Exception casting valid JSON with Java

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?

aren't you looking for String.valueOf(long) ?

itwasntme
– itwasntme

2015-05-29 16:19:31 +00:00
Commented May 29, 2015 at 16:19 — itwasntme
– itwasntme, Commented May 29, 2015 at 16:19

Ali Gajani · Accepted Answer · 2015-05-29 16:30:05Z

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.

edited May 29, 2015 at 16:30

answered May 29, 2015 at 16:24

Ali Gajani

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Comments

willeM_ Van Onsem · Accepted Answer · 2015-05-29 16:39:57Z

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")));

43iscoding · Accepted Answer · 2015-05-29 16:22:08Z

1

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"));

answered May 29, 2015 at 16:22

43iscoding

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Comments

Robert · Accepted Answer · 2015-05-29 16:35:52Z

1

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").

edited May 29, 2015 at 16:35

answered May 29, 2015 at 16:18

Robert

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5 Comments

Roy Hinkley Over a year ago

A string should be able to hold ANY long or Int. Why the exception?

Robert Over a year ago

In Java not really, they have different classes. You can convert between them, for e.g. Integer.valueOf(<string>) or Integer.toString(<int>).

willeM_ Van Onsem Over a year ago

@AndroidAddict: because java doesn't support overloaded cast operators. In C# you can. But a (String) conversion in Java only means downcasting (not converting) and since you cannot downcast a Long to a String, it will error.

Roy Hinkley Over a year ago

I want to represent the long value in a string. I do it all the time, why can't I do it here? I am really confused on this!

Robert Over a year ago

You can represent an integer in a string, try jsonObject.getString("_id")

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