Java : Parsing a Json formated String to Java

Question

I do a Action on Middle ware and if its success i get the Value as

String result = ["RESULT","DELETE","OK"]

And in Case if the Operation is Failed i get the resposne as

String result = ["RESULT","DELETE","ERROR"]

I need to know if the Operation is success Or Fail so for this i have done this

public class Test {

    public static void main(String args[]) {

        String result = "[\"RESULT\",\"DELETE\",\"ERROR\"]";

        if (!result.contains("ERROR")) {
            System.out.println("success");
        } else {
            System.out.println("Failed");
        }

    }

This is working fine , but not sure if this has any negative impact / or in cases the code may Fail .

Please suggest if there is a better approach .

Use a proper library to parse Json. E.g. Jackson.

bmargulies
– bmargulies

2013-02-19 13:33:55 +00:00
Commented Feb 19, 2013 at 13:33 — bmargulies
– bmargulies, Commented Feb 19, 2013 at 13:33

pcalcao · Accepted Answer · 2013-02-19 13:33:46Z

3

Your code can fail if, for instance, you get a success message containing ERROR (not likely, but can happen).

You should use a library to parse the result into a List/Array, look here on StackOverflow for a ton of solutions for parsing Json Strings to Objects in Java (Jackson is a library to do this, for instance).

You should also validate against a pre-set number of hypothesis, for instance, creating an enum for the possible result types, and checking if it's one of them.

answered Feb 19, 2013 at 13:33

pcalcao

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Comments

Raman · Accepted Answer · 2013-02-19 13:41:09Z

0

I would suggest to use object presentation instead of array here:

{
 "RESULT": {
   "operation": "DELETE",
   "status" : "ERROR"
 }
}

There is you can find a lot of tools to parse JSON in Java objects: http://www.json.org/

In other case it will be hard to extend set of codes. E.g.

String result = "[\"RESULT\",\"CREATE\",\"USER\", \"LOGIN\", \"ERROR\", \"SUCCESS\"]";

Was there error? or user with login ERROR was successfully created?

edited Feb 19, 2013 at 13:41

answered Feb 19, 2013 at 13:35

Raman

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Comments

Achintya Jha · Accepted Answer · 2013-02-19 13:40:50Z

-1

String[] result = {"RESULT","DELETE","OK"};
    if(Arrays.asList(result).contains("OK"))
        System.out.println("Ok");

Try Arrays.asList() method to check given arrays contains that element or not.

edited Feb 19, 2013 at 13:40

answered Feb 19, 2013 at 13:33

Achintya Jha

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Pawan Over a year ago

Type mismatch: cannot convert from String to String[]

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