在 Java 中构建 JSONObject 而不抛出异常

Question

Is there a way in Java to build a JSONObject without having to deal with an exception? Currently I'm using the default constructor, which however forces me to put try/catch blocks in the code. Since in the rest of the code I'm using the "opt" version of get, checking if the return value is null, is there a way to build the object in the same way? (that is, some kind of constructor that returns null if it can't build the json from a string).

Example:

         try {
               JSONObject temp = new JSONObject(someString);
         } catch (JSONException e) {
               e.printStackTrace();
         }

What I would like to do:

JSONObject temp = ??????(someString);
if(temp != null) {...}

Answer 1

You need to create the method manually, as any parser will most probably throw one or the other Exception in case of any discrepancy.Try something like this:-

 public JSONObject verifyJSON(String inputString){
        JSONObject temp;
        try {
            temp = new JSONObject(inputString);
      } catch (JSONException e) {
          temp = null;
            e.printStackTrace();

      }
        return temp;
    }

Then you can do :-

JSONObject temp = verifyJSON(someString);
if(temp != null) {...}