Question

I am starting with the json:

{
    "Key1" : "Value1",
    "Key2" : "Value2"
}

I am then hard-coding this json in a string:

String json = "{ \"Key1\" : \"Value1\", \"Key2\" : \"Value2\" }";

Next I attempt to parse the json:

JSONObject content = null;
try {
    content = new JSONObject(json);
} catch (JSONException e) {
    e.printStackTrace();
    return null;
}       

String key1 = content.optString("Key1", null);

If I look at the hashmap created from the call to JSONObject, it looks correct:

{Key2=Value2, Key1=Value1}

But when I look at the value of the string key1 in the debugger, I get this:

[V, a, l, u, e, 1, U, U, U, U, U, U, U, U, U, U]

Where U appears to be unicode character 25A1 (White Square).

I’ve also tried the generic get(“Key1”) method, casting the result to a string and I get the same behavior?!?

Answer 1

I do not see an issue with your code, as this slight modification appears to work:

public static void main(String[] args)
{
    String json = "{ \"Key1\" : \"Value1\", \"Key2\" : \"Value2\" }";

    JSONObject content = null;
    try
    {
        content = new JSONObject(json);
    }
    catch (JSONException e)
    {
        e.printStackTrace();
        return;
    }

    String key1 = content.optString("Key1", null);
    System.out.print(key1 + "end!");
}

The console output says this: Value1end!

When looking at the code in the debugger, I saw this:

If I were to venture a guess, it may just be additional buffer space the debugger is using to hold the String. Have you tried seeing what your code thinks it is?

"Value1".equals(key1); //should return true if everything is working correctly.