Why does JSONObject.optString add garbage to my strings?

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?!?

Maybe I'm way off here but defining a JSON object like that seems painful with all the escape characters. Have you tried using myJsonObject.put(string key, object value) to see what is stored? — Jack
– Jack, Commented Aug 12, 2011 at 18:44
@Jack It is painful but it was just a simple hack to troubleshoot. — Error 454
– Error 454, Commented Aug 12, 2011 at 18:51
Ahh gotcha - Yes I have noticed that sometimes when debugging a bunch of garbage is appended to my strings. — Jack
– Jack, Commented Aug 12, 2011 at 18:57

nicholas.hauschild · Accepted Answer · 2011-08-12 19:02:54Z

3

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:

enter image description here

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.

edited Aug 12, 2011 at 19:02

answered Aug 12, 2011 at 18:47

nicholas.hauschild

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2 Comments

Error 454 Over a year ago

My evaluation is also true. Perhaps it is debugging garbage, I was concerned that once I parse a more complicated object that my arrays and sub-ojbects would become tainted. I may have been too paranoid.

nicholas.hauschild Over a year ago

Sometimes it is best to just code it up and see what happens :)

Baz1nga · Accepted Answer · 2011-08-12 18:43:38Z

0

optString(java.lang.String key) Get an optional string associated with a key.

you should use

getString(java.lang.String key) Get the string associated with a key.

Or am i missing something?

For more: http://www.json.org/javadoc/org/json/JSONObject.html

answered Aug 12, 2011 at 18:43

Baz1nga

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