting process of the following line:

String s_id = ((String) jsonObject.get("_id"));

jsonObject.get("_id") returns - as specified by the error - a java.lang.Long , and you later cast it to String .

You can solve this, for instance with:

String s_id = jsonObject.get("_id").toString();

In Java the (String) is not a conversion , but a downcast. Since a String is not a subclass of Long , it will error. You can however call .toString() that will transform the Long in a textual representation.

---

The reason:

System.out.println("print value -  _id: "+jsonObject.get("_id"));

works is because if you "add" an object to a string, the toString method is called automatically. Implicitly you have written:

System.out.println("print value -  _id: "+Objects.toString(jsonObject.get("_id")));

Answer 2

You have to use the .toString() method to convert Long to String.

String strLong = Long.toString(jsonObject.get("_id")));

Reference :

Returns a String object representing this Long's value.

Also, the reason println outputs the value in the console is because PrintStream.println has an overload that takes an Object , and then calls its toString method.

Answer 3

The value of _id is correctly identified as Long (-1 in the example) when you use jsonObject.get("_id") .

If you want a String use jsonObject.getString("_id").

Answer 4

Why do you except the "_id" field to be a string?

Its value is -1 (opposed to "-1") so it is actually a Long instead of String.

However, if you need string, you can cast it via

String s_id = String.valueOf(jsonObject.get("_id"));