Re: JSONParser.parseStrict() with \"

Hi Thomas,

I don't have control over the string value that gets passed in to me. If I don't get \" in return, the JSON will be invalid. I'm having similar problem to this thread. http://stackoverflow.com/questions/13420115/padding-quotes-in-jsonobject

Thank you

On Monday, March 25, 2013 3:57:19 AM UTC-7, Thomas Broyer wrote:

On Monday, March 25, 2013 11:23:51 AM UTC+1, rkulisas wrote:

//josn is {"name":"item_name","index":0,"text":"Kindle Fire HD 8.9\"..."}

JSONValue value = JSONParser.parseStrict(json);

JSONObject obj;

JSONString name, text;

JSONNumber nxd;

if ((obj = value.isObject()) == null ) return null;

if ((value = obj.get("name")) != null) 

if ((name = value.isString()) != null)

   field.setName(name.stringValue());

if ((value = obj.get("text")) != null) 

if ((text = value.isString()) != null){

System.out.println("VALUE:" + value.toString());  //VALUE:"Kindle Fire HD 8.9\" ..."

System.out.println("TEXT:" + text.stringValue()); //TEXT:Kindle Fire HD 8.9" ...

field.setText(text.stringValue());

}

Am I supposed to use JSONValue.toString() to get my string value instead? Why JSONString.stringValue() doesn't parse properly?

The question is more: why do you want the \ to be there?

It's in your JSON because JSON uses " to markup string literals and thus has to escape " that appears within the string value, just like you do in Java or JavaScript or so many other languages (C, etc.)

stringValue() *does* return the correct value, it's just that you're expecting something else.

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