Java : checking the value in Java from a JSON Format

Question

I have a JSON as shown below

{
    "qty": "1",
    "dswer": 0,
    "bag": {
        "dm": "2",
        "rate": "---",
        "ghy": "3013-04-05T00:00:00.000-05:00",
        "dee": "301304",
        "desc": "SSAA APR 05 2013 --- BGG"
    }
}

My requirement is that if the rate attribute in above Json is --- then treat it that as as InValid Request .

I dont want to assign this JSON to a Java class , for various reasons

So i was trying this way

package com;

import java.util.ArrayList;
import java.util.List;

public class Test {

    public static void main(String args[]) {


        String str = "{\r\n" + 
                "    \"qty\": \"1\",\r\n" + 
                "    \"dswer\": 0,\r\n" + 
                "    \"bag\": {\r\n" + 
                "        \"dm\": \"2\",\r\n" + 
                "        \"rate\": \"---\",\r\n" + 
                "        \"ghy\": \"3013-04-05T00:00:00.000-05:00\",\r\n" + 
                "        \"dee\": \"301304\",\r\n" + 
                "        \"desc\": \"SSAA APR 05 2013 --- CALL\"\r\n" + 
                "    }\r\n" + 
                "}";

        if(str.contains("rate:---"))
        {
            System.out.println("InValid");
        }
        else
        {
            System.out.println("Valid");
        }


    }

}

But its always displaying as Valid .

Please tell me how to solve this ??

Thanks

BenMorel · Accepted Answer · 2014-05-22 14:45:44Z

3

if(str.contains("rate:---")) // is false because
if(str.contains("\"rate\": \"---\"")) // would be true

edited May 22, 2014 at 14:45

BenMorel

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answered Apr 29, 2013 at 10:04

jlordo

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Comments

Rahul · Accepted Answer · 2013-04-29 10:05:54Z

2

Your if condition needs to be this:-

if(str.contains("\"rate\": \"---\""))

Since in your JSON, both rate & --- are enclosed within double-quotes, you need to add those in your if condition by escaping those.

Even though this would work, I strongly recommend you parse your JSON properly using either org.json.JSONObject or com.google.gson.JsonObject.

answered Apr 29, 2013 at 10:05

Rahul

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