How to display the JSON value in a java servlet

Question

JSONObject jsonObj = new JSONObject();

 jsonObj.put("name","raja");

Declare the jsonobject var(jsonObj) and gave value("name","raja") using above code.

 System.out.println(jsonObj.name);

Just print the name but it shows the error.

To get a value from json object, use jsonObject.get(key), so it should be jsonObject.get("name") in your example. — Abhay Jain, Jun 16 '18 at 12:53

score 1 · Answer 1 · answered Jun 16 '18 at 13:01

1

try this:

System.out.println(jsonObj.getString("name"));

answered Jun 16 '18 at 13:01

System.out.println(jsonObj.get("name")); it works well. Thank u @Abhay Jain – raja Jun 16 '18 at 13:08
yes it gets you the object but in your case you know the type of your object that is a string – Jun 16 '18 at 13:10
how to exit the loop while running using angular js. – raja Jun 18 '18 at 09:43
this question you have to ask it in the tag angular not here. to exit a loop is simple add a boolean initialises the false and do your treatment when you want to exit changes this boolean has true (it's for a while loop) please specify which loop you use – Jun 18 '18 at 09:50

1 Answers1